In a recent hockey game, one player (m=100kg) was moving at a velocity of 8 m/s when he ran into another player (m=300kg) who was at rest (v=0m/s). After the collision, the two players were stuck together (moving with the same velocity). What was their velocity after the collision?

Question

Physics

Frost

15 June, 08:34

In a recent hockey game, one player (m=100kg) was moving at a velocity of 8 m/s when he ran into another player (m=300kg) who was at rest (v=0m/s). After the collision, the two players were stuck together (moving with the same velocity). What was their velocity after the collision?

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Answers (1)

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Mylee Padilla · Answer 1

The combined velocity of players after the collision, V = 2.67 m/s

Explanation:

Given data,

The mass of the first hockey player, m₁ = 100 kg

The velocity of the first hockey player, u₁ = 8 m/s

The mass of second hockey player, m₂ = 300 kg

The velocity of the second player, u₂ = 0 m/s

According to the law of conservation of momentum,

m₁u₁ + m₂u₂ = MV

V = m₁u₁ / M (∵ u₂ = 0)

Substituting the given values,

V = 100 x 8 / 300

= 2.67 m/s

Hence, the combined velocity of players after the collision, V = 2.67 m/s