# Physics Tutorial: Law of Conservation of Momentum and Kinetic Energy

In this Physics tutorial, you will learn:

• What is the Law of Conservation of Momentum?
• In which situations the Law of Conservation of Momentum is applied?
• The meaning of the Law of Kinetic Energy Conservation
• What are the limitations of the Law of Kinetic Energy Conservation?
• How to use the Laws of Conservation in calculating any missing velocity or mass
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6.7Law of Conservation of Momentum and Kinetic Energy

## Introduction

In the previous tutorial "Collision and Impulse", we stated that in absolutely elastic collisions, immediately after the impact occurs, an object has the same kinetic energy as it had before the collision. However, this does not seem the case in other types of collision. The reason of this inconsistency will be explained in this tutorial.

Also, we will see what happens with momentum in such cases. Does it remains numerically unchanged (is it conserved) or not?

Let's have a deeper look at such events to answer the above questions.

## Conservation of Momentum in Elastic Collisions

First, we'd like to state that in this tutorial only head-to-head collisions will be condiered. In these collisions, objects either move in the same directions as they were moving prior to the impact or they move in the opposite direction. No middle ways exist in these cases.

Suppose we have two balls of masses m1 and m2 respectively, moving towards each other at velocities v01 and v02 as shown in the figure.

When the balls collide with each other, they either move in the same or in the opposite direction, depending on the values of masses and velocities, i.e. on the magnitude of momentums. Sometimes, either one or both objects may stop as well. Whatever happens, they will experience a force of collision, which based on the Newton's Third Law of Motion (action-reaction principle), is written as F1 for the first ball while the reaction force of the second ball on the first ball is written as F2, where

### Equation 1

F1 = -F2

As a result, after the collision, the first ball will have the velocity v1 while the second ball the velocity v2.

The figure below shows the two balls after the head-to-head collision. For illustration purpose, after the collision they are taken as moving in opposite directions as they were moving before the collision, but no matter - they can move in whatever direction; the approach is always the same.

In addition to the change in direction, we don't expect the balls have the same magnitude of velocity as prior to the collison, so the arrows don't have the same lengths as in the first figure.

If we multily both sides of equation (1) by the time interval during which the collision occurs (this time interval is the same for both objects as they both experience the same event), we obtain

F1 × ∆t = -F2 ∆t

or

### Equation 2

J1 = -J2

From equation (2) it is obvious that impulses are also equal and opposite , just like the forces of interaction between the balls.

In the previous tutorial "Collision and Impulse. Types of Collision", we explained that impulse is equal to the change in momentum ∆p experienced by an object when it collides with another object. Therefore, we can write the equation (2) as

### Equation 3

∆p1 = -∆p2

We know that

∆p1 = p01-p1

where p01 and p1 are the initial and final momentums of the first ball, and

∆p2 = p02-p2

where p02 and p2 are the initial and final momentums of the second ball respectively.

Therefore, substituting the above two equations at eq. (3), we obtain

p01-p1 = -(p02-p2)
p01-p1 = -p02 + p2

### Equation 4

p01 + p02 = p1 + p2

The equation (4) is known as the mathematical expression of the Law of Conservation of Momentum. Its says that:

"The total momentum prior to the impact for two objects involved in a collision process is always equal to their total momentum after the impact, regardless the type of collision."

Giving that

p01 = m1 × v01
p02 = m2 × v02
p1 = m1 × v1

and

p2 = m2 × v2

we can write the equation (4) as

### Equation 5

m1 × v01 + m2 × v02 = m1 × v1 + m2 × v2

The equation (5) represents the long form of the equation (4), which is widely used in exercises.

### Example 1

A 300 g lab trolley moving at 20 cm/s collides with a 500 g trolley at rest as shown in the figure.

After the head-to-head collision, the first (the small) trolley changes direction and moves due left at 8 cm/s. What is the velcity (including the direction) of the second trolley after the collision?

### Solution 1

We can write the following clues based on the info provided in the problem:

m1 = 300 g = 0.3 kg
m2 = 500 g = 0.5 kg
v01 = 20 cm/s = 0.2 m/s
v02 = 0
v1 = - 8 cm/s = - 0.08 m/s (it is negative as the trolley turns back)
v2 = ?

Applying the equation (5) on conservation of momentum,

m1 × v01 + m2 × v02 = m1 × v1 + m2 × v2

we obtain after the substitutions,

0.3 × 0.2 + 0.5 × 0 = 0.3 × (-0.08) + 0.5 × v2
0.06 = -0.024 + 0.5 × v2
v2 = 0.06 + 0.024/0.5
= 0.168 m/s
= 1.68 cm/s

The positive result implies that the second trolley moves due right (as expected). Look at the figure:

## Conservation of Momentum in Inelastic Collisions

In inelastic collisions, both objects move at the same velocity after the collision as they are attached to each other. Therefore, we don't need to write v1 and v2 for the respective velocities after the collision but simply v which expressed the common velocity of the two objects after the collision. Therefore, the equation (5) becomes:

m1 × v01 + m2 × v02 = m1 × v + m2 × v

or

### Equation 6

m1 × v01 + m2 × v02 = (m1 + m2) × v

Equation (6) represents the mathematical expression of the Law of Convervation of Momentum in Inelastic Collisions.

### Example 2

A 40 kg boy jumps at 3 m/s on a 120 kg raft anchored at the shore as shown in the figure below. What is the velocity of the boat immediately after the boy jumps of it?

### Solution 2

This is a kind of inelastic collision as both the boy and the raft move together after he jumps on it. We can write the following clues based on the info provided in the problem:

m1 = 40 kg
m2 = 120 kg
v01 = 3 m/s
v02 = 0
v = ?

Applying the equation (6) on conservation of momentum for inelastic collisions

m1 × v01 + m2 × v02 = (m1 + m2) × v

we obtain after substitutions

40 × 3 + 120 × 0 = (40 + 120) × v
120 = 160 × v
v = 120/160 = 0.75 m/s

This result means both the boy and the raft move together due right at 0.75 m/s after he jumps on it.

## Conservation of Kinetic Energy in Collisions

Kinetic energy isn't precisely "conserved" throughout an elastic collision. When two objects are interacting, the total kinetic energy first decreases while the system potential energy increases. The difference between a perfectly (absolutely) elastic collision and all other collisions is that in a perfectly (absolutely) elastic collision, the entire energy that was stored as potential energy is recovered as kinetic energy at the instant the collision ends.

When colliding the objects will compress during the interaction and regain their initial shapes and dimensions exactly at the instant the interaction ends. This doesn't happen with ordinary objects on human scales. If we consider colliding steel spheres, we can find that about 95% of the initial kinetic energy is recovered at the end of the collision, and the result appears to be very close to perfectly elastic but it isn't instead.

It is possible to obtain perfectly elastic collisions when the interactions are produced through conserved fields such as gravitational or magnetic fields. Put repelling wide magnets on the ends of two carts, and they can collide elastically, as all the energy stored in the magnetic field is recovered at the end. The problem in this case is the friction between the carts and the surface that supports them. If friction is neglected, they can collide with no lost kinetic energy.

Therefore, we can conclude that Kinetic Energy is conserved only in perfectly elastic collisions. The equation involved in such a situation, is

### Equation 7

KE tot(0) = KEtot

where KE tot(0) is the initial kinetic energy of the system, i.e. the total initial kinetic energy of the two objects before the collision, while KEtot is their total kinetic energy after the collision.

Hence,

### Equation 8

KE0(1) + KE0(2) = KE1 + KE2

Writing the equation (7) for two objects as the ones shown in the figure below,

we obtain

### Equation 9

m1 × v201/2= m2 × v202/2+ m1 × v21/2+ m2 × v22/2

The equation (9) is the mathematical representation of the Law of Kinetic Energy Conservation for two objects, which says:

"Kinetic energy is conserved only in perfectly elastic collisions."

This does not occur in elastic collisions that are not perfect. To prove this, we can bring in mind situations when a light object such as a ball collides with a very heavy and unmoveable object such as a large rock or a wall. It is quite impossible for the ball to keep its original kinetic energy as some of it is lost during the collision. Therefore, the initial kinetic energy of the object is greater than its final kinetic energy. On the other hand, the unmoveable object had no kinetic energy either before or after the impact. Therefore, the total kinetic energy of the system decreases after the collision.

Likewise, kinetic energy is not conserved in non-elastic collisons as well. Let's explain this point through an example.

### Example 3

Two clay balls flying horizontally towards each other at 4 m/s and 3 m/s respectively move together after a head-to-head collision. The mass of the first ball is 200 g and that of the second ball is 500 g as shown in the figure.

1. Calculate the common velocity of the two objects after the collision
2. Prove that kinetic energy is not conserved in this case

### Solution 3

a. We have the following clues considering also the direction (we take let-to-right as positive):

m1 = 200 g = 0.2 kg
m2 = 500 g = 0.5 kg
v01 = 4 m/s
v02 = - 3 m/s (as it is moving from right to left)
v =

Applying the equation (6) on conservation of momentum for inelastic collisions

m1 × v01 + m2 × v02 = (m1 + m2) × v

we obtain after substitutions

0.2 × 4 + 0.5 × (-3) = (0.2 + 0.5) × v
0.8-1.5 = 0.7 × v
-0.7 = 0.7 × v
v = -1 m/s

This result means both obejcts move at 1 m/s due left after the collision.

b. Let's calculate the initial kinetic energy for each object and compare their sum to the kinetic energy of the whole system after the collison. If these results are equal, kinetic energy is conserved, otherwise it is not.

Kinetic energy of the first object before the collision is

KE0(1) = m1 × v201/2 = 0.2 × 42/2 = 1.6 J

and that of the second object before the collision is

KE0(2) = m2 × v202/2 = 0.5 × (-3)2/2 = 2.25 J

Therefore, the total energy of the two objects before the collision is

KE0(tot) = KE0(1) + KE0(2)
= 1.6 J + 2.25 J
= 3.85 J

On the other hand, we have for the kinetic energy of the system after the collision

KE(tot) = (m1 + m2 ) × v2/2
= 0.7 × (-1)2/2
= 0.35 J

As you see, kinetic energy is not conserved in this inelastic collision as 0.35 J < 3.85 J. Therefore, the law of kinetic energy conservation is not applied in inelastic collisions.

Often, the law of conservation of momentum is very helpful during mechanical events studied by means of energetic approach when no other options are available to calculate the velocity in any part of the system. Look at the example below.

### Example 4

A 20 g bullet hits horizontally a hanged wooden block. The block is initially at rest and its mass is M = 980 g. The bullet gets stuck in the block and the system bullet-block raises up to 20 cm above the initial position as shown in the figure. What was the initial velocity of the bullet? (Take g = 10 m/s2).

### Solution 4

The event contains three stages: the initial stage before the bullet hits the block (instant 0), the intermediate stage in which the bullet has just stuck in the block (instant 1) and the final stage in which the system block-bullet raises at h = 20 cm = 0.02 m above the original level (instant 2). In this instant, the system does not possess any kinetic energy but only gravitational potential energy.

We have M = 980 g = 0.98 kg, m = 20 g = 0.02 kg, M + m = 0.98 kg + 0.02 kg = 1 kg and h = 20 cm = 0.2 m.

Let's start from this last stage. We have:

GPE2 = (M + m) × g × h
= (0.98 + 0.02) × 10 × 0.2
= 2 J

This value represents the mechanical energy of the system, which at this stage is only potential.

In stage 2, when the bullet has just stuck in the block, the system possesses only kinetic energy. Based on the law of mechanical energy conservation, this kinetic energy of the system block-bullet is numerically equal to the gravitational potential energy we found at the previous stage. Thus,

KE1 = GPE2 = 2 J

Such an approach helps us calculate the common velocity v of the system after the collision (which obviously is inelastic). Thus, we have

KE1 = (M + m) × v2/2
2 = 1 × v2/2
v2 = 4
v = 2 m/s

Now, using the Law of Conservation of Momentum for inelastic collisions, we can find the initial velocity of bullet v0(bullet). Thus, since the block initially was at rest, then v0(Block) = 0. Therefore, we have

mbullet × v0(bullet) + MBlock × v0(Block) = (mbullet + MBlock) × v

Substituting the known values, we obtain

0.02 × v0(bullet) + 0.98 × 0 = (0.02 + 0.98) × 2
0.02 × v0(bullet) = 2
v0(bullet) = 2/0.02
= 100 m/s

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