# Homework Help: Finding velocity and work of a collision

1. Oct 27, 2016

### Jrlinton

1. The problem statement, all variables and given/known data
Two 2.5 kg bodies, A and B, collide. The velocities before the collision are $\vec{v}_A = \left( 50 \hat{i} + 50 \hat{j} \right)~m/s$ and $\vec{v}_B = \left( 38 \hat{i} + 2.3 \hat{j} \right)~m/s$. After the collision, $\vec{v}'_A = \left( 10 \hat{i} + 9.4 \hat{j} \right)~m/s$. What are (a) the x-component and (b) the y-component of the final velocity of B? (c) What is the change in the total kinetic energy (including sign)?

2. Relevant equations

3. The attempt at a solution
So i decoupled the x and y components and found the components of the velocity of B seperately
For x
2.5*50+2.5*38=25*10+2.5*v
v_x=78 m/s

For y
2.5*50+2.5*2.3=2.8*9.4+2.5*v
v_y=42.9 m/s

I then stayed with the decoupled approach to find the work done in each direction
for x:
initial KE= 1.25*50^2+1.25*38^2=4930J
final KE= 1.25*10^2+1.25*78^2=7730J

for y:
initial KE= 1.25*50^2+1.25*2.3^2= 3131.61J
final KE= 1.25*9.4^2+1.25*42.9^2=2410.96J

Subtracted the initial KE from final KE to find work (ignore the unnecessary parentheses as they give me a sense of organization)
(7730J+2410.96J)-(4930J+3131.61J)
2080.35 J
This cannot be right can it? How can the system gain KE?

Last edited by a moderator: Oct 27, 2016
2. Oct 27, 2016

### Jrlinton

It seems like the velocities haven't shown up. Here is a direct screenshot of the problem:

3. Oct 27, 2016

### TomHart

This is a strange problem. How can you have object A moving in the +x direction at 50 m/s, and object B moving in the +x direction at 38 m/s, then, after the collision, object A is moving in the +x direction at only 10 m/s - unless energy was imparted to object A by some other means than the collision?

One other possibility is that sometimes I just get confused when I look at some of these problems.

4. Oct 27, 2016

### Jrlinton

The only way I can see that happening is if B has a larger mass than A, but it is stated that their mass are equal. Anyway I used the solutions and they were correct. Unless I am missing something, this is a very peculiar problem indeed.

5. Oct 27, 2016

### TomHart

Even if B was infinite mass, the velocity of A could not change that much. If B was an infinite mass, and A and B had a perfectly elastic collision, A would bounce off of B with, at most, equal and opposite x component of the velocity. So since the difference is initially +12 m/s, the final difference would be -12 m/s - or an absolute velocity of (38-12) = 26 m/s.

Maybe B had some type of explosive reactive armor. That must be it. :)

6. Oct 27, 2016

### Staff: Mentor

More prosaically, perhaps the problem's values were auto-generated by software (so everyone gets a unique question) and the author didn't put in sufficient reality checking.