SITEMAP   School Physics Notes: Forces & motion 6.4 Complex momentum calculations

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Forces and Newton's Laws of Motion 6.4 Worked out problem solving questions involving more complex momentum calculations for exam practice

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6.4 Problem solving questions involving more complex momentum calculations

You must study Part 6.2 before attempting these questions

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.

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

(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)

(ii) What is the momentum of the gun?

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

(c) Why is it an advantages to the gunner to fire with a heavier 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. The green car (1), of mass 1000 kg, crashes into the stationary blue car (2) of mass 800 kg.

After the collision, the two cars are locked together.

(a) Calculate the velocity of the combined wrecked cars immediately after the condition.

(b) Calculate the kinetic energy of all three objects in the diagram.

(c) Calculate the difference, if any, of the total initial kinetic energies of the cars and that of the wreck,

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

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

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

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.

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 in terms of 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'! (a) Calculate the momentum of both cars.

(b) Calculate the momentum p of the 'wreck'

(c) What is the immediate velocity of the wreck after the impact and in what direction? (in terms of diagram)

(d) Is kinetic energy conserved in this collision? Reminder of diagram

A reminder of the data and the final velocity was 1.67 m/s to the right!

(e) If the impact time is 0.10 seconds calculate the decelerations of the two cars.

(f) Calculate the impact force experienced by each car.

(g) Comment on your answer to (f) in terms of Newton's 3rd Law of motion.

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.

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

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

Keywords, phrases and learning objectives for elastic/inelastic collisions and momentum

Be able to solve problem involving more complex momentum calculations.

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Worked out ANSWERS to the collision-momentum calculation questions - problem solving more complex situations

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: mgunvgun = mbulletvbullet

mgunvgun = -4.0 = 2.0 x vgun,  vgun = 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. The green car (1), of mass 1000 kg, crashes into the stationary blue car (2) of mass 800 kg.

After the collision, the two cars are locked together.

(a) Calculate the velocity of the combined wrecked cars immediately after the condition.

From the law of conservation of momentum

total momentum before collision = total momentum after collision

m1v1 + m2v2 = m3v3   (m3 = m1 + m2)

(1000 x 20) + (800 x 0) = (1000 + 800) x v3

20 000 + 0 = 1800v3, v3 = 20 000/1800 = 11.1 m/s (3sf)

(b) Calculate the kinetic energy of all three objects in the diagram.

KE = ½mv2

KEgreen car = ½ x 1000 x 202,  =  200 000 J

KEblue car = ½ x 800 x 02 = 0 J

KEwreck = ½ x 1800 x 11.12,  = 110 889 =  111 000 J  (to 3 sf)

(c) Calculate the difference, if any, of the total initial kinetic energies of the cars and that of the wreck,

and comment on the results of your calculations.

Kinetic energy is NOT conserved.

The loss of kinetic energy = 200 000 - 110 889 = 89 111 = 89 100 J  (to 3 sf)

Lots of sound and thermal energy created!

The loss of kinetic energy involves several changes of energy stores.

There is a loud bang on impact, so some energy is lost as sound.

The cars are compressed and the increase in potential energy plus friction effects, increases the thermal energy store of the wreck - which then cools down.

So, in the end, all the lost kinetic energy of the cars energy stores increases the thermal energy store of the surroundings - lost KE somewhat dissipated!

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.

mwhitevintitial + mredvinitial = mwhitevfinal + mredvfinal

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

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

Therefore: mwhitevintitial = mredvfinal

and since the masses are equal, mwhite = mred, 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'! (a) Calculate the momentum of both cars.

(i) green car momentum = 1000 x 10 = +10 000 kg m/s   (to the right)

(ii) blue car momentum = 500 x 15 = -7 500 kg m/s   (to the left)

(you must assign a positive and negative signs, vector quantity, opposite directions)

(b) Calculate the momentum p of the 'wreck'

From the law of conservation of momentum

pwreck = pgreen car + pblue car

pwreck = +10 000 + -7 500 = +2 500 kg m/s

(c) What is the immediate velocity of the wreck after the impact and in what direction? (in terms of diagram)

total mass = 1000 + 500 = 1500 kg

pwreck = mwreck x vwreck

2500 = 1500v, v = 2500 / 1500 = 1.6r = 1.67 m/s to the right (in terms of diagram and to 3 sf too!)

Note: Although the speed of the blue car is greater than the green, it has a greater  momentum, so the resultant movement of the 'wreck' is to the right.

(d) Is kinetic energy conserved in this collision?

(i) You need to calculate the kinetic energies of all the objects concerned

KE = ½mv2

KEgreen car = ½ x 1000 x 102,  =  50 000 J

KEblue car = ½ x 500 x 152 = 56 250 J

KEwreck = ½ x 1500 x 1.672,  = 110 889 =  2 092 J  (to 3 sf)

(ii) Kinetic energy is NOT conserved.

In fact, most of the kinetic energy of the colliding cars is lost on impact. Reminder of diagram

A reminder of the data and the final velocity was 1.67 m/s to the right!

(e) If the impact time is 0.10 seconds calculate the decelerations of the two cars.

(velocities are vector quantities, watch the signs, + to right, - to the left).

a = ∆v / ∆t = (v - u) / ∆t   (u = initial velocity, v = final velocity)

agreen car = (1.6r - 10) / 0.1 = -83.3 m/s2   (3 sf, r meaning recurring)

ablue car = (1.6r - -15) = 16.6r / 0.1 = 166.6r = 167 m/s2   (3 sf, r meaning recurring)

(f) Calculate the impact force experienced by each car.

F = ma

Fgreen car = 1000 x -83.3r = 83 333 = -83 300 N  (3 sf)

Fblue car = 500 x 166.7 = 83 333 = 83 300 N  (3 sf)

(g) Comment on your answer to (f) in terms of Newton's 3rd Law of motion.

The two forces are equal numerically, but act in opposite directions.

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

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 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.

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