Two particles collide. How do you find the kinetic energy lost in the impact?

Work out the total kinetic energy of both particles before the collision and again after it, then subtract the second from the first. Each particle contributes according to its own mass and its own speed, and the two contributions are added at each stage.

True or False? Kinetic energy lost in a collision has been destroyed.

False. Total energy is always conserved, and the kinetic energy has been transferred into other forms such as heat and sound. Only the kinetic energy of the objects has fallen, which is why the word loss is used rather than destruction.

True or False? The total kinetic energy of a system can increase as a result of an interaction.

True. In an explosion a cannon and its ball both start at rest with no kinetic energy at all, and both end up moving. The extra kinetic energy came from chemical energy in the propellant, so total energy is still conserved.

Why does a collision in which the particles coalesce lose the most kinetic energy?

Because they end with a single common velocity, so none of the kinetic energy is left in relative motion between them. Any other outcome leaves the two moving apart at some speed, and that separation carries kinetic energy with it.

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

Treat each collision separately , applying momentum and restitution to one impact at a time. Use a fresh letter for the velocities produced by each stage, so that speeds after the second collision are never confused with those after the first.

True or False? Objects accelerate between two successive collisions on a smooth horizontal surface.

False. On a smooth horizontal surface nothing acts along the line of motion between impacts, so each object travels at a constant speed . That is why distance divided by speed is enough to find the time between one collision and the next.

A LevelFurther MathsEdexcelFurther Mechanics 1FlashcardsElastic CollisionsElastic Collisions in 1D

Elastic Collisions in 1D (Edexcel A Level Further Maths: Further Mechanics 1): Flashcards

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FrontNewton's Law of Restitution
Define the coefficient of restitution between two objects.

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Cards in this collection (20)

Define the coefficient of restitution between two objects.
The coefficient of restitution $e$ is the ratio of their speed of separation to their speed of approach when they collide.
Being a ratio it is dimensionless, and its value depends on the materials the two objects are made from.
Objects with velocities $u_{1}$ and $u_{2}$ collide and leave with velocities $v_{1}$ and $v_{2}$ . Fill in the numerator of Newton's law of restitution:
$e = \frac{______}{u_{1} - u_{2}}$
The completed law is:
$e = \frac{v_{2} - v_{1}}{u_{1} - u_{2}}$
The order is reversed between the two differences, which is what keeps a separation and an approach both coming out positive.
What values can the coefficient of restitution take, and what happens at each extreme?
It satisfies $0 \leq e \leq 1$ .
At $e = 1$ the collision is perfectly elastic, and at $e = 0$ it is perfectly inelastic, which means the objects coalesce and move on together.
Two objects move along the same line at $5 {m s}^{- 1}$ and $2 {m s}^{- 1}$ . What is their speed of approach?
It depends on the directions: $3 {m s}^{- 1}$ if they travel the same way, and $7 {m s}^{- 1}$ if they travel towards each other.
The speed of approach is the rate at which the gap between them closes, so the two speeds subtract in one case and add in the other.
You know the masses and initial speeds of two colliding objects and the value of $e$ . Which two equations give you both final velocities?
Conservation of momentum and Newton's law of restitution, solved simultaneously.
Each on its own gives one equation in the two unknown final velocities, so together they determine both.
True or False?
Because $e$ is a ratio of speeds, you can ignore the signs of the velocities when using it.
False.
The formula uses velocities, which are negative for an object travelling in the negative direction, so the signs decide whether the two differences come out correctly.
A sign error here usually produces a value outside the permitted range, which is a useful check that something has gone wrong.
A final velocity comes out in terms of $e$ . How do you find the range of possible values of $e$ ?
Combine $0 \leq e \leq 1$ with what the situation tells you about direction, since each final velocity must be positive or negative according to the way that object actually moves.
The two conditions together give an inequality in $e$ .
How does the restitution formula simplify when an object rebounds off a fixed wall?
It becomes $e = \frac{v}{u}$ , the rebound speed divided by the approach speed.
The wall has no velocity of its own, so there is nothing to subtract on either side of the ratio.
A ball is dropped from height $h$ onto a horizontal plane and rebounds to height $H$ . How do you find $e$ ?
Turn each height into a speed using the constant acceleration results, then take the ratio of the two.
The $2 g$ cancels, leaving $e = \sqrt{\frac{H}{h}}$ , so the ball rebounds to a fixed fraction of its previous height every time.
Two particles collide. How do you find the kinetic energy lost in the impact?
Work out the total kinetic energy of both particles before the collision and again after it, then subtract the second from the first.
Each particle contributes according to its own mass and its own speed, and the two contributions are added at each stage.
True or False?
Kinetic energy lost in a collision has been destroyed.
False.
Total energy is always conserved, and the kinetic energy has been transferred into other forms such as heat and sound.
Only the kinetic energy of the objects has fallen, which is why the word loss is used rather than destruction.
For which value of $e$ is kinetic energy conserved in a collision?
Only when $e = 1$ .
For any smaller value some kinetic energy is transferred away in the impact, and in reality every collision has $e < 1$ , although some are modelled as perfectly elastic.
True or False?
The total kinetic energy of a system can increase as a result of an interaction.
True.
In an explosion a cannon and its ball both start at rest with no kinetic energy at all, and both end up moving.
The extra kinetic energy came from chemical energy in the propellant, so total energy is still conserved.
Why does a collision in which the particles coalesce lose the most kinetic energy?
Because they end with a single common velocity, so none of the kinetic energy is left in relative motion between them.
Any other outcome leaves the two moving apart at some speed, and that separation carries kinetic energy with it.
How do you handle a problem with more than one collision?
Treat each collision separately, applying momentum and restitution to one impact at a time.
Use a fresh letter for the velocities produced by each stage, so that speeds after the second collision are never confused with those after the first.
Objects $A$ and $B$ move in the same direction after colliding, with $A$ behind. When will they collide again?
Only if $A$ is still moving faster than $B$ , so that it catches up.
If $A$ is the slower of the two, or the two move apart, there is no second collision between them.
After colliding with $A$ , object $B$ hits a wall and rebounds. What makes a further collision with $A$ possible?
The wall reverses the direction of $B$ , so it now travels back towards $A$ .
They meet again provided $B$ is closing on $A$ , which needs its rebound speed to exceed the speed of $A$ in that same direction.
True or False?
Objects accelerate between two successive collisions on a smooth horizontal surface.
False.
On a smooth horizontal surface nothing acts along the line of motion between impacts, so each object travels at a constant speed.
That is why distance divided by speed is enough to find the time between one collision and the next.
$A$ and $B$ collide, then $B$ travels to a wall and rebounds. Where is $A$ when $B$ starts back?
Not where the first collision happened, because $A$ has been moving the whole time $B$ was travelling to the wall.
Find how long $B$ took to reach the wall, then multiply that time by the speed of $A$ to see how far it has gone.
Two objects a distance $x$ apart approach each other at speeds $v_{A}$ and $v_{B}$ . How long until they meet?
The gap closes at a rate of $v_{A} + v_{B}$ , so the time taken is $\frac{x}{v_{A} + v_{B}}$ .
They meet at the point dividing $x$ in the ratio $v_{A} : v_{B}$ , measured from where each of them started.

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