# Momentum of Cube: Magnitude and In-Between Speed

• dl447342
In summary, the conversation discusses the problem of finding the initial and final momentum of a ball in a collision with a truck. Part A and B provide equations for calculating the final momentum, but there is a missing piece of information about the collision. The conversation also mentions the fact that the collision is perfectly elastic and that the average velocities of the ball and truck are equal. The summary also includes a step-by-step explanation of how to find the in-between speed of the truck where the ball bounces back with zero velocity.
dl447342
Homework Statement
We will now shine a beam of light on a light sail and use conservation of energy and momentum to calculate the force exerted on the sail. We will start with our cube of light (length ##l##, cross sectional area A, energy E, and momentum ##p_* = E_*/c##) moving at the speed of light towards a perfectly reflecting sail (cross sectional area A and rest mass M, initially moving with speed V). We will assume perfect reflection of the cube of light off the light sail (“elastic collision”). After reflecting off the moving sail, why would we expect the magnitude of the momentum of the cube of light to be less than ##p_*##?

a) A perfectly elastic ball bounces back off a very massive truck at rest. Is there any decrease in the magnitude of the momentum of the ball?

b) A perfectly elastic ball bounces off a very massive truck in motion away from the ball. Think about the extreme case that the truck is moving only slightly slower than the ball. Show that there must be some speed of the truck where the ball bounces back with zero velocity.
Relevant Equations
Momentum conservation: Let ##V, V', v, v'## denote the initial and final speeds of the truck and the initial and final speeds of the ball, and let ##m, M## denote the masses of the ball and truck where ##m << M##. Then ##mv + MV = - mv' + MV'##.
A) and b) should be useful for solving the initial question.

If the truck is at rest initially, the magnitude of the momentum of the ball becomes ##|mv'|=|MV' - mv|##, but this may or may not be less than the magnitude ##mv##, depending on how large ##V'## is. ##V' = \frac{m(v+v')}{M}## in this case, so it should be about zero. But ##M## is also much larger than ##m##, so I'm not sure what to conclude.

B) If ##V\approx v##, then the final momentum of the ball is ##mv' \approx M(V' - v) - mv##, and using the fact that ##V' = \frac{Mv+m(v+v')}{M}## is just slightly larger than ##v##, I get a similar problem as in part A).

Also, how can I show that there must be an in between speed of the truck where the ball bounces back with zero velocity?

There’s a piece of information about the collision that you are not using.

haruspex said:
There’s a piece of information about the collision that you are not using.
Yeah I guess that's the fact that the collision is perfectly elastic. So the average speed for the ball is the same as the average speed of the truck.

dl447342 said:
Yeah I guess that's the fact that the collision is perfectly elastic. So the average speed for the ball is the same as the average speed of the truck.
Wait, what? Average speeds? Do you maybe mean average velocities?

jbriggs444 said:
Wait, what? Average speeds? Do you maybe mean average velocities?
Yes I mean average velocities. That was a typo. But can you expand on that? I set ##v -v' = V'## for part a) but I'm not sure what to do next.

dl447342 said:
Yes I mean average velocities. That was a typo. But can you expand on that? I set ##v -v' = V'## for part a) but I'm not sure what to do next.
Can you explain why ##v - v'## should be equal to ##V' - V##? It sounds like you were going for conservation of momentum but forgot about mass.

Or maybe you were going for averages but forgot that averages involve sums, not differences.

Ok, so to simplify things I'll redefine ##v## and ##v'## so that they can be negative (e.g. they are velocities). Conservation of momentum says ##mv + MV = mv' + MV' (1)## in both parts A and B, where ##v## and ##v'## are the initial and final speeds of the ball and ##V## and ##V'## are the initial and final speeds of the truck. Then ##mv' = M(V-V') + mv##. By perfect elasticity, the change in kinetic energy is zero, so ##0=1/2 m(v'-v)(v'+v) + 1/2 M(V'-V)(V' + V).## But ##M(V'-V) = -m(v'-v),## so letting ##\Delta p_m = m(v'-v)## we obtain ##\Delta p_m((v'+v) - (V'+V)) = 0## and since ##\Delta p_m \neq 0##, we get ##v'+v = V'+V.## When M is initially at rest, ##v' = V'+V - v = V' + 0 -v = V' - v## is negative as ##V' = \frac{m(v-v')}{M}## is very close to zero by (1). But when ##V## increases, ##v' = V'+V -v## eventually becomes positive, so there is an in-between speed ##V_i## of the truck so that ##v'=0\Rightarrow mv' = 0##.

## What is momentum?

Momentum is a measure of an object's motion, calculated by multiplying its mass and velocity.

## How is momentum related to the speed of a cube?

Momentum and speed are directly proportional. This means that as the speed of a cube increases, its momentum also increases.

## What is the magnitude of momentum?

The magnitude of momentum is the absolute value of the product of an object's mass and velocity. It is a measure of the strength of the object's motion.

## How is momentum conserved in a system?

Momentum is conserved in a system when the total momentum of all objects before and after a collision or interaction remains the same. This is known as the law of conservation of momentum.

## What is in-between speed and how does it relate to momentum?

In-between speed is the speed at which an object travels between two points in time. It is important to consider in-between speed when calculating momentum, as it can impact the overall change in momentum of an object.

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