Conservation of Momentum

1. Mar 14, 2009

dherm56

A cart (m1 = 110 kg) is moving to the right along a track at v1i = 17 m/s when it hits a stationary cart (m2 = 390 kg) and rebounds with a speed of v1f = 7 m/s in the opposite direction.
a) With what speed does the 390 kg cart move after the collision?

I used conservation of momentum, m1v1=m2v2, because there are no net external forces acting on the system. However, my answer of 1.97 is incorrect. Is there more to this problem that I'm not seeing?

2. Mar 14, 2009

LowlyPion

Conservation of momentum means that the sum of the momenta of the objects before collision equal the sum afterward. You are missing a term in the after collision side of your equation

3. Mar 14, 2009

dherm56

I changed my equation to m1v1i + m2v2i = m1v1f + m2v2f

There is no initial velocity for m2 so that term cancels out. However, I was still unable to obtain the final velocity of m2

4. Mar 14, 2009

djeitnstine

Since there is no change in momentum, then $$p_i=p_f$$. $$m_1v_{i}+m_2v_{i}=m_1v_{2f}+m_2v_{2f}$$. Note that $$v_i$$ of m2 is 0

edit: you guys got it before I posted.

5. Mar 14, 2009

dherm56

The conservation of momentum concept I understand. However, my answer of 2.82 is not correct.

I got the correct answer. I forgot that vf of m1 is negative in reference to the lab frame.

Thank you for the help

Last edited: Mar 14, 2009