Finding Displacement in a Box Collision with Coefficient of Restitution

In summary: If so, could you please explain how it would be used in this equation?Centre of mass is used to determine the velocity of an object.
  • #1
Jzhang27143
38
1

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.
 
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  • #2
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 :)
 
Last edited:
  • #3
Are you familiar with the concept "centre of mass"?
 

1. What is the "2014 F = MA #6 Box Collisions" experiment?

The "2014 F = MA #6 Box Collisions" experiment is a physics experiment that involves studying the principles of momentum and energy conservation in collisions between two boxes on a frictionless surface.

2. What is the purpose of this experiment?

The purpose of this experiment is to demonstrate the conservation of momentum and energy in collisions, as well as to calculate the final velocities of the boxes after the collision using the equations of motion.

3. What materials are needed for this experiment?

The materials needed for this experiment include two boxes, a frictionless surface, a ruler or measuring tape, and a motion sensor or other device to track the motion of the boxes.

4. How is this experiment performed?

To perform this experiment, the two boxes are placed on the frictionless surface with a small distance between them. One box is given an initial velocity and collides with the stationary box. The motion of the boxes is tracked and the final velocities are calculated using the equations of motion.

5. What are the key concepts that can be learned from this experiment?

This experiment helps to understand the concepts of momentum, energy, and their conservation in collisions. It also demonstrates the application of the equations of motion to calculate the final velocities of the boxes after the collision.

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