Asked by wanda
A 0.471-kg hockey puck, moving east with a speed of 8.17 m/s, has a head-on collision with a 0.795-kg puck initially at rest. Assuming a perfectly elastic collision, what will be the speed of the 0.795-kg puck after the collision?
Answers
Answered by
Henry
Given:
M1 = 0.795kg, V1 = 0.
M2 = 0.471kg, V2 = 8.17 m/s.
M1*V1 + M2*V2 = M1*V3 + M2*V4.
0.795*0 + 0.471*8.17 = 0.795*V3 + 0.471*V4,
Eq1: 0.795*V3 + 0.471*V4 = 3.85.
Conservation of KE Eq:
V3 = (V1(M1-M2) + 2M2*V2)/(M1+M2).
V3 = (0(M1-M2) + 2*0.471*8.17)/(0.795+0.471) = 6.08 m/s = Velocity of M1 after the collision.
M1 = 0.795kg, V1 = 0.
M2 = 0.471kg, V2 = 8.17 m/s.
M1*V1 + M2*V2 = M1*V3 + M2*V4.
0.795*0 + 0.471*8.17 = 0.795*V3 + 0.471*V4,
Eq1: 0.795*V3 + 0.471*V4 = 3.85.
Conservation of KE Eq:
V3 = (V1(M1-M2) + 2M2*V2)/(M1+M2).
V3 = (0(M1-M2) + 2*0.471*8.17)/(0.795+0.471) = 6.08 m/s = Velocity of M1 after the collision.
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