A wagon is rolling forward on level ground. Friction is negligible. The person s

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 93.6 kg. The mass of the rock is 0.343 kg. Initially the wagon is rolling forward at a speed of 0.475 m/s. Then the person throws the rock with a speed of 16.6 m/s. Both speeds are relative to the ground. Calculate the speed of the wagon after the rock is thrown directly forward (enter first) in one case and directly backward in another.

Explanation / Answer

let,

total mass m1=93.6 kg

mass of the rock m2=0.343 kg

initial wagan farward speed Vw=0.475 m/sec

speed of the rock Vr=16.6 m/sec

after rock is thrown,speed of the rock is V

a)

if the rock is thrown forward:

by using law of conservation of momentum,

m1*vw=m1*v+m2*vr

(93.6*0.475)=(93.6*V)+(0.343*16.6)

====> v=0.414 m/sec

speed of the rock is V=0.414 m/sec

b)
if the rock is thrown backward:

by using law of conservation of momentum,

m1*vw=m1*v-m2*vr

(93.6*0.475)=(93.6*V)-(0.343*16.6)

====> v=0.536 m/sec

speed of the rock is V=0.536 m/sec

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