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?

Question

Isabel Owens

Physics

15 June, 04:40

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?

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Corinne Key · 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)