Question

A 2.0 kilogram cart moving due east at 6.0 meters per second collides with a 3.0 kilogram cart moving due west. the carts stick together and come to rest after the collision. what was the initial speed of the 3.0 kilogram cart?

Answer 1

Hello

The total momentum must conserve before and after the collision.

1) Calling $m_1=2~Kg$ and $m_2=3~Kg$ the masses of the two carts, and $v_1=6~m/s$ and $v_2$ the velocities of the two carts before the collision, the initial momentum is
$p_1 = m_1 v_1 - m_2v_2$
where the negative sign means the two carts are moving into opposite directions.

2) The two carts after collision are at rest, this means the total momentum after collision is 
$p_2=0$

3) Since the momentum must be conserved, we can write 
$p_1=p_2$
and  substituting
$m_1 v_1 - m_2v_2=0$
from which we can get the velocity of the second cart:
$v_2= \frac{m_1}{m_2}v_1= \frac{2Kg}{3Kg} 6~m/s = 4~m/s$