Momentum of Cube: Magnitude and In-Between Speed

[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?

 

[haruspex]

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

 

[dl447342]

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.

 

[jbriggs444]

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?

 

[dl447342]

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.

 

[jbriggs444]

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.

 

[dl447342]

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 $v'=0\Rightarrow mv' = 0$.