A man of mass 60kg is standing on a platform of mass 140kg kept on a smooth horizontal surface. The man starts moving on the platform with a velocity 2m/s relative to the platform. Find the recoil velocity of the platform.
A
1m/s
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B
0.5m/s
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C
0.6m/s
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D
2m/s
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Solution
The correct option is C0.6m/s Given, mass of man, (m1)=60kg mass of platform, (m2)=140kg
Let the man move at a speed u towards right and the platform recoil at a speed v towards left, relative to the smooth surface.
Thus, speed of the man relative to platform is u+v=vr ⇒u=vr−v ...(1)
Here, man and platform form a system. There is no external horizontal force on the system. So, linear momentum of the system remains constant in horizontal direction.
Initially, both man and platform were at rest. So, applying conservation of linear momentum (taking rightward +ve) 0=−m2v+m1u ⇒m2v=m1u ⇒m2v=m1(vr−v) (from (i)) ⇒v=m1vrm1+m2 ⇒v=60×260+140=0.6m/s