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...

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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)

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