Conservation of Momentum

Show two balls with masses of m₁ and m₂ are moving in opposite directions in a straight line with masses of m1 and m2 are moving in opposite directions in straight line with velocities of v₁ and v₂, respectively. Having collided, their velocities become v₁' a v₂'.

The sum of momentum of both balls before collision is

p = p₁ + p₂ = m₁v₁+ m₂v₂

The sum of momentum of both balls after collision is

p' = p₁' + p₂' = m₁v₁' + m₂v₂'

Based on the law of momentum conservation p=p'. thus

m₁v₁+ m₂v₂ = m₁v₁' + m₂v₂'

The law of momentum conservation holds only if the sum of external forces working on the colliding bodies is equal to zero.

Example 4.2

A trained coach with a mass of 10.000 kg is running with the speed of 24 m/s to the right. The train coach collides with a similar coach which is initially at rest. If the two coaches are then connected as the result of the collision, what is their velocity after the collision?

Answer :
The sum of momentum before collision is
p₂ = m₁v₁+ m₂v₂ = (10.000 kg) (24 m/s) + (10.000 kg) ( 0 m/s)

= 2,4 x 10⁵ kg.m/s (the direction of the initial total momentum is to the right)

The law of momentum conservation

Having collided, the two coaches merge and move together with velocity v'. the sum of momentum before the collision is equal to the sum of momentum after the collision and the momentum becomes:

p₂ = (m₁+ m₂)v'

Based on the law of momentum conservation, the sum of momentum before collision is equal to the sum of momentum after collision; p₂ = p₁
(m₁+ m₂)v' = p₁ ⇒ v' = p₁/(m₁+ m₂)
= 2,4 x 10⁵ kg.m/s (10.000 kg + 10.000 kg) = 12m/s.
After a collision, both coaches are moving together to the right with a velocity of 12 m/s.

Exercise 4.2

1) The velocity of a bullet when it is fired is 200 m/s. If the masses of the bullet and the rifle are 10 g and 5 kg respectively, then what is the recoil velocity of the rifle toward the shoulder of the man when the bullet is fired?2. Bodies A and B have masses of 3 kg and 2) kg respectively. Body A is moving to the right with velocity of 5 m/s and B is moving to the left with velocity of 10 m/s in such a way that they are colliding with each other. If after the collision the two bodies are joined together then what is the velocity of the two bodies after collision.

Working Principles of a Rocket

There is wrong assumption that the forward movement of a rocket is due to the reaction of the thrust face done by the rocket to the air. However, this concept does not satisfy when the rocket is in outer space since there is no air to push the rocket.

The working principle of a rocket is similar to the working principle of a rifle firing a bullet. Consider that the rifle in a position to move freely without friction as illustrated in Figure 4.8. When the bullet is fired, the rifle will move backward with a certain velocity. In this situation, the law of momentum conservation holds.

The acceleration experienced by the rifle is similar to the acceleration acquired by the rocket. The acceleration of the is obtained from the thrust done by the rocket. Every molecule of gas can be assumed as on small bullet fired by the rocket. In that system, the total momentum of the rocket is always equal to the total momentum of the gas; as long as there is no external force that influences the rocket (the external forces that influence the rocket are neglected).

The force of gravity, which is acting as the external force, will decrease the momentum of the rocket (remember, a force can change the momentum of a body). Consider that the initial velocity of the rocket is v and its mass is m. The rocket ejects gas with the mass of Δm and velocity v_g so that the velocity of the rocket eject increases to v + Δv. The velocities of rocket and gas are measured relatively toward the same frame of reference, for example; toward the earth. Based on the values of velocities, the initial and final momentum of the rocket could be calculated as follows ( p₁and p₂, respectively);

P₁ = mvP₂ = (m – Δm) (v + Δv)

The final momentum of the gas is p_g = - Δmv_g, where the negative sign shows that the gas velocity is opposite the rocket velocity.
Base on the law of momentum conservation, it is obtained

mv = (m – Δm) (v + Δv) – Δmv_g

mv = mv – Δmv + m Δv – ΔmΔv - Δmv_g

0 = – Δmv + m Δv - Δmv_g
The value Δm.Δv is neglected since it is much smaller compared to the other terms, thus we obtain the change of the rocket's velocity as follows

Exercise 4.4

Determine the escape velocity needed by a rocket to leave the earth's surface. (g = 9.8 m/s², earth's radius is 6,400 km).
A rocket is flying with the rate of gas ejection of 300 kg/s. If the mass of the rocket is 450 kg and its relative velocity toward the gas is 300 km/hours, determine the average acceleration of the rocket.
A block with a mass of 1.5 kg is placed on a horizontal plane. The friction coefficient of the block with the horizontal plane is 0.2 A bullet with a mass of 10 g is fired horizontally, hits the block and buried in it. The block is displaced by 1 m. If g = 10 m/s², then determine the speed of the bullet when it collides with the block.
Block A which mass is 3 kg, is at rest on a floor. The kinetic friction coefficient of the floor is 0.2. Block A is collided by block B, which mass is 2 kg and moving at the speed of 20 m/s. After the collision, block A and B are moving together. Determine the distance covered before both of them stop.
A ball with the mass of 60 g is colliding with the wall perpendicularly. The ball is then turning back with the speed of 40 m/s, which is the same speed as the ball's speed before it collided with the wall. What is the momentum change experienced by the ball during the collision?
A bullet with 10 gram of mass hits the block with 1.5 kg of mass on a surface of a horizontal plane, with the friction coefficient of 0.2. Due to the shot, the block is displaced by 1-meter. Determine the speed of the bullet just before it hits the block.

Multiple Choice

1) A bullet with 10-gram mass is fired at the velocity of 300 m/s to 1.49 kg mass of block which is freely hung on a 1.6-meter long rope. If the earth's gravitational acceleration is 10 m/s², the rope will deflect from the vertical position with an angle of...

a. 30^o
b. 37^o
c. 45^o
d. 60^o
e. 67^o

2) A 200 gram mass of marble is rolling on a frictionless floor so that its momentum is 4 kg.m/s. The marble's speed is....

a. 5 m/s

b. 10 m/s

c. 15 m/s
d. 20 m/s
e. 25 m/s

3) A 100 gram mass of the ball, which slides at speed of 100 m/s is struck back so that its speed becomes 76 m/s. If the duration of the contact between the ball and the stick is 0.02 second, the force done by the stick is ...........
a. 100 N
b. 120 N
c. 176 N
d. 24 N
e. 2 N

4) A car with 800 kg mass, which moves at speed 2 m/s, is braked by friction force of 32 N. The distance covered by the braking car, before it finally stops, is....
a. 50 m
b. 25 m
c. 15 m
d. 5 m
e. 10 m

5) A 2 kg particle moves in a straight path at positive x-axis direction with the velocity of 3 m/s. If a 6 N force has been working on the particle for 3 s, after 3 s ...........
a. its speed will be 3 m/s
b. its velocity direction is toward the negative x-axis
c. the particle never stops
d. its velocity keeps on increasing
e. the direction of the particles is unchanged

You Need to Know

The Function of Helmet

One of the equipment of a motor rider is the helmet. The inner layer of a helmet is made of a soft material such as foam rubber sponge. This layer has a function to lengthen the time of impact to the head so that it reduces the contact force to the head during the shock. What happens if the helmet does not have soft layer? When shock happens, the interaction force with the head will take place shortly and it can cause a headache.

Conservation of Momentum