Question

A 92-kg rugby player running at 7.5 m/s collides in midair with a 112-kg player moving in the opposite direction. After the collision each player has zero velocity.
What was the initial speed of the 112-kg player before the collision?

Answer 1

By the condition of momentum conservation we can say that since there is no external force along horizontal direction so we will have

$m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}$

here we know that

$m_1 = 92 kg$

$v_{1i} = 7.5 m/s$

$m_2 = 112 kg$

also we know that

$v_{1f} = v_{2f} = 0$

now from above equation we have

$92(7.5) + 112 v = 0 + 0$

now we have

$v = 6.16 m/s$

so the speed of the other player must be 6.16 m/s