A person of mass M having a good traction with the surface of support can provide a rock of mass m = 0.02 M with horizontal velocity of 5.00 m/s . If the same effort is generated while standing on a very slippery surface, define velocity of the same rock relatively to the person.

Given, Mass of person = M kg Mass of rock = 0.02M kg Velocity of stone , v= 5m/s Assuming velocity of man to be v' after throwing the stone, By using law of conservation of momentum, we get, 0=mv+Mv'⇒v'=-mMv [m is mass of rock]⇒v'=-0.02MM(5)⇒v'=-0.1 msSo, velocity of rock with respect to the person is given by, vrp=vr-vp =v-v' =5-(-0.1) =5.01 ms

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A person of mass Mhaving a good traction with the surface of support can provide a rock of mass m= 0.02 M with horizontal velocity of 5.00 mis. If the same effort is generated while standing on a very slippery surface, define velocity of the same rock relatively to the person.

A person of mass Mhaving a good traction with the surface of support can provide a rock of mass m= 0.02 M with horizontal velocity of 5.00 mis. If the same effort is generated while standing on a very slippery surface, define velocity of the same rock relatively to the person.

Glencoe Physics: Principles and Problems, Student Edition

1st Edition

ISBN:9780078807213

Author:Paul W. Zitzewitz

Publisher:Paul W. Zitzewitz

Chapter6: Motion In Two Dimensions

Section6.3: Relative Velocity

Problem 27PP

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