True or False? A heavy lorry parked at the kerb has a large momentum.

False. Momentum is mass multiplied by velocity , and a parked lorry has velocity zero, so its momentum is zero. Mass alone gives an object no momentum at all: it has to be moving.

Define direct collision .

A direct collision is one in which the two objects are travelling along the same straight line when they meet. Everything then happens in one dimension, so each velocity can be described by a single number with a sign.

In what three ways can two objects be moving before they collide directly?

One of them may be stationary, with the other moving towards it. Otherwise they may be moving in the same direction with the faster one behind, or in opposite directions towards each other.

Two objects coalesce in a collision. What does that mean, and how do you handle it?

They merge into a single object, which moves off with one common velocity. Afterwards you treat them as one particle whose mass is the sum of the two masses, so the equation has a single term on the right instead of two.

True or False? If two objects move towards each other, at least one must have its direction changed by the collision.

True. They cannot pass through each other, so they cannot both carry on in the directions they were going. At least one must be turned back, unless they coalesce or are both brought to rest, when neither carries on at all.

When does conservation of momentum hold?

Only when there are no external forces acting on the objects during the collision. The forces the two exert on each other are internal to the pair and are equal and opposite, so whatever momentum one gains the other loses, and the total is unchanged.

You do not know which way an object moves after a collision. How should you set up the equation?

Assume it moves in the positive direction, and put a positive symbol for its velocity into the equation. If the answer comes out negative, that tells you it is really moving the other way, so nothing is lost by guessing wrongly.

How do you handle a problem with more than one collision?

Take each collision separately and apply conservation of momentum to that one collision alone. The velocities coming out of one collision are the velocities going into the next, so they have to be carried through in order.

One of the objects hits a wall at right angles. What happens to it, and what will the question have to tell you?

Its direction is reversed , so it comes back towards the other object. The question has to give you something extra, such as the change in momentum or the loss in kinetic energy, from which its new speed can be worked out.

True or False? When a ball bounces off a wall, momentum is conserved for the ball on its own.

False. The wall exerts an external force on the ball, and conservation of momentum requires there to be no external forces. That is why a wall cannot be handled by a momentum equation alone, and the question has to supply something else instead.

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Momentum & Collisions (Cambridge (CIE) A Level Maths: Mechanics): Flashcards

Exam code: 9709

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Define momentum.

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Cards in this collection (20)

Define momentum.
The momentum of a particle is the product of its mass and its velocity, so momentum $= m v$ .
Its direction is the same as the direction in which the particle is moving.
What is the SI unit of momentum, and what force unit is it equivalent to?
Momentum is measured in ${kg m s}^{- 1}$ , being a mass in kilograms multiplied by a velocity in ${m s}^{- 1}$ .
That is equivalent to the newton second, $N s$ .
Fill in the two missing words:
A particle moving in the negative direction has a $______$ momentum, because momentum is a $______$ quantity.
A particle moving in the negative direction has a negative momentum, because momentum is a vector quantity.
A momentum only means anything once a positive direction has been chosen, and reversing that choice reverses every sign in the problem.
A dog of mass $15 kg$ runs at $6 {m s}^{- 1}$ . Find its momentum.
The momentum is $15 \times 6 = 90 {kg m s}^{- 1}$ , in the direction the dog is running.
A momentum is not fully described by its size alone, so the direction belongs in the answer.
True or False?
A heavy lorry parked at the kerb has a large momentum.
False.
Momentum is mass multiplied by velocity, and a parked lorry has velocity zero, so its momentum is zero.
Mass alone gives an object no momentum at all: it has to be moving.
A ball's momentum changes from $- 3 {kg m s}^{- 1}$ to $4 {kg m s}^{- 1}$ . Find the change in its momentum.
The change is $4 - (- 3) = 7 {kg m s}^{- 1}$ .
Because momentum carries a sign, a reversal of direction makes the change larger than either momentum on its own, which is why subtracting the sizes and getting $1$ is wrong.
Define direct collision.
A direct collision is one in which the two objects are travelling along the same straight line when they meet.
Everything then happens in one dimension, so each velocity can be described by a single number with a sign.
In what three ways can two objects be moving before they collide directly?
One of them may be stationary, with the other moving towards it.
Otherwise they may be moving in the same direction with the faster one behind, or in opposite directions towards each other.
Two objects coalesce in a collision. What does that mean, and how do you handle it?
They merge into a single object, which moves off with one common velocity.
Afterwards you treat them as one particle whose mass is the sum of the two masses, so the equation has a single term on the right instead of two.
True or False?
If two objects move towards each other, at least one must have its direction changed by the collision.
True.
They cannot pass through each other, so they cannot both carry on in the directions they were going.
At least one must be turned back, unless they coalesce or are both brought to rest, when neither carries on at all.
Complete the principle of conservation of momentum, where $u$ is a velocity before the collision and $v$ one after it:
$m_{1} u_{1} + m_{2} u_{2} =______+______$
The completed principle is:
$m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$
In words, the total momentum after the collision is the same as the total momentum before it.
When does conservation of momentum hold?
Only when there are no external forces acting on the objects during the collision.
The forces the two exert on each other are internal to the pair and are equal and opposite, so whatever momentum one gains the other loses, and the total is unchanged.
You do not know which way an object moves after a collision. How should you set up the equation?
Assume it moves in the positive direction, and put a positive symbol for its velocity into the equation.
If the answer comes out negative, that tells you it is really moving the other way, so nothing is lost by guessing wrongly.
Particles $P$ of mass $3 kg$ and $Q$ of mass $5 kg$ move towards each other at $4 {m s}^{- 1}$ and $2 {m s}^{- 1}$ and collide, after which $P$ rebounds at $1 {m s}^{- 1}$ . Find the velocity of $Q$ .
Taking $P$ 's original direction as positive, $3 (4) + 5 (- 2) = 3 (- 1) + 5 v$ .
That gives $2 = 5 v - 3$ , so $v = 1 {m s}^{- 1}$ , and the answer is positive, which means $Q$ has had its direction reversed too.
How do you handle a problem with more than one collision?
Take each collision separately and apply conservation of momentum to that one collision alone.
The velocities coming out of one collision are the velocities going into the next, so they have to be carried through in order.
Two objects are moving along the same straight line after earlier collisions. Fill in two of the conditions for them to collide again:
if they are moving in the same direction, they collide again when the one in front is $______$
if they are moving in $______$ directions towards each other, they always collide again
If they are moving in the same direction, they collide again when the one in front is slower.
If they are moving in opposite directions towards each other, they always collide again.
The third case is one of them being stationary with the other moving towards it.
One of the objects hits a wall at right angles. What happens to it, and what will the question have to tell you?
Its direction is reversed, so it comes back towards the other object.
The question has to give you something extra, such as the change in momentum or the loss in kinetic energy, from which its new speed can be worked out.
$A$ of mass $0.1 kg$ moving at $10 {m s}^{- 1}$ meets $B$ of mass $0.2 kg$ coming the other way at $2 {m s}^{- 1}$ , and afterwards $A$ still moves forwards at $1 {m s}^{- 1}$ . Find the velocity of $B$ .
Taking $A$ 's direction as positive, $0.1 (10) + 0.2 (- 2) = 0.1 (1) + 0.2 v$ .
That gives $0.6 = 0.1 + 0.2 v$ , so $v = 2.5 {m s}^{- 1}$ , and $B$ has been turned round to travel in $A$ 's direction.
After all the collisions $A$ is moving at $1 {m s}^{- 1}$ and $B$ , ahead of it, at $0.25 {m s}^{- 1}$ in the same direction. Will they meet again?
Yes: they are travelling the same way and the one in front is the slower, so $A$ gradually catches $B$ up.
Comparing the two speeds is the whole of the decision, and no further momentum equation is needed to reach it.
True or False?
When a ball bounces off a wall, momentum is conserved for the ball on its own.
False.
The wall exerts an external force on the ball, and conservation of momentum requires there to be no external forces.
That is why a wall cannot be handled by a momentum equation alone, and the question has to supply something else instead.

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