Question

Two hockey players are traveling at velocities of v1 = 14 m/s and v2 = -16 m/s when they undergo a head-on collision. After the collision, they grab each other and slide away together with a velocity of -3.9 m/s. Hockey player 1 has a mass of 115 kg. What is the mass of the other player?

Answer 1

Conservation of momentum says:
m1u1 + m2u2 = m1v1+m2v2
115(14) + m2(-16) = (115+m2)(-3.9)
129 - 16m2 = -448.9 - 3.9m2
577.5 = 12.1m2
47.7272...b= m2
m2 = 47.2 (this seems a little strange, so you might have to check my maths, but the process should be correct)