Two men of the same mass m=50kg jump from a stationary buggy of mass M=50kg simultaneously, with a velocity of u=20m/s towards left. Find the velocity of the buggy after they jump.
A
20m/s towards left
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B
403m/s towards right
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C
10m/s towards right
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D
203m/s towards left
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Solution
The correct option is B403m/s towards right Given,
Masses of men =m=50kg each
Mass of the buggy =M=50kg
Velocity of each man w.r.t buggy u=20m/s (towards left)
Let v= velocity of buggy after men jump off (towards right)
There is no external force. Hence momentum will be conserved. Taking leftward direction +ve, Pi=Pf ⇒0=m(u−v)+m(u−v)−Mv ⇒2m(u−v)=Mv ⇒2mu=(2m+M)v ⇒v=(2×50×202×50+50) ⇒v=403m/s towards right