School Physics notes: Momentum - elastic and non-elastic collision calculations

6. Elastic and non-elastic collisions, conservation of momentum - concepts and calculations and Newton's 2nd law of motion

Doc Brown's school physics revision notes: GCSE physics, IGCSE physics, O level physics,  ~US grades 8, 9 and 10 school science courses or equivalent for ~14-16 year old students of physics

What is the momentum of a an object?

What is an elastic collision?   What is an inelastic collision?

How do we solve problems and do calculations involving momentum?

(a) What is the momentum of an object and how do we calculate it?

(b)

(c)

(d) Problem solving questions involving more complex momentum calculations

(e)

(f) More advanced calculations involving momentum and Newton's 2nd/3rd Laws

(a) What is the momentum of an object and how do we calculate it?

All moving objects have what we call momentum.  (sometimes denoted by a lower case p)

The faster an object moves, the bigger its momentum.

The bigger the mass of a moving object, the bigger its moment.

The momentum of an object is directly proportional to its velocity AND its mass.

Momentum (p) is a vector quantity, it has both size and direction

Therefore the equation for momentum is ...

momentum (kilogram metre per second) = mass (kilogram) × velocity (metre per second)

momentum p (kg m/s) = mass m (kg) x velocity v (m/s)

p = m × v

Example of simple calculations using the formula for momentum

Q1.1  A 120 kg sprinter is travelling at 9.0 m/s, what is the momentum of the runner?

momentum p = m × v

p = 120 x 9 = 360 kg m/s

Q1.2 If a 20 g bullet has a momentum of 8 kg m/s, calculate its velocity.

p = m × v,  v = p/m     (don't forget to change the g to kg!)

velocity = 8/(20/1000) = 400 m/s

Q1. 3 What mass must an object have if it has a momentum of 1.5 x 106 kg m/s and a velocity of 30 m/s?

p = m × v,  m = p/v

mass = 1.5 x 106/30 = 5.0 x 104 kg

Q1.4 What is the velocity of a 1500 kg car travelling north with 30 000 kg m/s of momentum?

p = m x v, v = p / m

v = 30 000 / 1500 = 20 m/s in a northerly direction

Q1.5 -

TOP OF PAGE and sub-index

(b) The Law of Conservation of Momentum

Here we will consider collisions between two objects in a closed system.

Here a closed system here means no other external forces affect the situation e.g. a collision between two objects - the event.

If an external force like friction is involved, total momentum cannot be conserved.

The total momentum of an event in a closed system is the same before and after the event (e.g. a collision between two objects).

This is called the 'Law of Conservation of Momentum' and you can use it do lots of calculations!

e.g. total momentum of two colliding objects = total moment of objects after collision

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

substituting m and v for the mass and velocity gives ...

m1v1 + m2v2 = m1v3 + m2v4

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

(you are assuming there is no change in mass, no bits have flown off!)

and if two objects stick together' after the collision then: p1 + p2 = p3

substituting m and v for the mass and velocity gives ...

m1v1 + m2v2 = m3v3   (where m3 = m1 + m2)

where v1 and v2 are the initial velocities and v3 and m3 (m3 = m1 + m2) are the final velocity and mass after the collision.

In other words the large object formed by collision has the momentum equal to the two momentums of the colliding objects added together.

Momentum is conserved for both elastic and inelastic collisions.

For a perfect elastic collision, no kinetic energy is lost - kinetic energy conserved.

In an elastic collision, the total energy in the kinetic energy stores of the colliding objects is the same as before and after the collision.

You will not have to solve problems for elastic collisions - the maths is too difficult for GCSE physics, with two sets of equations, for momentum (mv) and kinetic energy (E = ½mv2), to solve e.g. for the resultant velocities!

For an inelastic collision, kinetic energy is not conserved - kinetic energy is lost in some form e.g. heat or sound.

In an inelastic collision, some of the moving objects kinetic energy stores are lost and transferred to other energy stores of the objects themselves or the environment.

This is because the atoms are bashed together increasing their potential energy store (compressed for a fraction of a second). They 'relax' to their normal state by losing the energy as heat  (thermal energy) or sound.

For inelastic collisions you can solve a variety of problems using the principle of 'conservation of momentum'.

TOP OF PAGE and sub-index

(c) Other examples of momentum effects

The above examples involve collision but there are all sorts of momentum situations e.g.

When a gun is fired the bullet goes in the forward direction (positive momentum), but the gun recoils in the complete opposite direction (negative momentum).

From Newton's 3rd law, you get an equal and opposite force reaction.

The bullet goes one way and the gun the other way from the force of the explosion.

numerically, ignoring +/- signs: mgunvgun = mbulletvbullet   (see Q2.2 below)

Note that the combined momentum = zero, which it is at the start and immediately after the gun is fired.

Before a rocket is fired it has zero velocity and zero momentum.

When the rocket's propellant is fired, the rocket goes one way (positive momentum) and the exhaust gases go in the opposite direction (negative momentum).

So, in conserving momentum, we can say ..

numerically, ignoring +/- signs: mrocketvrocket = mexhaust gasesvexhaust gases   (see Q2.2 below)

Because the rocket produces gases with great force and velocity, the rocket must respond according to Newton's 3rd law of motion - equal and opposite forces in dramatic action!

-

TOP OF PAGE and sub-index

(d) Problem solving questions involving more complex momentum calculations

Q2.1 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)

Q2.2(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.

Q2.3 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!

Q2.4 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)

Q2.5 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.

Q2.6 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!

Q2.7 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.

Q2.8 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.

Q2.9

TOP OF PAGE and sub-index

(e) Change in Momentum and the consequences of Newton's Second Law of Motion

When a resultant force acts on an object for any length of time it will cause a change in momentum by changing the velocity.

The resultant force causes a change in speed or direction in the direction of the resultant force.

The resultant force causes a change in momentum in the direction of the resultant force.

A way of stating Newton's 2nd law is to say the change in momentum is proportional to the size of the resultant force and the time interval during which the force is acting on the object.

This statement is justified by following the three mathematical steps below.

A resultant force on an object will cause it to accelerate or decelerate.

force = mass x acceleration

F = ma   (the mathematical expression of Newton's 2nd Law of Motion)

NOW acceleration is the change in velocity in a specific change in time.

a = ∆v / ∆t  (the usual formula for acceleration)

a = acceleration in m/s2, v in m/s,  time t in seconds s

If you combine these two equations you get

force = mass x change in velocity / time

F = m ∆v / ∆t = m(v-u) / ∆t   (where v and u are the final and initial velocities)

but, for a given object of mass m, m ∆v = ∆mv = ∆p = change in momentum

so can write Newton's 2nd law equation as

F =  ∆mv / ∆t = ∆p / ∆t   (force is equal to the rate of change of momentum)

These equations express force as the rate of change of momentum

and ∆p = ∆mv = F x ∆t   (remember ∆t is the time over which the resultant force acts)

F x ∆t is sometimes called the impulse.

This means that a force applied to an object over any period of time must change that object's velocity.

From the point of view of solving calculation problems you need to be pretty familiar with all this maths!

In calculations you can then use Newton’s 2nd law of motion (F = ma) as follows:

force (newton, N) = change in momentum (kilogram metre per second, kg m/s) / time (second, s)

F = ∆mv / ∆t = (mv - mu) / t,   where = mass of object kg, u = initial velocity m/s, v = final velocity m/s

this means the rate of change of momentum is a directly related to the resultant force or force applied,

so you can say the force equals the rate of change of momentum,

the equation can be written as    F = mΔv / Δt    since m is constant for a given object

which can be expressed as ...

change of momentum (kg m/s) = resultant force (N) × time for which it acts (s)

∆p = ∆mv = F x ∆t   (the product of the resultant force x time is sometimes called the impulse)

(if the force is varying, then the average force is used in the calculation, but I don't think this is needed for GCSE)

Consequences of the force equalling the rate of change of momentum

For any moving object the faster the change in momentum (Δmv) occurs, the greater the force (F) involved, and the shorter the time taken (∆t).

You should be able to see this from the equation ∆mv = F x ∆t

for a given mv, increase in F means a decrease in ∆t,

for a given mv, decrease in F means an increase in ∆t,

This has serious implications for e.g. car crashes.

The faster the crash happens the bigger the force on the car and its occupants.

This sudden change momentum change in such a short time results in a great impact force.

The smaller Δt for a given momentum change, the greater the force involved and the greater the chance of injury.

If you can make the rapid deceleration occur over a longer time (increasing Δt) you reduce the resultant force and decrease the chance of serious injury.

That is what crumple zones are for in the design of modern cars.

In terms of F = ma, you are trying to reduce a and so decrease F.

(For more details on this issue see

TOP OF PAGE and sub-index

(f) More advanced calculations involving momentum and Newton's 2nd/3rd Laws

Q3.1 What explosive force is required to accelerate a 20 g bullet in a gun from 0 m/s to 300 m/s in 0.15 seconds.

20 g = 20/1000 = 0.02 kg, u = 0 m/s, v -= 300 m/s, ∆t = 0.15 s

F = ∆mv / ∆t = (mv - mu) / t

F = {(0.02 x 300) - (0.02 x 0)} / 0.15

F = (0.02 x 300) / 0.15 = 6.0 / 0.15 = 40 N

It is important that you can solve these sorts of problems in different ways, depending on what information you get and how you are expected to deduce things from it e.g.

Using the same information you can solve this problem using F = ma directly.

a = 300 / 0.15 = 2000 m/s2, m = 0.02 kg, F = ma, F = 0.02 x 2000 = 40 N

This is the approach I took on my stopping distances, impact forces - example calculations

Q3.2 What force must a tennis player generate with a tennis racket exert on a 60 g tennis ball to accelerate it from rest to a speed of 50 m/s in 10 milliseconds (10 ms).

mass of tennis ball = 60 g = 60/1000 = 0.06 kg, time force applied = 10 ms = 10/1000 = 0.01 s

∆p = ∆mv = F x ∆t  (this comes from Newton's 2nd Law equation F = m∆v/∆t)

mv changes from zero to 0.06 x 50 = 3.0 kg m/s

therefore: 3.0 = F x 0.01,  F = 3.0 / 0.01 = 300 N

Q3.3 An object experiences a change in momentum of 30 kg m/s.

If the change takes place in 5 seconds, what force on the object produced this change in momentum?

F = ma = m∆/∆t = ∆mv / ∆t = 30 / 5 = 6 N

Q3.4 A force of 500 N is applied to an object to change its momentum by 2000 kg m/s.

For how long was the force applied?

F = ma = m∆v/∆t = ∆mv/∆t

∆t = ∆mv / F = 2000 / 500 = 4 seconds

Q3.5 A 0.80 g lead pellet fired from an airgun experiences a force of 4.8 N in a time interval of 25 milliseconds - the time it takes for the pellet to travel down the barrel.

(a) Calculate the speed at which the pellet leaves the gun.

F = ma = m∆v / ∆t = m(v-u) / ∆t

m = 0.8 / 1000 = 0.0008 kg,  ∆t = 25 / 1000 = 0.025 s (25 milliseconds), v = ?, u = 0 m/s

F = mv / ∆t

v = ∆t x F / m = 0.025 x 4.8  / 0.0008 = 150 m/s

(b) If the airgun recoils at a speed of 0.05 m/s, what is the mass of the airgun?

You can solve this part of the problem two ways

(i) From the law of conservation of momentum

momentum of airgun = momentum of pellet

m x 0.05 = 0.0008 x 150

m = 0.0008 x 150 / 0.05 = 2.4 kg

(ii) From Newton's 3rd Law, which tells you that the force on the gun is equal in size to the force on the lead pellet.

F = ma, m = F / a

m =? kg, F = 4.8 N, a = ∆v / ∆t = 0.05 / 0.025 = 2.0 m/s

m = 4.8 / 2.0 = 2.4 kg

Q3.6

For lots more questions like these, see section

Note on impulse - a vector quantity

impulse = force x time

(time x the force is applied in the direction of action in Ns)

impulse = mass x change in momentum = mΔv

TOP OF PAGE and sub-index

Motion and associated forces notes index (including Newton's Laws of Motion)

gcse physics

gcse physics revision notes

3. Acceleration, friction, drag effects and terminal velocity experiments gcse physics revision notes

gcse physics revision notes

gcse physics revision notes IGCSE revision notes momentum calculations elastic inelastic collisions KS4 physics Science notes on momentum calculations elastic inelastic collisions GCSE physics guide notes on momentum calculations elastic inelastic collisions for schools colleges academies science course tutors images pictures diagrams for momentum calculations elastic inelastic collisions science revision notes on momentum calculations elastic inelastic collisions for revising physics modules physics topics notes to help on understanding of momentum calculations elastic inelastic collisions university courses in physics careers in science physics jobs in the engineering industry technical laboratory assistant apprenticeships engineer internships in physics USA US grade 8 grade 9 grade10 AQA GCSE 9-1 physics science revision notes on momentum calculations elastic inelastic collisions GCSE notes on momentum calculations elastic inelastic collisions Edexcel GCSE 9-1 physics science revision notes on momentum calculations elastic inelastic collisions for OCR GCSE 9-1 21st century physics science notes on momentum calculations elastic inelastic collisions OCR GCSE 9-1 Gateway  physics science revision notes on momentum calculations elastic inelastic collisions WJEC gcse science CCEA/CEA gcse science

TOP OF PAGE and sub-index               Website content © Dr Phil Brown 2000+. All copyrights reserved on revision notes, images, quizzes, worksheets etc. Copying of website material is NOT permitted. Exam revision summaries & references to science course specifications are unofficial.

Doc Brown's Physics TOP OF PAGE and sub-index