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A particle of mass m having collided with a stationary particle of mass M deviated by an angle pi//2 whereas the particle M recoiled at an angle theta=30^@ to the direction of the initial motion of the particle m. How much (in per cent) and in what way has the kinetic energy of this system changed after the collision, if M//m=5.0? |
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Answer» Solution :Since, no external impulsive force is effective on the system `"M+m"`, its total momentum along any direction will REMAIN conserved. So from `P_x=const.` `m U=Mv_1 cos theta` or, `v_1=(m)/(M)(u)/(costheta)` (1) and from `p_y=const` `mv_2=Mv_1sin theta` or, `v_2=M/mv_1sintheta=u tan theta`, [using (1)] Final kinetic energy of the system `T_f=1/2mv_2^2+1/2Mv_1^2` And INITIAL kinetic energy of the system `=1/2m u^2` So, % change `=(T_f-T_i)/(T_i)xx100` `=(1/2m u^2tan^2theta+1/2Mm^2/M^2(u^2)/(cos^2theta)-1/2m u^2)/(1/12 m u^2)xx100` `=(1/2u^2tan^2theta+1/2m/Mu^2sec^2theta-1/2u^2)/(1/2u^2)xx100` `=(tan^2theta+m/Msec^2theta-1)xx100` and putting the values of `theta` and `m/M`, we GET % of change in kinetic energy `=-40%` 
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