Question

A small, 300g cart is moving at 1.20m/s on an air track when it collides with a larger, 5.00kg cart at rest. After the collision, the small cart recoils at 0.820m/s . What is the speed of the large cart after the collision?

Answer 1

This question relates to momentum. The total amount of momentum before and after the collision is the same. The momentum of an object is equal to its mass times its velocity. So, try to find the total momentum before the collision first, because the total after the collision will be the same.

Answer 2

Okay, so the momentum (p) of the small cart is mass times velocity right? p = .36

Answer 3

Not sure what to do next :-(

Answer 4

so for the large cart: p = mv

(.36) = (5)(v)
.36 / 5 = v
v = .072

Answer 5

but that is incorrect... what am I doing wrong?

Answer 6

Ok, lets just look at the momentum before the collision. The large cart is at rest so it has no momentum. Also, lets make the mass of the small cart .3 kg to keep our units consistent. I'm also going to assume the cart is moving to the right initially. Also, let the large cart be called "L" and the small cart "S". So the momentum before the collision is:
\[p_i=m_s v_s=(.3 kg) (1.2 m/s) = .36 kg m/s right\]

Answer 7

Yep... it's trucking along.. about to collide with L....

Answer 8

I have the small carts momentum after the collision at p = .246 kgm/s left..

Answer 9

The total momentum after the collision is going to be the momentum of the large cart plus the momentum of the small cart. Remember momentum is a vector so we need to consider direction. We're going to let "towards the right" be positive (it doesn't actually matter which way we consider positive, but we have to be consistent). Now, the collision happens.

When they say the small cart "recoils" i believe they mean the small cart hits the large cart, then goes towards the left, which is the negative direction. So it's momentum is:
\[p_s=m_s v_s = (.3 kg)(-.82 m/s)=-.246 kg m/s\]Remember, the left is negative which is why the velocity is negative.

We know that -.246 plus the momentum of the large cart must equal the total which is .36. So:
\[p_L + (-.246 kg m/s) = .36 kg m/s\]We can solve for the momentum of the large cart after the collision by adding .246 to both sides:
\[p_L=.606 kg m/s\]Almost there. We now know how much momentum the large cart had. Since momentum is mass times velocity we just divide this by its mass to get its velocity:
\[v_L=p_L / m_L = (.606 kg m/s)/(5kg) = .1212 m/s\]

Answer 10

Thank you, it's making sense now.

Answer 11

Glad I could help. I think the hardest part about these problems is being careful with direction. It doesn't matter which direction you call positive, as long as you keep it the same throughout a problem. Remember that momentum is conserved and to consider direction and these questions will become easier. And as with anything, practice makes perfect with these problems.

Answer 12

Yes, that makes sense. I wasn't sure how to setup the equation though. I'm going to post one more similar problem with time involved.. if you have time, maybe you can take a look.

Answer 13

Sorry but I'm signing off for the night, if you haven't gotten any help on it by tomorrow I'd be glad to take a look.