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Negative Constant Velocity Graph: What, How, Examples

January 21, 2024February 5, 2022 by Manish Naik

The article discusses a negative constant velocity graph with solved problems and examples.

The motion is depicted when we plot its displacement or position-time graph. For an object moving opposite to another, when we plot a graph by taking a displacement or position on a y-axis and a time taken on the x-axis, it displays the negative constant velocity of an object moving in opposite.

The constant negative velocity is due to an object’s displacement or position decreasing with time. The position is the exact location of an object in a given time, whereas the displacement is the position different from one time to another.

The negative constant velocity graph reveals how an object moves in the opposite or reverse direction when no external force acts. Therefore, the graph’s slope seems to be a straight line beginning at the maximum displacement and minimum time values. It ended at a minimum displacement and maximum time values. Hence, the slope also supports us calculate the value of an object’s negative velocity.

What is Negative Constant Velocity Graph — **Negative Constant Velocity Graph**

Suppose the car moves on the opposite road to other vehicles. From the displacement-time graph, calculate the negative constant velocity.

Given:

From graph,

s₁ = 80m

s₂ = 0m

t₁ = 0sec

t₂ = 10 sec

To Find:

v =?

Formula:

v^→= s₂-s₁/t₂-t₁ = Δs/Δt

Solution:

The value of the negative constant velocity of moving car is calculated as,

v^→ = s₂-s₁/t₂-t₁

Substituting all values,

v^→= 0 -80/10-0

v^→ = -8

The car moves with a negative constant velocity of -8m/s.

What is Negative Constant Velocity Graph?

The negative constant velocity graph signifies an object moving in a negative direction.

When an object moves negative or opposite, its displacement time graph displays a slightly cross and steeper straight line slope indicates the ‘negative constant velocity’. The more stepper the graph’s slope is, the faster an object moves in a negative direction with negative constant velocity.

If an object rest, then the displacement – time graph illustrated the zero motion as a horizontal straight line slope. If an object accelerates, the displacement – time graph illustrates the changing motion as a curved slope.The constant velocity conceived when zero acceleration on an object is due to no external forces being exerted, represented by displacement –time graph as straight-line slope.

Since the velocity is measured as relative motion between two objects, the constant velocity is termed ‘negative’ when two objects move opposite each other.

The perfect example of constant negative velocity is the two escalators moving in the opposite direction. i.e., one is going up, and the other is down. For the escalator going up, the people standing on the escalator going down seem to go down with constant negative velocity as its displacement decreases. Similarly, the escalator going down and the people standing on the escalator going up seem to be going with a constant negative velocity.

Negative Constant Velocity Example — **Negative** Constant Velocity Example
(credit: shutterstock)

Suppose a person chooses to go down on the escalator going up; it moves with constant negative velocity with respect to the escalator going down. At the same time, the person moves with constant positive velocity with respect to the escalator going down.

Read more about Relative Motion.

From the data, draw the displacement and time graph for both escalators going up and down with respect to each other.

Escalator is moving Up with respect to Escalator moving Down.
Displacement	80	70	60	50	40	30	20	10	0
Time	0	1	2	3	4	5	6	7	8

Escalator is moving Down with respect to Escalator moving Up
Displacement	60	55	48	42	35	30	25	20	11	7	0
Time	0	1	2	3	4	5	6	7	8	9	10

Calculate the negative constant velocity for both escalators is moving opposite to each other. Comment on the result.

Given:

For escalator going Up

s₁ = 80m

s₂ = 0m

t₁ = 0sec

t₂= 8sec

For escalator going Up

s₁ = 60m

s₂ = 0m

t₁ = 0sec

t₂ = 10sec

Solution:

The negative constant velocity of the escalator going up w.r.t. escalator going down is calculated as,

v^→= s₂-s₁/t₂-t₁

Substituting all values,

v^→ = 0 -80/8-0

v^→ = -10 ……………(1)

The negative constant velocity of the escalator going down w.r.t. escalator going up is calculated as,

v^→ =s₂-s₁/t₂-t₁

Substituting all values,

v^→= 0 -60/10-0

v^→ = -6……… (2)

From (1) and (2), we observe that the escalator going up w.r.t. the escalator going down is moving more rapidly with a negative constant velocity of 10m/s.

Therefore, the slope of the escalator going down w.r.t. escalator going up is more down than the escalator going up w.r.t. escalator going down.

Suppose two trains are running opposite on different tracks. After running a certain distance, both trains stop at the station for some time and start running opposite each other.

For data below, draw the negative constant velocity graph for train A running opposite to train A.

Displacement	50	35	25	15	0	0	-15	-25	-40
Time	0	1	2	3	4	5	6	7	8

Calculate the initial and final negative constant velocity value.

Solution:

The section OA represents the train moving with negative constant velocity w.r.t. another train.

The initial negative constant velocity of train A is calculated as,

v^→ = s₂-s₁/t₂-t₁

Substituting all values,

v^→= 0 -50/4-0

v^→ =-50/4

v^→ = -12.5m/s

The horizontal line in section AB shows that both trains stop for about a second. Hence, the value of the negative constant velocity of train A zero in section AB.

The section BC represents the train again starting moving with negative constant velocity w.r.t. another train.

The final negative constant velocity of train A is calculated as,

v^→ = s₂-s₁/t₂-t₁

Substituting all values,

v^→ = -40/8-5

v^→= -40/3

v^→= -13.3m/s

Train A’s final negative constant velocity is greater than the initial negative constant velocity.

Also Read:

What Is Constant Negative Velocity: Why, How, When and Problem Examples

January 21, 2024January 31, 2022 by Manish Naik

The article discusses about what is constant negative velocity with some solved problem examples.

The constant negative velocity is defined by values of displacement. If it increases with time, an object drives straight with constant positive velocity. But if it decreases with time, an object drives in reverse or opposite direction with constant negative velocity.

Since velocity is a vector quantity, it can be positive or negative depending on its direction, determined by its displacement. When no net force applies, an object moves with constant speed in the same direction, either straight or opposite. In the previous article, we have thoroughly discussed about constant positive velocity when an object moves in a straight direction.

In the displacement time graph plot for the decreasing displacement per unit time for an object moving in the opposite direction, we obtain a straight line vertical slope. The slope starts from a higher displacement value and ends at a lower displacement value, symbolizing the constant negative velocity.

**Constant Negative Velocity Slope in Displacement Time Graph**

The graph also helps us to calculate the value of constant negative velocity from its slope.

v^→ = s₂-s₁/t₂-t₁

Calculate the constant negative velocity of the car from the below displacement vs. time graph.

Given:

From graph,

s₁ = 60m

s₂= 0m

t₁ = 0sec

t₂ = 6sec

To Find:

v^→ =?

Formula:

v^→ =s₂-s₁/t₂-t₁= Δs/Δt (s^→)

Solution:

The constant negative velocity of the car is calculated as,

v^→= s₂-s₁/t₂-t₁

Substituting all values,

v^→=0 -60/6-0

v^→ = -10

The constant negative velocity of the car is -10m/s.

How is Constant Velocity Negative?

The constant velocity is negative as an object’s displacement decreases with time.

The constant velocity becomes negative when the object’s displacement decreases with time. It occurs because an object moves in the opposite direction after traveling a certain distance straight with constant positive velocity.

Since there is no external force, an object cannot accelerate in any other direction, and it moves with a constant velocity. But since it is moving, its displacement varies with time. Hence, the constant velocity value is either positive or negative, depending upon the displacement is either increasing or decreasing.

The displacement time graph illustrates both constant positive and constant negative velocity when an object moves forward and backward.

Suppose a car moves forward on an inclined slope of a hill, and its motion is called constant positive energy. But if the driver cannot accelerate the car on an inclined plane, then the car starts moving backward, and its motion is termed as its constant negative velocity.

Draw a displacement time graph from the following data that includes displacement by the moving car on an inclined plane per unit time:

Displacement(meter)	Time (secs)
0	0
20	1
40	2
60	3
80	4
100	5
100	6
90	7
80	8
70	9
60	10

Also, calculate the constant positive velocity and constant negative velocity of the car.

Given:

For constant positive velocity-

s₂: 80m

s₁: 0s

t₂: 4m

t₁: 0s

For constant negative velocity:

s₂: 60m

s₁: 90m

t₂: 10s

t₁: 7s

Formula:

v^→= s₂-s₁/t₂-t₁

Solution:

The constant positive velocity of car is calculated as,

v^→= s₂-s₁/t₂-t₁

v^→ = 80-0/4-0

v^→ = 80/4

v^→ = 20m/s

The car moves on an inclined plane with a constant positive velocity of 20m/s.

The constant negative velocity of car is calculated as,

v^→ = s₂-s₁/t₂-t₁

v^→ =90-60/10-7

v^→ =30/3

v^→ = 10m/s

The car moves on an inclined plane with a constant negative velocity of 10m/s.

When is Constant Velocity Negative?

The constant velocity negative is when an object moves in the opposite direction.

The positive and negative signs of velocity reveal the direction of an object’s motion. So whenever an object moves in the opposite direction, it drives with constant negative velocity. But this negative direction is defined by the coordinate system to specify the position.

An object’s final displacement may be either zero, positive or negative, larger, smaller, or the same as initial displacement. Hence, its constant velocity also is zero, positive or negative, when no force is applied.

The displacement time graph shows different shapes of slopes for different values of negative velocities. That means the slope appears more down for the high value of constant negative velocity than the low value of negative velocity, which demonstrates which object is moving faster than the other.

Usually, we estimate the velocity as positive. In contrast, we do not express the velocity in a negative value. Hence, the constant negative velocity is expressed as the relative motion between two objects. One object’s forward direction is the opposite to the others when both move oppositely. The relative motion between two oppositely moving objects is specified by its coordinate system (x,y,z,t).

Suppose two athletes, A and B, run opposite each other with constant velocity. For athlete A, athlete B is running with constant negative velocity. Hence we say athlete B is running with a constant negative velocity with respect to athlete A. Whereas, for athlete B, athlete A is running with a constant negative velocity. So we can say it is a constant negative velocity of athlete A with respect to athlete B.

Read more about Relative Motion.

Constant Negative Velocity Example

The constant negative velocity example depicts the opposite motion of an object which is listed below:

Inclined Plane
Automatic Door
Steering Wheel
Old Telephone
Stretched Spring
Stretched Rubber
Moving Car
Running Trains

Constant Negative Velocity Examples — **Constant** Negative Velocity Examples

Inclined Plane

When pushing the box upon an inclined slope-like ramp, it slides with constant positive velocity as it pushes it in a positive forward direction. Suppose we leave the box midway through the slope; it slides backward on an inclined ramp in a negative direction. So we can say it slides oppositely with constant negative velocity.

Automatic Door

It is another example where an object moves with constant velocity in a positive direction when we push it. But if we leave the door after we pass, it automatically starts to move in the opposite direction to being closed. While this opposite automatic motion, the door moves with constant negative velocity.

Steering Wheel

It is a constant negative velocity example in terms of rotational motion. While taking a left or right or U-turn, we rotate the steering wheel fully in a clockwise or anticlockwise direction. In such a case, steering wheels move with constant positive angular velocity. To turn the vehicle, we need to rotate the wheel in the opposite direction again, where it moves with constant negative angular velocity.

Dialing Old Telephone

An old-fashioned example depicts an object moving with constant angular motion. While dealing with the telephone, we rotate the dialer in the clockwise direction. Hence its constant angular velocity is positive.

But when we leave the dealer after one rotation, it automatically comes to its original position. The old telephone’s automatic opposite motion portrays that its constant angular velocity is negative.

Stretched Spring

When we stretch the spring forward, its displacement increases with time; hence it moves with constant positive velocity. But the spring has the restoring force that enables it to regain its original position after releasing the stretch. Hence, the spring again moves in the opposite direction to regain its original position due to spring force. Since it moves in the opposite direction, its constant velocity is negative.

Stretched Rubber

Like the last example, rubber also stretches with constant positive velocity in a positive direction. But the rubber material also has the restoring force, which allows it to regain its original position. Hence, after we remove the stretch on the rubber, it moves in the opposite direction regain due to elastic force. Therefore, its constant velocity is negative.

Moving Car

Suppose a boy is standing at the side of the road. If the car is moving away from the boy, then the car moves in a positive direction with the constant positive velocity with respect to the car itself and also with respect to the boy on the road.

But if the car moves toward the boy, then the car moves with constant positive velocity to itself only; but it moves with the constant negative velocity with respect to the boy.

Running Trains

When two trains run on separate railway tracks, their motion is relative. Therefore, one’s constant positive velocity appears as the constant negative velocity to another train and vice-versa.

Also Read:

Constant Velocity in Displacement Time Graph: What, Why, Facts, Problems

January 21, 2024January 27, 2022 by Manish Naik

The article discusses some facts about constant velocity in displacement time graph.

The displacement time graph, known as the position-time graph, represents an object’s motion. When we plot the moving object’s displacement on a y-axis as a dependent variable and time on an x-axis as an independent variable, its straight-line slope reveals an object moves with constant velocity.

Calculate the constant velocity of the moving car from below displacement vs. time graph.

Given:

From graph,

s₁= 0m

s₂ = 60m

t₁ = 0sec

t₂ = 6sec

To Find:

v ^→ =?

Formula:

v ^→= s₂-s₁/t₂-t₁ = Δs^→/Δt^→

Solution:

The value of constant velocity of moving is calculated as,

v ^→= s₂-s₁/t₂-t₁ = Δs^→/Δt^→

Substituting all values,

v ^→ = 60-0/6-0

v ^→ = 10

The car moves with constant velocity of 10m/s.

What is Constant Velocity in Displacement Time Graph?

The constant velocity in the displacement time graph is the straight-line slope.

The displacement time graph displays various slopes based on the motion. But the slope emerges as a steep straight line signifies the constant velocity. The steeper or more vertical the graph’s slope line, the faster an object travels with constant velocity as no net force is acted.

From the displacement – time graph above, calculate the following:

How far does the car travel in 8 secs?

What is the constant velocity of the car initially and laterally?

What is the average speed of the car?

Formula:

v ^→ = s₂-s₁/t₂-t₁
v=s_f-s_i/t_f-t_i

Solution:

1) The car travelled 60m in 8 seconds.

2) (i) The constant velocity of the car initially is calculated as,

v ^→ = s₂-s₁/t₂-t₁

Subsisting values from the slope of graph,

v ^→ = 50-0/5-0

v ^→ = 50/5

v ^→ = 10

The constant velocity of the car initially is 10m/s

(ii) The constant velocity of the car latterly is,

v ^→ = s₂-s₁/t₂-t₁

Subsisting values from the slope of graph,

v ^→= 100-50/10-7

v ^→= 50/3

v ^→ = 16.6

The constant velocity of the car latterly is 16.6m/s

3)The average speed of the car is calculated as,

v=s_f-s_i/t_f-t_i

Substituting all values from the slope of graph,

v = 100-0/10-0

v = 10

The car’s average speed is 10m/s.

When Velocity is Constant in Displacement Time Graph?

The velocity is constant in the displacement time graph when its displacement changes.

The correlation between displacement and velocity gives rise to the diverse slope shapes in the displacement time graph. It depends on the positive or negative value of the constant velocity as per an object’s direction.

Let’s illustrate the displacement-time graph for high and low constant velocities for the car moving to the right. We observe the slope of the low constant velocity graph is more down than the high constant velocity graph. Both slopes of constant velocities is positive as the displacement changes in an increasing manner. That’s how we can imagine which object is moving faster than the other from the displacement time graph.

**Constant Positive Velocities**
(credit: shutterstock)

On the other hand, if an object’s displacement decreases with time, it shows the constant negative velocity in the displacement time graph.

We have examined the subsequent facts in the displacement time graph:

Greater the slope, the greater the constant velocity.
If displacement increases, sloping upwards as time increases, and constant velocity is positive.
If displacement decreases, sloping downwards as time increases, and constant velocity is negative.
If displacement is zero, the slope is horizontal, and the constant velocity is zero.

Read more about Relative Motion.

Analyze and comment on different sections in the displacement time graph below. Also, calculate the constant positive velocity and constant negative velocity from the graph.

**Constant Positive Velocity and Constant Negative Velocity**

Solution:

1) The section OA represents the constant positive velocity as displacement is increasing.

The positive constant velocity is calculated as

v ^→ = s₂-s₁/t₂-t₁

Substituting values from the section AB of the graph,

v ^→ = 6-0/2-0

v ^→ =3

The constant positive velocity is 3m/s.

2) The displacement does not change from section A to B for 4 seconds; that means an object remains stopped at 2m and its constant velocity is zero.

3) The section BC represents the constant negative velocity as displacement is decreasing.

The negative constant velocity is calculated as

v ^→ = s₂-s₁/t₂-t₁

Substituting values from the section BC of the graph,

v ^→ = 0-6/2-0

v ^→ = -6/2

v ^→ =-3

The constant negative velocity is -3m/s.

4) At point C, displacement is zero; that means the moving object with constant velocity returns to its original position and becomes stationary.

Also Read:

Is Horizontal Speed Constant: Why, How, When, Problems

January 21, 2024January 24, 2022 by Manish Naik

The article discusses about is horizontal speed constant, along with its problem examples.

The horizontal speed is constant throughout the path as no force acts on the projectile in the horizontal direction to alter its acceleration. At the same time, the projectile travels downwards due to the gravity force, which defines the projectile motion along with constant horizontal speed.

The projectile travels vertically and horizontally at the exact time. Firstly, it happens because the air resistance force or air drag on the projectile is neglected as it is minimal. Hence, there is no force acting on the projectile in the horizontal direction to change its horizontal component of speed. As a result, the projectile traveled a constant horizontal distance (x) in unit time with constant horizontal speed (v_H).

x = v_Ht …………. (*)

The horizontal speed equals the initial velocity u at the projectile launch. Hence,

v_H = u …………… (1)

If a cannonball fired from the cannon horizontally from the cliff, it takes 5 sec to reach the ground. Also, it falls on the ground at 20m from the base of the cliff. Calculate both the horizontal and vertical speeds of the cannonball after the fire?

Given:

t = 5s

x = 20m

g = 9.8m/s²

To Find:

v_H =?
v_V =?

Formula:

x = v_Ht
v_V = u_V + gt

Solution:

The horizontal speed is calculated as,

x = v_Ht

v_H = x/t

Substituting all values,

v_H = 20 / 5

v_H = 4

The horizontal speed of the cannonball is 4m/s

The vertical speed is calculated from the kinematics equation of motion,

v_V = u_V + gt

Since initial vertical velocity is zero. i.e., u_V= 0.

Substituting all values,

v_V = 0 + 9.8 x 5

v_V = 49

The vertical speed of the cannonball is 49m/s.

Why is Horizontal Speed Constant?

The horizontal speed is constant in projectile motion due to no forces in the horizontal direction.

The horizontal speed implies that the projectile moves in a straight path without changing its direction. No net forces are acting in a horizontal direction after the projectile is launched. Hence, the projectile’s horizontal speed stays constant throughout the trajectory motion due to zero acceleration.

A football moves 20m along the ground at 10 sec after a player kicks it. When a player kicks the football at 60°, it changes velocity from 5m/s to 10m/s; it reaches the ground in 5 sec after moving in a parabolic path. Calculate the horizontal speed and vertical velocity in both cases.

Given:

Case 1:

x = 20m

t = 10sec

Case 2:

θ = 60°

u = 5m/s

u_V = 10m/s

t = 5sec

g = 9.8m/s²

To Find:

v_H =?
v_V =?

Formula:

x = v_Ht
v_V = u_V + gt

Solution:

1) As the football moves along the ground, the launch angle θ is zero.

Hence, in case 1, the football moves with constant horizontal speed, and the vertical velocity v_V is zero.

The horizontal speed of football is calculated as,

x = v_Ht

v_H= x/t

Substituting all values,

v_H= 20/10

v_H= 2

The football moves along the ground with a constant horizontal speed of 2m/s.

2) As the football launches at 60°, instead of horizontal speed, the football moves with a horizontal velocity equal to the initial velocity of football u = 5m/s.

The vertical velocity of the football is calculated as,

v_V = u_V + gt

Substituting all values,

v_V = 10 + 9.8 x 5

v_V = 10 + 49

v_V = 59

The vertical velocity of the football launched at 60° is 59m/s.

How is Horizontal Speed Constant?

The horizontal speed is constant only when the object moves in a parabolic path.

The projectile motion in parabolic trajectory defined by horizontal and vertical velocity is only controlled by the downward gravity force, which sets up only vertical acceleration. Since there is no net force to set up the horizontal acceleration, the projectile moves with constant horizontal speed.

Read more about Relative Motion.

Earlier, we discussed the launch angle and horizontal speed relation. We will learn more about the time in which the projectile remains in the air. The best combination for the time in the air when the projectile moves with horizontal speed is launching at 45°, which provides the maximum horizontal distance to the projectile. The basketball and javelin thrower is a projectile motion example that proves this sentence true.

**Horizontal Speed in Basketball**
(credit: shutterstock)

We have determined earlier the formula about horizontal distance x. The vertical distance travelled by projectile is calculated using second kinematics equation of motion.

y = u_yt + 1/2 gt²

But u_V is zero.

y = gt²/2 ……….. (4)

Substituting value of t from equation (*), t = x/v_x , we get

y = g(x/v_x)² / 2

y = gx/2v_x² ……………… (5)

The above equation is the parabola equation, which predicts how much the projectile travels vertical and horizontal after launch. It shows the parabolic trajectory of the projectile, which launched parallel to horizontal.

Let’s calculate the magnitude of the resultant velocity of the projectile using the Pythagoras theorem.

v² =v_x² + v_y²

From equation (1) and (4).

v= √(u²+g²t²) ……………. (6)

Now let’s calculate the formula about how the projectile remains in the air from the second kinematics equation of motion for vertical distance traveled (4),

t = √(2y/g) ………… (7)

That means, if we determine the time for a projectile in the air (t), we can also calculate the horizontal distance travelled (x) as the constant horizontal speed (v_H) is the same as the initial velocity at which the projectile launched during the trajectory.

Therefore, the horizontal distance (x) by the projectile with constant horizontal speed (v_H) is given by,

x = v_Ht

Using equation (1) and (7), we get

x = u √(2y/g) ………………(8)

**Horizontal and Vertical Distance in Projectile Motion**
(credit: shuttertstock)

Suppose a basketball is thrown at angle 30°, which moves initially at 5m/s and then reaches the basket in 4sec after parabolic trajectory.

Calculate the vertical distance travelled by the basketball.

Calculate the horizontal distance travelled by the basketball.

Calculate the horizontal speed of the basketball.

Calculate the resultant velocity of the basketball.

Given:

θ = 30°

u = 5m/s

t = 4s

g = 9.8m/s²

To Find:

y =?
x =?
v_H =?
v =?

Formula:

y = gt²/2
x = u √(2y/g)
x = v_Ht
v = √(u²+g²t²)

Solution:

1)The vertical distance travelled by the basketball is calculated as,

y =gt²/2

Substituting all values,

y = 9.8*4²/2

y = 78.4

The vertical distance travelled by basketball, launched at 30°, is 78.4m.

2) The horizontal distance travelled by the basketball is calculated as,

x = u √(2y/g)

Substituting all values,

x = 5 √(2*74.8/9.8)

x = 5√15.26

x = 5×3.9

x = 19.5

The horizontal distance travelled by basketball, launched at 30°, is 19.5m.

3)The horizontal speed of basketball is calculated as,

x = v_Ht

v_H = x/t

Substituting all values,

v_H = 19.5/4

v_H = 4.87

The basketball moves with a constant horizontal speed of 4.87m/s.

4)The resultant velocity of basketball is calculated as,

v = √(u²+g²t²)

Substituting all values,

v = √(5²+9.8²*4²)

v = √(25+1536.64)

v = √1561.64

v = 39.51

The resultant velocity of a basketball, launched at 30°, is 39.51m/s.

Also Read:

Horizontal Speed Vs Horizontal Velocity: Comparative Analysis

January 21, 2024January 20, 2022 by Manish Naik

The article discusses comparative analysis and insight about Horizontal Speed Vs Horizontal Velocity.

Horizontal Speed	Horizontal Velocity
It is a scalar quantity that is either zero or positive.	It is a vector quantity that is either zero, negative, or positive.
It concerns both linear and projectile motion.	It concerns only projectile motion.
The body travels in the horizontal direction.	The projectile travels in both horizontal and vertical directions at the same time.
It implies how fast the body is traveling after being launched.	It implies which direction the projectile is traveling after being launched.
It corresponds to the distance travelled by the body.	It corresponds to the displacement of a projectile.
Since no force acted horizontally, the body travelled a constant distance in unit time.	Since no force acted horizontally, the projectile has a constant displacement in unit time.
It is denoted by ‘v’ (italic v) and expressed in meter/second.	It is denoted by ‘v’ and expressed in kilometers/hour.
It is calculated as distance (d) per time. v = d/t	It is calculated as displacement (s) per time. v = s/t

How to differentiate Horizontal Speed and Horizontal Velocity?

Let’s discuss some projectile examples that differentiate between the projectile’s horizontal speed and horizontal velocity.

Suppose the airplane moves at a speed of 100m/s at an altitude of 1000m, dropping the box with a 10km/hr velocity towards the ground. In such a case, we note that the airplane moves in a straight direction with a constant horizontal speed of 100m/s. In distinction, the box falls in a curved trajectory when dropped from the airplane.

The falling box displays projectile motion, and it falls to the right with horizontal velocity and downward with vertical velocity due to gravity force. We also observe that the falling box is initially horizontal x-direction and not vertical y-direction. So, v_y = 0. The horizontal velocity maintains its initial value v_x = 100km/hr throughout the drop.

Is Horizontal Velocity Constant: Why, When, How, Problem Examples

January 21, 2024January 16, 2022 by Manish Naik

The article discusses about is horizontal velocity constant or why horizontal velocity constant in projectile motion, along with its problem examples.

After the applied force sets the parabolic path in projectile motion, only the gravity force acts downward to a projectile in the vertical direction. So an object moves downward with varying vertical velocity but constant horizontal velocity.

When is Horizontal Velocity Constant?

The horizontal velocity during projectile motion is constant.

As per definition, the projectile motion concerns a single gravity force that provokes a net force toward the earth’s center. Since no horizontal force acts in projectile motion, its horizontal velocity stays constant and progresses towards the earth’s center due to gravity force.

To understand the concept of downward acceleration with constant horizontal velocity, consider an example of a boy throwing a ball in a straight direction with high velocity. Let’s say the gravity is negligible in this case, then how does the ball move in the absence of gravity?

Newton’s law says, “an object in motion continues to move in motion unless some force acts on it“. That means; a ball moves in a straight path with a constant velocity, which agrees with the law of inertia.

What is Inertia? — **Law of Inertia in Projectile Motion**
(credit: shutterstock)

Now in the same case, what if gravity exists? The gravity force influences the ball’s vertical motion, causing a vertical acceleration, agreeing with the free-falling object accelerating at a rate of acceleration due to gravity g. Because of gravity force, the ball will drop vertically below its straight path after parabolic trajectory as per the attribute of the projectile motion.

Suppose that boy applies a force of 20N to throw the ball. The ball achieves a velocity of 10m/s and persists in moving with the same velocity horizontally. At the same time, the vertical velocity achieves the velocity at 9.8m/s each second due to gravity force. That means the ball has a vertical acceleration of about 9.8 m/s but no horizontal acceleration.

**Horizontal Velocity Constant**
(credit: shutterstock)

The downward acceleration results in a lowered trajectory from the straight line the projectile ball travels if the gravity force is negligible. The ball possesses its constant horizontal velocity in the existence of gravity which acts to drive the exact vertical velocity as before.

However, the gravity or the vertical force works perpendicular to the horizontal velocity of the projectile. Since the perpendicular components are independent of each other, the gravity force cannot impact the ball’s horizontal velocity. That’s why any projectile moves with downward acceleration but constant horizontal velocity.

Why is Horizontal Velocity Constant?

The horizontal velocity of the projectile is constant because of its inertia.

If there is no force, an object’s state will not modify as per the law of inertia. A projectile’s tendency to remain in a motion with constant velocity resulted in the horizontal motion of the projectile. Therefore, the projectile drives with a stable horizontal velocity.

**Why is Horizontal Velocity Constant** (credit: shutterstock)

What Is Change in Momentum: How to Find, Facts and Problems, Examples

January 20, 2024January 13, 2022 by Manish Naik

The article discusses about what is change in momentum and formulas about how to find change in momentum.

The change in momentum is a difference in motion. When an object is in motion, strikes, or collides with another object, the exerted force accelerates an object by varying its motion. The change in momentum is calculated using the Impulse formula or conservation of momentum.

The moving object is said to be in momentum. The amount of momentum gained by an object is proportional to its mass and velocity together. Hence, the change in one quantity can cause a change in momentum. That means, if you increase or decrease an object’s mass, then its momentum changes. Similarly, the momentum also changes when you increase or decrease its velocity.

The change in velocity means an object is accelerating, and we have learned that acceleration is caused by force. So greater the acceleration caused by force, the greater its change in momentum!!!

Suppose an object is at rest or does not have momentum, then it requires sufficient force to overcome the friction so that an object moves with some momentum. If an object is already in momentum and force is applied in the opposite direction, its momentum will decrease. But If force applies in the same direction, then momentum will increase.

A large force exerted over a short time causes a significant change in momentum. If the force is small but exerted over a long time, a significant change in momentum also occurs. That means when a force applies to an already gained momentum object for a certain period of time, its momentum changes.

Change in Momentum Examples

The change in momentum examples explains how momentum changes when a force is acted upon.

Long Jump
Hitting a Ball
Driving a Vehicle
Football or Rugby
Playground Slide

Momentum Change Examples — **Change in Momentum Examples**

Long Jump

To attain momentum before jumping, the athlete runs a certain distance. Once the athlete gains some momentum after running, they apply force on the ground to jump that changes its momentum, and the athlete jumps forward. After the jump, the athlete needs to apply a large force again on the ground to stop their motion – which changes their momentum again.

Change in Momentum Jumping Example — **Momentum Change in Long Jump**
(credit: shutterstock)

Hitting a Ball

The ball gains momentum when thrown by the bowler during sports like cricket or baseball. When a batter hits the ball with a bat, the applied force to the ball changes its momentum, and then a ball moves in the direction of an applied force. The ball’s momentum changes again when it is stopped or caught by the fielder.

Change in Momentum Hitting Ball Example — **Momentum Change in Hitting Ball**
(credit: shutterstock)

Driving Vehicle

Driving a vehicle such as a car or truck involves a continuous change in momentum. We apply a force on its accelerating paddle to speed up the vehicle. To stop the vehicle suddenly, we apply a force on its brake paddle, which changes the vehicle’s momentum as per applied force.

Change in Momentum Driving Example — **Momentum Change in Driving Vehicle** (credit: shutterstock)

Football or Rugby

In football or rugby, the striker accumulates maximum momentum when he dribbles or carries the ball near the goal coast by bypassing other players. The defenders near the goal coast then try to stop or change the striker’s momentum by tackling him.

Playground Slides

When children start sliding on the playground slide from height, they achieve momentum downwards. But the friction present on the playground slide surface changes the momentum of sliding children by opposing their motion, preventing them from falling at the end of the slide.

Change in Momentum Formula

The change in momentum formula is calculated using Newton’s second law of motion and kinematics equations of motion.

Newton’s second law displays that an object accelerates when a force applies. Since the force applied for a specific time interval changes an object’s motion, it also causes a change in momentum. The product of force and time interval is termed as ‘impulse’, which measures the change in momentum.

The first kinematics equation of motion is,

v_f = v_i + at

v_f– v_i = at

As per Newton’s second law, F = ma, a = F/a

v_f-v_i=(F/m)*t

m(v_f– v_i) = Ft

mΔv=Ft

Whereas RHS in the above equation is termed as ‘Impulse’ (denoted by J) and LHS is the formula of Momentum Change (ΔP)

Therefore, we can also write as,

J=Ft=mΔv=ΔP…………. (*)

Impulse = Change in Momentum

The equation (*) is also known as ‘Impulse –Momentum change equation’.

The more significant the impulse, the more significant the momentum change.

The strength of force applied to an object depends on how long it acts. The concept of impulse quantifies the effect of force.

Suppose a striker having a mass of 40kg dribbles the football at 10m/s. When he approaches the defender in 0.10sec, he dribbles the football at high speed, such as 15m/s, to pass the defender successfully.

**Momentum Change in Football**
(credit: shutterstock)

What is the initial momentum of the striker?

What is the final momentum of the striker when he is approached by the defender?

What is the change in momentum of the sticker?

How much force is applied by the defender to stop the striker?

Calculate Impulse applied by the defender.

Given:

m = 40kg

v_i= 10m/s

v_f = 15m/s

t = 0.10s

To Find:

P_i =?
P_f =?
ΔP =?
F =?

Formula:

P = mv
ΔP = P_f –P_i
Ft = ΔP

Solution:

The initial momentum of a striker is calculated as,

P_i = mv_i

P_i = 40 x 10

P_i= 400

The initial momentum of striker is 400kg.m/s

The final momentum of a striker is calculated as,

P_f= mv_f

P_f= 40 x 15

P_f = 600

The final momentum of striker is 600kg.m/s

The change in momentum of a striker is calculated as,

ΔP = P_f –P_i

ΔP = 600 – 400

ΔP = 200

The change in momentum of a striker is 200kg.m/s.

The force applied by the defender to stop the striker is calculated using the impulse-momentum Change formula.

Ft = ΔP

Substituting all values,

F(0.10) = 200

F = 200/0.10

F = 2000

The force applied by the defender is 2000N.

The impulse by defender is calculated as,

J = Ft

J = 2000 x 0.10

J = 200

The impulse by defender on striker is 200N.s.

How to Calculate Momentum Change?

The momentum change is calculated using the law of conservation of momentum.

When an external force acts on an object, we can compute its momentum change by impulse formula. But when there is no external force, the total momentum of colliding objects remains the same. That’s how we can calculate the momentum change due to collision using conservation of momentum.

**How to Calculate Momentum Change** (credit: shutterstock)

Suppose two objects have a momentum P₁ and P₂ due to their masses m₁ and m₂ and velocity u₁ and u₂. Due to collision, their momentum changes to P_1’ and P_2’ since their velocity changes to v₁ and v₂.

As per conservation of momentum,

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

Since we want to calculate change in momentum, rearranging terms m1 on LHS and m2 on RHS,

m₁u₁ – m₁v₁ = m₂u₂ – m₂v₂

P_1f– P_1i = P_2f-P_2i

ΔP₁ = ΔP₂

Momentum change in object 1 = Momentum change in object 2

When two balls have masses 5kg and 3kg moving towards each other at 8m/s and 15m/s, respectively, and after the collision, if the first ball moves away with a velocity of 5m/s.

Calculate the change in momentum of the first ball

Calculate the change in velocity of the second ball after the collision.

Given:

m₁ = 5kg

m₂ = 3kg

u₁ = 8m/s

u₂ = 15m/s

v₁ = 5m/s

To Find:

ΔP₁ =?
v₂ =?

Formula:

ΔP₁ = P_1f– P_1i
m₁u₁ – m₁v₁ = m₂u₂ – m₂v₂

Solution:

The change in momentum of first ball is calculated as,

ΔP₁ = P_1f– P_1i

ΔP1 = m₁u₁ – m₁v₁

Substituting all values,

ΔP₁ = 5 x 8 – 5 x 5

ΔP₁ = 40 – 25

ΔP₁ = 25

The momentum change of first ball is 25kg.m/s

The velocity change of second ball after collision is calculated as,

m₁u₁ – m₁v₁ = m₂u₂ – m₂v₂

Substituting all values,

25 = 3 x15 – 3v₂

25 = 45 – 3v₂

v₂= -20/3

v₂ = -6.66

The velocity change of the second ball is -6.6m/s.

Also Read:

How to Calculate Momentum of a System: Various Problems and Facts

January 21, 2024January 9, 2022 by Manish Naik

In the world of physics, understanding the concept of momentum is crucial. Momentum is a fundamental quantity that helps us describe and analyze the motion of objects and systems. By calculating momentum, we can gain insights into the behavior of physical systems, such as the motion of particles, collisions, and even rotating objects. In this blog post, we will explore how to calculate momentum of a system, delve into special cases, and even touch upon the concept of angular momentum. So, let’s get started!

Calculating Momentum of a System

The Formula for Calculating Momentum

The formula for calculating momentum is simple yet powerful. Momentum is defined as the product of an object’s mass and its velocity. Mathematically, it can be expressed as:

$\text{Momentum (p)} = \text{Mass (m)} \times \text{Velocity (v)}$

where both mass and velocity are vector quantities. In terms of units, momentum is measured in kilogram-meters per second (kg·m/s).

How to Use the Momentum Formula

how to calculate momentum of a system — Image by Tdadamemd – Wikimedia Commons, Wikimedia Commons, Licensed under CC BY-SA 3.0.

To calculate the momentum of an object, you need to know its mass and velocity. Let’s consider a simple example. Suppose we have a car with a mass of 1500 kg, moving at a velocity of 30 m/s. To determine its momentum, we can use the formula:

$\text{Momentum (p)} = \text{Mass (m)} \times \text{Velocity (v)}$

Plugging in the values, we get:

$\text{Momentum} = 1500 \, \text{kg} \times 30 \, \text{m/s}$

Calculating this, we find that the momentum of the car is 45,000 kg·m/s.

Worked out Examples on Calculating Momentum

Let’s work through a few more examples to solidify our understanding.

Example 1:
Suppose a truck with a mass of 2000 kg is moving at a velocity of 20 m/s. What is its momentum?

Using the formula, we can calculate the momentum as:

$\text{Momentum} = 2000 \, \text{kg} \times 20 \, \text{m/s}$

The momentum of the truck is therefore 40,000 kg·m/s.

Example 2:
Consider a collision between a car and a stationary object. The car has a mass of 1000 kg and is initially traveling at 15 m/s. During the collision, the car comes to a stop. What is the change in momentum of the car?

To find the change in momentum, we need to calculate the final momentum and subtract the initial momentum. The final momentum can be calculated using the formula:

$\text{Momentum} = \text{Mass} \times \text{Velocity}$

For the car at rest, its final momentum is given by:

$\text{Final Momentum} = 1000 \, \text{kg} \times 0 \, \text{m/s} = 0 \, \text{kg·m/s}$

The initial momentum of the car is:

$\text{Initial Momentum} = 1000 \, \text{kg} \times 15 \, \text{m/s} = 15,000 \, \text{kg·m/s}$

Therefore, the change in momentum is:

$\text{Change in Momentum} = \text{Final Momentum} - \text{Initial Momentum} = 0 - 15,000 = -15,000 \, \text{kg·m/s}$

The negative sign indicates that the momentum of the car has decreased.

Special Cases in Calculating Momentum

How to Determine the Initial Momentum of a System

When dealing with systems of objects, it is essential to consider the initial momentum of the system before any external forces act upon it. The initial momentum of a system can be calculated by simply adding up the individual momenta of each object in the system.

For example, imagine a system consisting of two objects: a car with a mass of 1000 kg moving at 20 m/s and a truck with a mass of 2000 kg moving at 15 m/s. To determine the initial momentum of the system, we add the momenta of the two objects:

$\text{Initial Momentum} = \text{Momentum of Car} + \text{Momentum of Truck} = (1000 \, \text{kg} \times 20 \, \text{m/s}) + (2000 \, \text{kg} \times 15 \, \text{m/s})$

Calculating this, we find that the initial momentum of the system is 45,000 kg·m/s.

How to Measure the Final Momentum of a System

The final momentum of a system can be determined in a similar manner to the initial momentum. We add up the individual momenta of each object in the system after any external forces have acted upon it.

Continuing with our previous example, let’s say the car collides with the truck, and both objects come to a stop. The final momentum of the system would then be:

$\text{Final Momentum} = \text{Momentum of Car} + \text{Momentum of Truck} = (1000 \, \text{kg} \times 0 \, \text{m/s}) + (2000 \, \text{kg} \times 0 \, \text{m/s})$

The final momentum of the system is zero, indicating that the objects have come to a complete stop.

Calculating the Total Momentum of a System Before and After Collision

In scenarios involving objects colliding or interacting with each other, we can examine the conservation of momentum. According to the law of conservation of momentum, the total momentum of a system before a collision is equal to the total momentum after the collision, as long as no external forces act on the system.

To calculate the total momentum before and after a collision, we sum up the individual momenta of all objects in the system.

For instance, consider a collision between a car and a truck. The car has a mass of 1000 kg and is initially traveling at 20 m/s, while the truck has a mass of 2000 kg and is initially moving at 15 m/s. The total momentum before the collision is:

$\text{Total Initial Momentum} = \text{Momentum of Car} + \text{Momentum of Truck} = (1000 \, \text{kg} \times 20 \, \text{m/s}) + (2000 \, \text{kg} \times 15 \, \text{m/s})$

After the collision, let’s assume the car and truck come to a stop. The total momentum after the collision is:

$\text{Total Final Momentum} = \text{Momentum of Car} + \text{Momentum of Truck} = (1000 \, \text{kg} \times 0 \, \text{m/s}) + (2000 \, \text{kg} \times 0 \, \text{m/s})$

Remarkably, the total initial momentum is equal to the total final momentum, confirming the conservation of momentum.

Angular Momentum of a System

Understanding Angular Momentum

Image by Olivier Cleynen – Wikimedia Commons, Wikimedia Commons, Licensed under CC0.

In addition to linear momentum, we can also consider angular momentum when dealing with rotating objects or systems. Angular momentum is a measure of an object’s rotational motion and depends on its moment of inertia and angular velocity.

How to Calculate Angular Momentum of a Single Disk System

For a single disk rotating around a fixed axis, the angular momentum can be calculated using the following formula:

$\text{Angular Momentum (L)} = \text{Moment of Inertia (I)} \times \text{Angular Velocity (ω)}$

The moment of inertia depends on the mass distribution of the object and is specific to each shape. The angular velocity is the rate at which the object rotates.

How to Calculate Angular Momentum of a Two Disk System

When dealing with a system of objects, such as two disks rotating around a fixed axis, we can calculate the total angular momentum by summing up the individual angular momenta of each object.

Let’s say we have two disks with different masses, moments of inertia, and angular velocities. The total angular momentum of the system would be:

$\text{Total Angular Momentum} = \text{Angular Momentum of Disk 1} + \text{Angular Momentum of Disk 2}$

Calculating the angular momentum for each disk and adding them up will give us the total angular momentum of the system.

Worked Out Examples on Calculating Angular Momentum

To reinforce our understanding of angular momentum, let’s work through a couple of examples.

Example 1:
Consider a disk with a moment of inertia of 0.5 kg·m² and an angular velocity of 4 rad/s. What is its angular momentum?

Using the formula for angular momentum, we can calculate:

$\text{Angular Momentum} = \text{Moment of Inertia} \times \text{Angular Velocity}$

Plugging in the values, we get:

$\text{Angular Momentum} = 0.5 \, \text{kg·m²} \times 4 \, \text{rad/s}$

Calculating this, we find that the angular momentum of the disk is 2 kg·m²/s.

Example 2:
Suppose we have two disks in a system. Disk 1 has a moment of inertia of 0.3 kg·m² and an angular velocity of 5 rad/s, while Disk 2 has a moment of inertia of 0.2 kg·m² and an angular velocity of 3 rad/s. What is the total angular momentum of the system?

To find the total angular momentum, we add the individual angular momenta of each disk:

$\text{Total Angular Momentum} = \text{Angular Momentum of Disk 1} + \text{Angular Momentum of Disk 2}$

Calculating this, we find:

$\text{Total Angular Momentum} = (0.3 \, \text{kg·m²} \times 5 \, \text{rad/s}) + (0.2 \, \text{kg·m²} \times 3 \, \text{rad/s})$

Simplifying this, we determine that the total angular momentum of the system is 3 kg·m²/s.

How can you calculate the momentum of a system before a collision? Calculating momentum before a collision.

To calculate the momentum of a system before a collision, you need to determine the momentum of each individual object involved in the collision and then add them together. The momentum of an object is given by the product of its mass and velocity. By calculating the momentum of each object before the collision and adding them together, you can determine the total momentum of the system before the collision occurs. This process is explained in more detail in the article Calculating momentum before a collision.

Numerical Problems on how to calculate momentum of a system

Problem 1:

A system consists of two objects with masses of 5 kg and 8 kg, respectively. The velocity of the first object is 4 m/s to the right, while the velocity of the second object is 6 m/s to the left. Calculate the total momentum of the system.

Solution:

The momentum of an object is given by the equation:

$p = m \cdot v$

where $p$ is the momentum, $m$ is the mass, and $v$ is the velocity.

For the first object:
$p_1 = m_1 \cdot v_1 = 5 \, \text{kg} \cdot 4 \, \text{m/s} = 20 \, \text{kg m/s}$

For the second object:
$p_2 = m_2 \cdot v_2 = 8 \, \text{kg} \cdot (-6 \, \text{m/s}) = -48 \, \text{kg m/s}$

The total momentum of the system is the sum of the individual momenta:
$p_{\text{total}} = p_1 + p_2 = 20 \, \text{kg m/s} + (-48 \, \text{kg m/s}) = -28 \, \text{kg m/s}$

Therefore, the total momentum of the system is -28 kg m/s to the left.

Problem 2:

A system consists of three objects with masses of 2 kg, 3 kg, and 4 kg, respectively. The velocities of the objects are 5 m/s to the right, 2 m/s to the right, and 4 m/s to the left. Calculate the total momentum of the system.

Solution:

For the first object:
$p_1 = m_1 \cdot v_1 = 2 \, \text{kg} \cdot 5 \, \text{m/s} = 10 \, \text{kg m/s}$

For the second object:
$p_2 = m_2 \cdot v_2 = 3 \, \text{kg} \cdot 2 \, \text{m/s} = 6 \, \text{kg m/s}$

For the third object:
$p_3 = m_3 \cdot v_3 = 4 \, \text{kg} \cdot (-4 \, \text{m/s}) = -16 \, \text{kg m/s}$

The total momentum of the system is the sum of the individual momenta:
$p_{\text{total}} = p_1 + p_2 + p_3 = 10 \, \text{kg m/s} + 6 \, \text{kg m/s} + (-16 \, \text{kg m/s}) = 0 \, \text{kg m/s}$

Therefore, the total momentum of the system is 0 kg m/s.

Problem 3:

A system consists of two objects with masses of 6 kg and 9 kg, respectively. The velocity of the first object is 3 m/s to the left, while the velocity of the second object is 7 m/s to the right. Calculate the total momentum of the system.

Solution:

For the first object:
$p_1 = m_1 \cdot v_1 = 6 \, \text{kg} \cdot (-3 \, \text{m/s}) = -18 \, \text{kg m/s}$

For the second object:
$p_2 = m_2 \cdot v_2 = 9 \, \text{kg} \cdot 7 \, \text{m/s} = 63 \, \text{kg m/s}$

The total momentum of the system is the sum of the individual momenta:
$p_{\text{total}} = p_1 + p_2 = -18 \, \text{kg m/s} + 63 \, \text{kg m/s} = 45 \, \text{kg m/s}$

Therefore, the total momentum of the system is 45 kg m/s to the right.

Also Read:

How to Calculate Momentum Before Collision: Elastic, Inelastic, Formula and Problems

January 20, 2024January 6, 2022 by Manish Naik

Master calculating momentum before a collision: Explore formulas, tackle problems, and understand elastic and inelastic collisions.

Before we dive into the calculations, let’s first understand what momentum is. Momentum is a fundamental concept in physics that describes the motion of an object. It is defined as the product of an object’s mass and velocity. In simpler terms, momentum tells us how much “oomph” an object has when it is in motion.

Now, let’s explore how to calculate momentum before a collision. In this blog post, we will discuss the momentum formula, provide a step-by-step guide on how to calculate momentum, and explore factors that affect momentum before a collision. We will also compare momentum before and after a collision and discuss the principle of conservation of momentum.

How to Calculate Momentum Before Collision

Explanation of the Momentum Formula

The momentum of an object can be calculated using the following formula:

$text{Momentum (p)} = text{Mass (m)} times text{Velocity (v)}$

The formula states that momentum is equal to the product of an object’s mass and velocity. Mass refers to the amount of matter an object contains, while velocity is the speed at which the object is moving in a specific direction.

Step-by-Step Guide on How to Calculate Momentum

To calculate momentum before a collision, follow these steps:

Identify the mass of the object (m) in kilograms.
Determine the velocity of the object (v) in meters per second.
Multiply the mass and velocity to find the momentum (p) of the object.

Let’s take a look at an example to better understand this process.

Worked Out Example: Calculating Momentum Before Collision

Let’s say we have a car with a mass of 1000 kg traveling at a velocity of 20 m/s. To calculate the momentum before a collision, we can use the formula:

$text{Momentum (p)} = text{Mass (m)} times text{Velocity (v)}$

Substituting the given values into the formula, we have:

$text{Momentum (p)} = 1000 , text{kg} times 20 , text{m/s}$

Simplifying the calculation, we find:

$text{Momentum (p)} = 20000 , text{kg} cdot text{m/s}$

So, the momentum before the collision is 20000 kg·m/s.

Factors Affecting Momentum Before Collision

how to calculate momentum before collision — Image by Fizped – Wikimedia Commons, Wikimedia Commons, Licensed under CC BY 3.0.

Now that we know how to calculate momentum before a collision, let’s explore the factors that can affect it.

Mass of the Object

The mass of an object plays a crucial role in determining its momentum. The greater the mass, the greater the momentum. This means that an object with a larger mass will be harder to stop or change its motion.

Velocity of the Object

The velocity of an object also has a significant impact on its momentum. The higher the velocity, the greater the momentum. An object moving at a faster speed will have a larger momentum compared to an object moving at a slower speed.

Direction of the Object’s Movement

The direction in which an object is moving is vital in determining its momentum. Momentum is a vector quantity, meaning it has both magnitude and direction. Two objects with the same mass and velocity but moving in opposite directions will have equal magnitudes of momentum but different directions.

Comparing Momentum Before and After Collision

Now that we understand how to calculate momentum before a collision, let’s explore how we can compare momentum before and after a collision. In this scenario, we will consider the principle of conservation of momentum.

How to Calculate Momentum After Collision

When two objects collide, their momentum can be calculated using the same formula we discussed earlier:

$text{Momentum (p)} = text{Mass (m)} times text{Velocity (v)}$

However, to calculate the momentum after a collision, we need to consider the masses and velocities of both objects involved in the collision.

Understanding the Principle of Conservation of Momentum

The principle of conservation of momentum states that the total momentum of a system of objects remains constant before and after a collision, provided no external forces act on the system. In simpler terms, the total momentum before the collision will be equal to the total momentum after the collision.

Worked Out Example: Comparing Momentum Before and After Collision

Let’s consider a collision between two objects: a ball with a mass of 0.5 kg and a velocity of 10 m/s, and another ball with a mass of 0.3 kg and a velocity of -5 m/s. The negative velocity indicates the opposite direction of motion. To compare the momentum before and after the collision, we can follow these steps:

Calculate the momentum before the collision using the individual masses and velocities of the objects.
Determine the momentum after the collision using the masses and velocities of both objects after the collision.

By applying the formula for momentum, we can calculate the momentum before the collision as follows:

$text{Momentum (p)} = text{Mass (m)} times text{Velocity (v)}$

For the first ball, the momentum before the collision is:

$text{Momentum (p1)} = 0.5 , text{kg} times 10 , text{m/s} = 5 , text{kg} cdot text{m/s}$

For the second ball, the momentum before the collision is:

$text{Momentum (p2)} = 0.3 , text{kg} times (-5) , text{m/s} = -1.5 , text{kg} cdot text{m/s}$

Now, let’s consider the momentum after the collision. The total momentum of the system after the collision will be the sum of the individual momenta of the balls. Let’s assume the first ball stops after the collision, and the second ball continues moving in the same direction. Therefore, the momentum after the collision can be calculated as:

$text{Total Momentum (p)} = text{Momentum (p1)} + text{Momentum (p2)}$

Substituting the values we calculated earlier, we have:

$text{Total Momentum (p)} = 5 , text{kg} cdot text{m/s} + (-1.5 , text{kg} cdot text{m/s}) = 3.5 , text{kg} cdot text{m/s}$

So, the total momentum after the collision is 3.5 kg·m/s.

How is momentum calculated before a collision and in a system, and how do they relate to each other?

Calculating momentum in a system is an essential concept in physics, and it involves determining the mass and velocity of each object within the system. By using the formula p = mv, where p represents momentum, m is the mass, and v is the velocity, we can calculate the momentum of individual objects before a collision and the total momentum of the system. The link Calculating momentum in a system provides detailed information on how to perform these calculations. Understanding the momentum before a collision and in a system allows us to analyze and predict the resulting motion and energy changes during interactions between objects.

Numerical Problems on how to calculate momentum before collision

Problem 1:Two cars, A and B, are traveling towards each other on a straight road. Car A has a mass of 800 kg and is traveling at a velocity of 20 m/s towards the east. Car B has a mass of 1200 kg and is traveling at a velocity of 10 m/s towards the west. Calculate the momentum of each car before the collision.

Solution:
Given:
Mass of car A, $m_A = 800$ kg
Velocity of car A, $v_A = 20$ m/s (towards the east)
Mass of car B, $m_B = 1200$ kg
Velocity of car B, $v_B = -10$ m/s (towards the west)

The momentum of an object is given by the product of its mass and velocity.

The momentum of car A before collision, $p_A = m_A cdot v_A = 800 cdot 20 = 16000$ kg·m/s (towards the east)

The momentum of car B before collision, $p_B = m_B cdot v_B = 1200 cdot (-10) = -12000$ kg·m/s (towards the west)

Therefore, the momentum of car A before the collision is 16000 kg·m/s towards the east, and the momentum of car B before the collision is 12000 kg·m/s towards the west.

Problem 2: Two objects, X and Y, are moving towards each other on a frictionless surface. Object X has a mass of 2 kg and is moving towards the east with a velocity of 4 m/s. Object Y has a mass of 3 kg and is moving towards the west with a velocity of 6 m/s. Calculate the total momentum before the collision.

Solution:
Given:
Mass of object X, $m_X = 2$ kg
Velocity of object X, $v_X = 4$ m/s (towards the east)
Mass of object Y, $m_Y = 3$ kg
Velocity of object Y, $v_Y = -6$ m/s (towards the west)

The total momentum before the collision is the sum of the individual momenta of the objects.

The momentum of object X before collision, $p_X = m_X cdot v_X = 2 cdot 4 = 8$ kg·m/s (towards the east)

The momentum of object Y before collision, $p_Y = m_Y cdot v_Y = 3 cdot (-6) = -18$ kg·m/s (towards the west)

Total momentum before collision, $p_{text{total}} = p_X + p_Y = 8 + (-18) = -10$ kg·m/s

Therefore, the total momentum before the collision is -10 kg·m/s.

Problem 3: A stationary object X with a mass of 5 kg is struck by an object Y with a mass of 2 kg. Object Y is initially moving towards the east with a velocity of 10 m/s. After the collision, object X and object Y move together with a velocity of 4 m/s towards the west. Calculate the initial velocity of object X before the collision.

Solution:
Given:
Mass of object X, $m_X = 5$ kg
Mass of object Y, $m_Y = 2$ kg
Initial velocity of object Y, $v_{Y,text{initial}} = 10$ m/s (towards the east)
Final velocity of both objects, $v_{text{final}} = -4$ m/s (towards the west)

Let the initial velocity of object X be $v_{X,text{initial}}$ .

According to the law of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision.

The momentum of object X before collision, $p_{X,text{initial}} = m_X cdot v_{X,text{initial}}$

The momentum of object Y before collision, $p_{Y,text{initial}} = m_Y cdot v_{Y,text{initial}}$

The momentum after the collision, $p_{text{final}} = text{total mass} cdot v_{text{final}}$

Since the total mass after the collision is the sum of the masses of object X and object Y, $m_{text{total}} = m_X + m_Y$

Using the law of conservation of momentum:

$p_{X,text{initial}} + p_{Y,text{initial}} = p_{text{final}}$

$m_X cdot v_{X,text{initial}} + m_Y cdot v_{Y,text{initial}} = (m_X + m_Y) cdot v_{text{final}}$

Substituting the given values:

$5 cdot v_{X,text{initial}} + 2 cdot 10 = 5 + 2) cdot (-4)$

$5 cdot v_{X,text{initial}} + 20 = 7 cdot (-4)$

$5 cdot v_{X,text{initial}} = -28 - 20$

$5 cdot v_{X,text{initial}} = -48$

$v_{X,text{initial}} = frac{-48}{5} = -9.6$ m/s

Therefore, the initial velocity of object X before the collision is -9.6 m/s towards the west.

Also Read:

How to Find Momentum After Collision: Elastic, Inelastic, Formula and Problems

January 20, 2024December 30, 2021 by Manish Naik

The article discusses different formulas and problems on how to find momentum after collision.

An object’s velocity changes during a collision due to external force from another object. The velocity change causes a change in momentum after collision. So, we can find the momentum after collision using the impulse formula, laws of conservation of momentum, and conservation of energy.

The momentum before the collision is P_i =mu. The momentum after collision is also found by estimating a change in an object’s velocity v after the collision. P_f = mv

Read more about Momentum.

Suppose a stationary pull ball having a mass of 8kg is hit by another ball. After the collision, the ball is in motion at 5m/s. Determine the pool ball’s momentum after the collision.

**How to Find Momentum After Collision**

Given:

m = 8kg

v = 5m/s

To Find: ∆P =?

Formula:

∆P = P_f – P_i

Solution:

The momentum of ball after collision is calculated as,

∆P = P_f – P_i

∆P = mv – mu

Since pool ball at rest, i.e., u=0

∆P = mv

Substituting all values,

∆P = 8 x 5

∆P = 40

The pool ball’s momentum after collision is 40kg⋅m/s.

How to Find Momentum after Collision Formula?

The momentum after collision is determined using the impulse formula.

When we speak about finding momentum after collision of only one object, we can calculate it using the impulse formula. Impulse is the momentum change after collision due to the external force. Since collisions occur rapidly, it is tough to calculate the external force applied and time separately.

Once we computed momentum before P_i and after collision P_f, we can find impulse in terms of external force by another object as,

“Impulse (∆P) is the product of external force F and time difference (∆t) in which change in momentum occurs.”

What is Impulse — **Impulse – Change in Momentum**

Mathematically,

∆P = F ∆t

P_f – P_i = F ∆t

Read more about Types of Forces.

A football kicked the football having a mass of 5kg on the frictionless ground surface with a force of 30N over 5 sec. What is the velocity and momentum of football after kicking?

Given:

m = 5kg

F = 30N

∆t = 5 sec

To Find:

v₂=?
P_f=?

Formula:

P = mv
∆P = F ∆t

Solution:

The momentum of football before kicking is,

P_i = m₁v₁

Since football is at rest. i.e., v₁=0

Therefore, P_i = 0

The momentum of football before kicking is zero.

The momentum of football after kicking is calculated using the Impulse formula.

∆P = F ∆t

P_f-P_i = F ∆t

Since Pi = 0

P_f = F ∆t

Substituting all values,

P_f = 30 x 5

P_f = 150

The momentum of football after kicking is 150kg⋅m/s

The velocity of football after kicking is,

m₂v₂ = 150

v₂ = 150/5

v₂ = 30

The velocity of football after kicking is 30m/s.

How to Find Total Momentum of Two Objects after Collision?

The total momentum of two objects after collision is estimated using the law of conservation of momentum.

When two objects collide, their respective momentum changes because of their velocities, but their total momentum after collision remains the same. The total momentum after collision is summed by adding all the respective momentums of colliding objects.

In a closed or isolated system, when two objects holding different masses and velocities collide, they may move with each other or away, depending on the types of a collision – such as inelastic collision or elastic collision.

Types of Collision — **Elastic and Inelastic Collision**
(credit: shutterstock)

After the collision, their momentum, which is the product of their masses and velocities, is also varied. But when talking about the total momentum of an isolated system, it remains unchanged. During the collision, whatever momentum one object loses is gained by another object. That’s how the total momentum of colliding objects is conserved.

Suppose momentum of object 1 is P₁ = m₁u₁

Momentum of object 2 is P₂ = m₂u₂

Momentum of both objects before collision is P_i = P₁ + P₂ = m₁u₁ + m₂+u₂

If there is no net force involved during the collision, then momentum after collision Pf of both objects remains the same as before the collision.

Therefore, As per law of conservation of momentum,

P_i = P_f

m₁u₁ + m₂+u₂ = m₁v₁ + m₂+v₂ ……………………. (*)

Notice velocities of both objects changed after collision from u to v. That shows their respective momentum after collision also gets changed.

For an isolated system,

“The total momentum after collision is exactly as before collision as per the law of conservation of momentum.”

Suppose two marble pebbles having masses 10kg and 5kg moving at 8m/sec and 12 m/sec respectively; collide with each other. After the collision, both pebbles move away from each other with the same masses. If one pebble moves away with a velocity of 10m/sec, what is the second pebble’s velocity?

Given:

m₁ = 10kg

m₂ = 5kg

u₁= 8m/sec

u₂= 12m/sec

v₁= 10m/sec

To Find: v₂ =?

Formula:

m₁u₁ + m₂+u₂ = m₁v₁ + m₂+v₂

Solution:

The law of conservation of momentum calculates the velocity of the second pebble,

For isolated systems when no net force acts,

m₁u₁ + m₂+u₂ = m₁v₁ + m₂+v₂

Note that second objects move opposite to the first object. Therefore, the momentum of the second object must be negative.

Substituting all values,

10 x 8 + (- (5 x12) = 10 x 10 + (-(5xv2)

80 – 60 = 100 -5v2

5v₂ = 100 -20

v₂ = 80/5

v₂ = 16

The velocity of the second pebble after the collision is 16m/sec.

Read more about Relative Velocity.

How to Find Momentum after Elastic Collision?

The momentum after elastic collision is estimated using the law of conservation of energy.

The total momentum is conserved during the collision. The kinetic energy of a respective object may change after the collision, but the total kinetic energy after elastic collision stays the same. So, we can find momentum after elastic collision utilizing the law of conservation of energy.

Elastic collision

When the collision between objects is elastic, the total kinetic energy is conserved.

As per law of conservation of energy,

Rearranging equation (*) by terms with m1 on one side and terms with m2 on other.

Now rearranging equation (#) by terms with m1 on one side and the terms with m2 on other and cancel ½ common factor,

Recognize the first term on the left hand side is ‘1’ in the above equation, we get.

………………. (1)

Substitute above equation into equation (*), to eliminate v₂, we get

Finally rearrange above equation and solve for velocity v₁ of object 1 after collision,

Substitute above equation into equation (1) velocity v₂ of object 2 after collision,

Read more about Kinetic Energy.

When a 10kg ball moving at 2m/s elastically collides with another ball having mass 2kg oppositely moving at 4m/s. Calculate the final velocities of both balls after the elastic collision.

Given:

m₁ = 10kg

m₂ = 2kg

u₁ = 2m/s

u₂ = -4m/s

To Find:

v₁ =?
v₂ =?

Formula:

Solution:

The velocity of ball 1 after elastic collision is calculated as,

Substituting all values,

v₁ = 0

That means, the elastic collision stopped the ball 1.

The velocity of ball 2 after elastic collision is calculated as,

Substituting all values,

v₂= 6 m/s

That means the elastic collision changes the velocity of the second ball to 6m/s.

How to Find Momentum after Inelastic Collision?

The momentum after collision is determined using the law of conservation of momentum.

The total momentum is conserved during the collision. But the total kinetic energy of the system is also changed like the kinetic energy respective object, and the collision is said to be inelastic. So, we can find momentum after inelastic collision using the law of conservation of momentum.

**How to Find Momentum after Inelastic Collision?** (credit: shutterstock)

If the collision is elastic, both objects move away from each other with different velocities v₁, v₂ in opposite directions.

But if the collision is inelastic, both objects move with one final velocity V in the same direction.

Therefore, the momentum P_f after inelastic collision becomes m₁V + m₂V or V(m₁+m₂)

So, the equation of conservation of momentum for inelastic collision is,

m₁u₁ + m₂+u₂= V(m₁+m₂)

The formula for final velocity after inelastic collision is,

V=(m₁u₁ + m₂+u₂)/(m₁+m₂)

Read more about Speed.

Two boys are playing on the playground slide in the park. The first boy having a mass of 20kg sliding at 10m/s on the slide. Since the first boy becomes slower at certain portions latterly collides with another boy having a mass of 30kg who slides down at 12 m/s. What will be the velocity of both boys who slide down together after collision?

Given:

m₁ = 20kg

m₂ = 30kg

u₁ = 10m/s

u₂ = 12m/s

To Find: V =?

Formula:

V=(m₁u₁ + m₂+u₂)/(m₁+m₂)

Solution:

The final velocity of both boys sliding after collision is calculated as,

V=(m₁u₁ + m₂+u₂)/(m₁+m₂)

Substituting all values,

V = 11.2

The final velocity of both boys sliding after an inelastic collision is 11.2m/s.

Quick Facts

What is momentum and why is it important in collisions?

A: Momentum is a fundamental concept in physics that describes the motion of an object. It is calculated by multiplying an object’s mass by its velocity. In collisions, momentum is important because it determines how objects interact and how their motion changes.

What is the law of conservation of momentum?

A: The law of conservation of momentum states that the total momentum of a system of objects remains constant if no external forces act on it. This means that the total momentum before a collision is equal to the total momentum after the collision.

How do you calculate the momentum of an object?

A: The momentum of an object is calculated by multiplying its mass (in kilograms) by its velocity (in meters per second). The formula for momentum is: momentum = mass × velocity.

What are elastic and inelastic collisions?

Property	Elastic Collisions	Inelastic Collisions
Kinetic Energy	Conserved. Total kinetic energy before and after the collision is the same.	Not conserved. Total kinetic energy after the collision is less than before.
Momentum	Conserved. Total momentum before and after the collision remains constant.	Conserved. Total momentum before and after the collision remains constant, just as in elastic collisions.
Colliding Objects	Objects bounce off each other with no permanent deformation or generation of heat.	Objects may stick together or deform, generating heat and possibly sound or light.
Examples	Billiard balls colliding, an atom striking a perfectly rigid surface.	Car crashes, a lump of clay hitting a wall and sticking to it.
Energy Conversion	No conversion of kinetic energy into other forms of energy.	Kinetic energy is partially converted into other forms of energy, such as heat, sound, or potential energy (in the case of deformation).
Mathematical Complexity	Relatively simple to calculate final velocities using conservation laws.	More complex due to the need to consider energy dissipation and possible sticking together of objects.
Post-Collision Velocities	Can be calculated precisely using conservation laws.	Less predictable; it often requires additional information about the energy conversion and possible sticking.
Coefficient of Restitution	Equal to 1 (perfectly elastic).	Less than 1, indicating some energy loss.
Real-World Occurrence	Rare—most real-world collisions have some degree of inelasticity.	Common, as most collisions convert some kinetic energy into other forms.

Momentum in Elastic Collsion

Momentum in InElastic Collsion

What is the impulse-momentum theorem?

A: The impulse-momentum theorem states that the change in momentum of an object is equal to the applied impulse, which is the product of the force applied to the object and the time interval over which the force is applied.

What happens to the momentum of two objects in a collision?

A: In a collision between two objects, the momentum of the system is conserved. This means that the total momentum before the collision is equal to the total momentum after the collision. If the objects stick together after the collision, they move together with a combined momentum.

What is Newton’s third law and how does it relate to momentum?

A: Newton’s third law states that for every action, there is an equal and opposite reaction. In the context of collisions, this means that the forces exerted by the objects on each other are equal and opposite, resulting in a change in momentum for both objects.

How do you find the momentum after a collision?

A: To find the momentum after a collision, you need to calculate the final momentum of the system. This can be done by adding up the individual momenta of the objects involved in the collision. The formula for momentum is: momentum = mass × velocity.

What happens to the kinetic energy in a collision?

A: In general, the kinetic energy is not conserved in a collision. In an elastic collision, however, the kinetic energy is conserved, meaning that it remains the same before and after the collision. In an inelastic collision, some of the kinetic energy is lost as heat, sound, or deformation.

Is momentum conserved in all types of collisions?

A: Yes, momentum is conserved in all types of collisions. Whether it is an elastic or inelastic collision, the total momentum of the system before the collision is equal to the total momentum after the collision.

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