Question

A wagon is rolling forward on level ground. Friction is negligible. The person sitting in the wagon is holding a rock. The total mass of the wagon, rider, and rock is 98.9 kg. The mass of the rock is 0.291 kg. Initially the wagon is rolling forward at a speed of 0.490 m/s. Then the person throws the rock with a speed of 16.1 m/s. Both speeds are relative to the ground. Find the speed of the wagon after the rock is thrown (a) directly forward in one case and (b) directly backward in another.

Answer 1

Part A.

When rock is thrown in forward direction, then Using Momentum conservation:

Pi = Pf

M*V = m1*v1 + m2*v2

M = Mass of rock + wagon = 98.9 kg

m1 = mass of rock = 0.291 kg

m2 = mass of wagon = 98.9 - 0.291 = 98.609 kg

V = Initial speed of rock + wagon = +0.490 m/sec

v1 = final speed of rock = +16.1 m/sec

v2 = final speed of wagon = ?

v2 = (M*V - m1*v1)/m2

v2 = (98.9*0.490 - 0.291*16.1)/98.609 = 0.444 m/sec

v2 = final speed of rock in forward direction = 0.444 m/sec

Part B.

When rock is thrown in backward direction, then Using Momentum conservation:

Pi = Pf

M*V = m1*v1 + m2*v2

M = Mass of rock + wagon = 98.9 kg

m1 = mass of rock = 0.291 kg

m2 = mass of wagon = 98.9 - 0.291 = 98.609 kg

V = Initial speed of rock + wagon = +0.490 m/sec

v1 = final speed of rock = -16.1 m/sec

v2 = final speed of wagon = ?

v2 = (M*V - m1*v1)/m2

v2 = (98.9*0.490 - 0.291*(-16.1))/98.609 = 0.539 m/sec

v2 = final speed of rock in forward direction = 0.539 m/sec

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