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- Physics Tutorial: Gravitational Potential Energy. Kepler Laws
- Physics Tutorial: Dynamics of Rotational Motion
- Physics Tutorial: Torque
- Physics Tutorial: Law of Conservation of Momentum and Kinetic Energy
- Physics Tutorial: Collision and Impulse. Types of Collision
- Physics Tutorial: Linear Momentum
- Physics Tutorial: Moment of Force. Conditions of Equilibrium
- Physics Tutorial: Centre of Mass. Types of Equilibrium
- Physics Tutorial: Momentum and Impulse in Two Dimensions. Explosions.
- Significant Figures Calculator
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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

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.

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 m_{1} and m_{2} respectively, moving towards each other at velocities v*⃗*_{01} and v*⃗*_{02} 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 F*⃗*_{1} for the first ball while the reaction force of the second ball on the first ball is written as F*⃗*_{2}, where

As a result, after the collision, the first ball will have the velocity v*⃗*_{1} while the second ball the velocity v*⃗*_{2}.

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

F*⃗*_{1} × ∆t = -F*⃗*_{2} ∆t

or

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

We know that

∆p*⃗*_{1} = p*⃗*_{01}-p*⃗*_{1}

where p*⃗*_{01} and p*⃗*_{1} are the initial and final momentums of the first ball, and

∆p*⃗*_{2} = p*⃗*_{02}-p*⃗*_{2}

where p*⃗*_{02} and p*⃗*_{2} are the initial and final momentums of the second ball respectively.

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

p*⃗*_{01}-p*⃗*_{1} = -(p*⃗*_{02}-p*⃗*_{2})

p*⃗*_{01}-p*⃗*_{1} = -p*⃗*_{02} + p*⃗*_{2}

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

p*⃗*_{01} = m_{1} × v*⃗*_{01}

p*⃗*_{02} = m_{2} × v*⃗*_{02}

p*⃗*_{1} = m_{1} × v*⃗*_{1}

p

p

and

p*⃗*_{2} = m_{2} × v*⃗*_{2}

we can write the equation (4) as

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

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?

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

m_{1} = 300 g = 0.3 kg

m_{2} = 500 g = 0.5 kg

v_{01} = 20 cm/s = 0.2 m/s

v_{02} = 0

v_{1} = - 8 cm/s = - 0.08 m/s (it is negative as the trolley turns back)

v_{2} = ?

m

v

v

v

v

Applying the equation (5) on conservation of momentum,

m_{1} × v*⃗*_{01} + m_{2} × v*⃗*_{02} = m_{1} × v*⃗*_{1} + m_{2} × v*⃗*_{2}

we obtain after the substitutions,

0.3 × 0.2 + 0.5 × 0 = 0.3 × (-0.08) + 0.5 × v*⃗*_{2}

0.06 = -0.024 + 0.5 × v*⃗*_{2}

v*⃗*_{2} = *0.06 + 0.024**/**0.5*

= 0.168*m**/**s*

= 1.68 cm/s

0.06 = -0.024 + 0.5 × v

v

= 0.168

= 1.68 cm/s

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

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 v*⃗*_{1} and v*⃗*_{2} 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:

m_{1} × v*⃗*_{01} + m_{2} × v*⃗*_{02} = m_{1} × v*⃗* + m_{2} × v*⃗*

or

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

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?

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:

m_{1} = 40 kg

m_{2} = 120 kg

v_{01} = 3 m/s

v_{02} = 0

v = ?

m

v

v

v = ?

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

m_{1} × v*⃗*_{01} + m_{2} × v*⃗*_{02} = (m_{1} + m_{2}) × v*⃗*

we obtain after substitutions

40 × 3 + 120 × 0 = (40 + 120) × v*⃗*

120 = 160 × v*⃗*

v*⃗* = *120**/**160* = 0.75 m/s

120 = 160 × v

v

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

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

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 KE_{tot} is their total kinetic energy after the collision.

Hence,

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

we obtain

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.

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.

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

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

m_{1} = 200 g = 0.2 kg

m_{2} = 500 g = 0.5 kg

v_{01} = 4 m/s

v_{02} = - 3 m/s (as it is moving from right to left)

v =

m

v

v

v =

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

m_{1} × v*⃗*_{01} + m_{2} × v*⃗*_{02} = (m_{1} + m_{2}) × 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

0.8-1.5 = 0.7 × v

-0.7 = 0.7 × v

v

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

KE_{0(1)} = *m*_{1} × v^{2}_{01}*/**2* = *0.2 × 4*^{2}*/**2* = 1.6 J

and that of the second object before the collision is

KE_{0(2)} = *m*_{2} × v^{2}_{02}*/**2* = *0.5 × (-3)*^{2}*/**2* = 2.25 J

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

KE_{0(tot)} = KE_{0(1)} + KE_{0(2)}

= 1.6 J + 2.25 J

= 3.85 J

= 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)} = *(m*_{1} + m_{2} ) × v^{2}*/**2*

=*0.7 × (-1)*^{2}*/**2*

= 0.35 J

=

= 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.

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).

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:

GPE_{2} = (M + m) × g × h

= (0.98 + 0.02) × 10 × 0.2

= 2 J

= (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,

KE_{1} = GPE_{2} = 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

KE_{1} = *(M + m) × v*^{2}*/**2*

2 =*1 × v*^{2}*/**2*

v^{2} = 4

v*⃗* = 2 m/s

2 =

v

v

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

m_{bullet} × v*⃗*_{0(bullet)} + M_{Block} × v*⃗*_{0(Block)} = (m_{bullet} + M_{Block}) × v*⃗*

Substituting the known values, we obtain

0.02 × v*⃗*_{0(bullet)} + 0.98 × 0 = (0.02 + 0.98) × 2

0.02 × v*⃗*_{0(bullet)} = 2

v*⃗*_{0(bullet)} = 2/0.02

= 100 m/s

0.02 × v

v

= 100 m/s

The **Law of Conservation of Momentum** for two colliding objects 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."**

The mathematical expression of the Law of Conservation of Momentum for two objects (1) and (2) colliding elastically, is

p*⃗*_{01} + p*⃗*_{02} = p*⃗*_{1} + p*⃗*_{2}

or

m_{1} × v*⃗*_{01} + m_{2} × v*⃗*_{02} = m_{1} × v*⃗*_{1} + m_{2} × v*⃗*_{2}

where m_{1} and m_{2} are the objects' masses, v_{01} and v_{02} are their respective initial velocities prior to collision and v_{1} and v_{2} are their velocities after the collision.

When the collision is inelastic, the Law of Conservation of Momentum is written as

m_{1} × v*⃗*_{01} + m_{2} × v*⃗*_{02} = (m_{1} + m_{2}) × v*⃗*

where v*⃗* is the common velocity of the two objects after the collision as they move together.

As for kinetic energy, it is conserved only in perfectly elastic collisions. The equation that represents mathematically the Law of Conservation of Kinetic Energy is

The two abovementioned laws are very fundamental in mechanics and they are known as the "Laws of Conservation".

**1)** Two identical billiard balls move towards each other at 4 m/s and 1 m/s respectively as shown in the figure.

After a head-to-head collision, the first ball moves at 2 m/s in the same direction as prior to the collision. What is the velocity of the second ball after the collision?

- 2.5 m/s due right
- 1.5 m/s due left
- 1.5 m/s due right
- 1.0 m/s due right

**Correct Answer: D**

**2)** A 2 t freight car is disconnected from its locomotive and stars moving slightly downhill at 1.2 m/s. It collides with another 3 t freight car at rest and afterwards, they move together.

What is the common velocity of the two freight cars after the collision? (1 t = 1000 kg)

- 0.48 m/s downhill
- 1.2 m/s downhill
- 0
- 0.8 m/s downhill

**Correct Answer: A**

**3)** A 40 g bullet moving horizontally at 400 m/s hits horizontally a 2960 g wooden block initially at rest. After the bullet is stuck on the wooden block, the system moves together along a horizontal surface whose friction coefficient is 0.4. How far can the system move in these conditions? For convenience, take g = 10 m/s2.

- 5.33 m
- 0.67 m
- 3.56 m
- 35.6 m

**Correct Answer: C**

We hope you found this Physics tutorial "Law of Conservation of Momentum and Kinetic Energy" useful. If you thought the guide useful, it would be great if you could spare the time to rate this tutorial and/or share on social media, this helps us identify popular tutorials and calculators and expand our free learning resources to support our users around the world have free access to expand their knowledge of physics and other disciplines. In our next tutorial, we expand your insight and knowledge of "Law Of Conservation Of Momentum And Kinetic Energy" with our Physics tutorial on Momentum and Impulse in Two Dimensions. Explosions.

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