Asked by Zach
A rock (m = 1.5 kg) moving at 4.2 m/s north collides with a ball (m = 0.061 kg) that is initially at rest. What is the velocity of the rock after the collision if the velocity of the ball after the collision was 8.4 m/s north?
Answers
Answered by
Damon
Initial north momentum = 1.5*4.2 = 6.3
which is also the final momentum
6.3 = .061 * 8.4 + 1.5 * v
v = 3.86 m/s
which is also the final momentum
6.3 = .061 * 8.4 + 1.5 * v
v = 3.86 m/s
Answered by
drwls
All motion remains in the north direction. Apply conservation of linear momentum in that direction.
Mrock*Vrock,initial = Mball*Vball + Mrock*Vrock,final
1.5*4.2 = 0.061*8.4 + 1.5*Vrock,final
Vrock,final = 3.86 m/s
The rock's direction remains north (positive)
Mrock*Vrock,initial = Mball*Vball + Mrock*Vrock,final
1.5*4.2 = 0.061*8.4 + 1.5*Vrock,final
Vrock,final = 3.86 m/s
The rock's direction remains north (positive)
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