INDEX physics notes: Conservation of momentum, elastic & non-elastic collisions, Newton's 2nd law calculations

Q1 The diagram below shows the sequence of events when a moving green ball collides with a stationary purple ball.

The green ball has a mass of 2.0 kg and is moving at 4.0 m/s prior to colliding with the stationary purple ball.

After the collision both balls are moving in a forward direction.

If the green ball is still moving forward at 1.5 m/s, calculate the velocity of the purple ball.

From the law of conservation of momentum

total momentum before collision = total momentum after collision

m₁v₁ + m₂v₂ = m₁v₃ + m₁v₄

(2 x 4) + (3 x 0) = (2 x 1.5) + (3 x v₄)

8 + 0 = 3 + 3v₄,  3v₄ = 8 - 3 = 5,  therefore V₄ = 5/3 = 1.7 m/s (2sf)

Q2(a) Why does a gun recoil (move backwards) when fired?

At the start both the gun and the bullet have zero moment.

After firing the bullet then has its own momentum in a forward direction from the shooter.

BUT, this must be balanced by a momentum in the opposite direction because total momentum must be conserved.

So, on firing, the gun itself must get its own equal but opposite momentum (remember that momentum is a vector quantity!).

Therefore the gun's equal and opposite momentum causes it to recoil backwards.

(b) A 10 g bullet accelerates to 400 m/s after being fired from a gun with a mass of 2.0 kg.

(i) What is the momentum of the bullet? (kg = g/1000)

p = m × v = (10/1000) x 400 = 4.0 kg m/s

(ii) What is the momentum of the gun?

From the law of conservation of momentum, the gun must have a momentum equal and opposite to that of the bullet.

Therefore gun momentum = -4.0 kg m/s

(the minus sign is important, particularly in more complex calculations, the gun recoils in the opposite direction to the bullet - Newton's 3rd Law in action).

(iii) What is the velocity of the recoiling gun?

The momentum of the gun and bullet must be equal

(from the law of conservation of momentum)

therefore numerically: m_gunv_gun = m_bulletv_bullet

m_gunv_gun = -4.0 = 2.0 x v_gun,  v_gun = 4.0/2.0 = -2.0 m/s

(again, note the minus sign, because the gun recoils in the opposite direction to the bullet).

(c) Why is it an advantages to the gunner to fire with a heavier gun?

From the equation: momentum = mass x velocity

The bigger the mass of the gun, for the same momentum, the lower the recoil velocity and the lesser impact on the person firing the gun.

Q3 The diagram below shows the sequence of events when a moving car crashes into a stationary car and they combine together and move forward.

Q4 Imagine a 0.2 kg ball moving at 5 m/s colliding with a 0.3 kg ball moving at 2 m/s in the same direction.

After the collision the 0.2 kg green ball stops moving and the 0.3 kg purple ball continues moving in the same direction.

Calculate the final velocity of the of the 0.3 kg purple ball.

First sort out all the masses and velocities

where v₁ and v₂ are the initial velocities and v₃ and v₄ the velocities after the collision

m₁ = 0.2 kg, initial velocity v₁ = 5 m/s, final velocity v₃ = 0 m/s

m₂ = 0.3 kg, initial velocity v₂ = 2 m/s, final velocity v₄ = ? m/s

total momentum of two of objects before collision = total moment of objects after collision

for two colliding objects, where p = momentum: p₁ + p₂ = p₃ + p₄

substituting m and v for the mass and velocity of the two objects gives ...

m₁v₁ + m₂v₂ = m₁v₃ + m₂v₄

Then substitute everything in the equation, rearrange and deduce the answer (v₄).

(0.2 x 5) + (0.3 x 2) = (0.2 x 0) + (0.3 x v₄)

1.0 + 0.6 = 0 + 0.3v₄,  0.3v₄ = 1.6

therefore the final velocity of the 0.3 kg ball = v₄ = 1.6/0.3 = 5.3 m/s (2 sf)

Q5 Imagine a 0.2 kg ball moving at 5 m/s hitting a 0.3 kg ball moving at 2 m/s in the same direction.

After the collision both balls keep moving in the same direction, but with different velocities.

If the 0.2 kg ball continues to move at a velocity of 1.5 m/s, calculate the final velocity of the 0.3 kg ball.

This is an almost identical problem to Q2.4, except there is no zero momentum term.

So, sorting out all the masses and velocities

where v₁ and v₂ are the initial velocities and v₃ and v₄ the velocities after the collision

m₁ = 0.2 kg, initial velocity v₁ = 5 m/s, final velocity v₃ = 1.5 m/s

m₂ = 0.3 kg, initial velocity v₂ = 2 m/s, final velocity v₄ = ? m/s

total momentum of two objects before collision = total moment of objects after collision

for two colliding objects, where p = momentum: p₁ + p₂ = p₃ + p₄

substituting m and v for the mass and velocity of the two objects gives ...

m₁v₁ + m₂v₂ = m₁v₃ + m₂v₄

where v₁ and v₂ are the initial velocities and v₃ and v₄ the velocities after the collision

Then substitute everything in the equation, rearrange and deduce the answer (v₄).

(0.2 x 5) + (0.3 x 2) = (0.2 x 1.5) + (0.3 x v₄)

1.0 + 0.6 = 0.3 + 0.3v₄,  0.3v₄ = 1.6 - 0.3 = 1.3

therefore the final velocity of the 0.3 kg ball = v₄ = 1.3/0.3 = 4.3 m/s (2 sf)

Note the final velocity of the 0.3 kg ball is less than in Q2.4 because in Q2.4 it received all the momentum from the collision.

Q6 Snooker players can hit the white ball 'dead centre' into a stationary red ball. The stationary red ball then flies off at the same velocity as the incoming white ball and the white ball is left in a stationary position where the red ball was.

Ignoring friction and assuming the balls have the same mass, explain this observation.

The law of conservation states the total momentum must be the same before and after the collision.

m_whitev_intitial + m_redv_initial = m_whitev_final + m_redv_final

 m_redv_initial is zero, since the red ball is initially stationary with a momentum of zero.

 m_whitev_final is zero, since the white ball is now stationary, also with a momentum of zero.

Therefore: m_whitev_intitial = m_redv_final

and since the masses are equal, m_white = m_red, then the final velocity of the red ball must equal the initial velocity of the white ball.

It should be noted that skilled snooker players can play all sorts of tricks with the 'physics' of snooker and often have a total disregard for the law of conservation of momentum!

Q7 Imagine a 1000 kg car travelling with a velocity of 10 m/s crashing head-on into a 500 kg car travelling in the opposite direction with a velocity of 15 m/s. Assume they 'crunch' together to form a single object - the 'wreck'!

Q8 A momentum question involving Newton's 3rd Law, conservation of momentum and conservation of kinetic energy.

A 0.20 kg green ball (m1) moving at 5.0 m/s collides in line with a 0.40 kg purple ball (m2) moving at 2.0 m/s.

Also, from high speed photography, the collision impact time was found to be 0.05 seconds.

After the collision, with both moving in the same direction, the 0.20 kg green ball continues moving forwards at a slower speed of 2.0 m/s and the purple ball also continues moving forwards at an increased speed of 3.50 m/s.

(a) Using Newton's 2nd Law, calculate the force involved in causing the changes in velocity.

F = ma = m∆v/∆t = m(v-u) ÷ 0.05, where F = force (N), m = mass (kg), v and u = final and initial velocities.

For the green ball: F = 0.2 x (2 - 5) ÷ 0.05 = -12 N

For the purple ball: F = 0.4 x (3.5 - 2) ÷ 0.05 = 12 N

(b) Do the results of your calculations in (a) comply with Newton's 3rd Law of Motion?

Yes they do. See notes on Newton's 3rd Law

Newton's Third Law of motion states that when two objects interact, the forces they exert on each other are equal in numerical value, act in opposite directions and are of the same type.

The impact forces are the same size and act in opposite directions.

The impact forces act on two different objects with same type of normal contact force.

Data reminder!

(c) Does the experiment comply with (i) the concept of a 'closed' system AND (ii) do the observations comply with the law of conservation of momentum?

(i) The collision does take place in a closed system - the experiment is a physical system that does not allow transfer of matter in or out of the system, and is so fast there is no time for energy to leave or enter the system.

(ii) momentum p = mv, where p in kg m/s, m in kg, v in m/s.

Momentums before collision:

for green ball p = 0.2 x 5 = 1.0 kg m/s

for purple ball p = 0.4 x 2 = 0.8 kg m/s

total momentum before collision = 1.0 + 0.8 = 1.8 k m/s

Momentums after collision:

for green ball p = 0.2 x 2 = 0.4 kg m/s

for purple ball p = 0.4 x 3.5 = 1.4 kg m/s

total momentum after collision = 0.4 + 1.4 = 1.8 k m/s

The total momentum is the same as before and after the collision, so the law of conservation of momentum is obeyed.

(d) Show by calculation whether kinetic energy is conserved or otherwise? Comment on your results.

KE = ½mv², where KE in J, m = mass of object in kg and velocity in m/s.

Kinetic energies before collision:

green ball KE = ½ x 0.2 x 5² = 2.5 J

purple ball KE = ½ x 0.4 x 2² = 0.8 J

total kinetic energy before collision = 2.5 + 0.8 = 3.3 J

Kinetic energies after collision:

green ball KE = ½ x 0.2 x 2² = 0.4 J

purple ball KE = ½ x 0.4 x 3.5² = 2.45 J

total kinetic energy after collision = 0.4 + 2.45 = 2.85 J (2.9 2sf)

You can see from the calculations that kinetic energy is not conserved in this collision of two objects.

This is an inelastic collision.

The atoms of the object are compressed and kinetic energy is converted into potential energy, thermal energy and sound.

INDEX physics notes: Conservation of momentum, elastic and non-elastic collisions, Newton's 2nd law calculations - problem solving