How fast must a 100 kg object be going in order for it to stop a 200 kg object traveling at 10 km/hr when the two objects collide head on?

Hello!
I suppose that both objects move along the same straight line towards each other with the uniform velocities. Also I suppose that they do not lose energy during the collision (this is called elastic collision).
Denote the first object as A with the mass m_A and the second as B with the mass m_B. Denote the magnitude of the A's speed before the collision as V_A (let it be directed to the right) and the magnitude of the B's speed as V_B (to the left). The speed of B after the collision is zero (it stops), the magnitude of the speed of A after the collision is V_e. I think it will be directed to the left.
Then consider the projection to the line of their movement and use the momentum conservation law:
m_A V_A - m_B V_B = -m_A V_e
and the kinetic energy conservation law:
m_A (V_A)^2/2 + m_B (V_B)^2/2 = m_A (V_e)^2/2.
Note the plus and minus signs.
This is the system of equations, the unknowns are V_e and V_A. Let's solve it.
The first equation is equivalent to  m_B V_B =m_A(V_A + V_e),
the second equation is equivalent to  m_B (V_B)^2 =m_A ((V_e)^2 - (V_A)^2).
Divide the latter by the former and obtain  V_B = V_e - V_A,  or  V_e = V_A + V_B. Substitute this to the first equation and obtain
m_B V_B = m_A(V_A+V_A+V_B)=m_A(2V_A+V_B).
This gives us  m_B V_B=2m_A V_A+m_A V_B, or
V_A = V_B (m_B-m_A)/(2m_A).
This is the final formula, and numerically it is  10*100/200 = 5 ((km)/h), a half of V_B.
(you can find V_e by yourself and make sure it is positive, so we guessed its direction correctly)
http://hyperphysics.phy-astr.gsu.edu/hbase/conser.html