Finding Displacement in a Box Collision with Coefficient of Restitution

[Jzhang27143]

Homework Statement

A cubical box of mass 10 kg with edge length 5 m is free to move on a frictionless horizontal
surface. Inside is a small block of mass 2 kg, which moves without friction inside the box. At
time t = 0, the block is moving with velocity 5 m/s directly towards one of the faces of the box,
while the box is initially at rest. The coefficient of restitution for any collision between the block
and box is 90%, meaning that the relative speed between the box and block immediately after a
collision is 90% of the relative speed between the box and block immediately before the collision.

After 1 minute, the block is a displacement x from the original position. Which of the following
is closest to x?

A) 0 m
B) 50 m
C) 100 m
D) 200 m
E) 300 m

http://www.aapt.org/physicsteam/2014/upload/exam1-2014-2-2-answers.pdf

Homework Equations

Conservation of Linear Momentum
x = vt (a = 0)

The Attempt at a Solution

I am not sure how to do this question. My first idea was to use the coefficient of restitution to find the velocity of the box at individual time intervals between collisions and use x = vt to find the box's displacement and the block's displacement relative to the box. However, there seem to be several time intervals and it seems to be very time consuming. Is there a faster way to do this question? After all, this is a contest question.

 

[Nathanael]

Interesting problem.

Is there a way you can generalize the speed of the box over each interval?

What is going to be different between odd and even intervals?

Can you generalize the length of each time interval?

What have you tried so far?

Edit:
I've overlooked the quick and simple method

Voko, on the other hand, has not :)

 

[voko]

Are you familiar with the concept "centre of mass"?