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Momentum is conserved in explosions such as when a riffle is fired. During the firing, the bullet receives an equal but opposite amount of momentum to that of the rifle.

Total momentum before collision = Total momentum after collision

Where; mg is mass of the rifle (or gun), Vg is velocity of the rifle which is also called recoil velocity. mb is mass of the bullet, Vb is velocity of the bullet.

For any explosion of bodies, the amount of momentum for one body is equal but opposite to that of another body.

The negative sign indicates that the momenta are in opposite directions.

**Example:1**

A bullet of mass 8g is fired from a gun of mass 500g. If the missile velocity of the bullet is 500ms-1. Calculate the recoil velocity of the gun.

**Solution**

The negative sign indicates that the recoil velocity, Vg is in opposite direction to that of the bullet.

**Example:2**

A bullet of mass 200g is fired from a gun of mass 4kg. If the muzzle velocity of the bullet is 400ms-1, calculate the recoil velocity.

**Solution**

**Example: 3**

A bullet of mass 12.0g travelling at 150ms-1penetrates deeply into a fixed soft wood and is brought to rest in 0.015s. Calculate

(i) How deep the bullet penetrates the wood [1.125m]

(ii) the average retarding force exerted by the wood on the bullet. [120N]

**Rocket and jet engines**

These work on the principle that in any explosion one body moves with a momentum which is equal and opposite to that of another body in the explosion. For the rocket and the jet engine, the high velocity hot gas is produced by the burning of fuel in the engine.

**Note: **Rockets use liquid oxygen while jets use oxygen from air.

**How a rocket engine work:**

**Principle: **the jet and rocket engines work on the principle that momentum is conserved in explosion.

**High velocity:** the high velocity of the hot gas results in the burning of the fuel in the engine.

**Large momentum:** the large velocity of the hot gas results in the gas to leave the exhaust pipe with a large momentum.

**Engine:** the engine itself acquires an equal but opposite momentum to that of the hot gas.

**Note:** when the two bodies collide and they move separately after collision but in opposite directions then.

m1u1+m2u2 = m1v1+m2(-v2)

m1u1+m2u2 = m1v1- m2v2

**Example:**

A body Q of mass 50g collides with a stationary body “P” of mass 4g. If a body “Q” moves backward with a velocity of 10ms-1 and a body “P”, moves forward with a velocity of 6ms-1. Calculate the initial velocity of a body Q.

**Solution**

Total momentum before collision = Total momentum after collision

**Example: 2**

A moving ball “P” of mass 100g collides with a stationary ball Q of mass 200g. After collision, P moves backward with a velocity of 2ms-1 while Q moves forward with a velocity of 5ms-1. Calculate the initial velocity of P.

**Solutions**

mQuQ+mPuP= mQvQ+mpvP

0.2(0)+ 0.1𝑢𝑃= 0.2(5)+0.1(−2)

**Example: 3.**

A body of mass 10kg moves with a velocity of 20ms-1.Calculate its momentum.

**Solution**

**Example:2**

A 20kg mass traveling at 5mls is accelerated to 8mls. Calculate the change in momentum of the body.

**Solution**

**Note:** The change in momentum is called Impulse.

**Example:3**

A one tonne car traveling at 20ms-1 is accelerated at 2ms-2 for five second. Calculate the;

(i) change in momentum

(ii) rate of change in momentum

(iii) Accelerating force acting on the body.

**Solution**

From above, the force applied is equal to the rate of change in momentum. This leads to Newton’s second law of motion.

**Example:4**

A van of mass 1.5 tonnes travelling at 20ms-1, hits a wall and is brought to rest as a result in 0.5seconds. Calculate the;

(i) Impulse

(ii) Average force exerted on the wall.

**Solution**

**Example:5**

A man of mass 60kg jumps from a high wall and lands on a hard floor at a velocity of 6ms.Calculate the force exerted on the man’s legs if;

(i) He bends his knees on landing so that it takes 1.2s for his motion to be stopped.

(ii) He does not bend his knees and it takes 0.06s to stop his motion.