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A particle moving on a smooth horizontal surface strikes a stationary wall. The angle of strike is equal to the angle of rebound & is equal to `37^(@)` and the coefficient of restitution with wall is `e = (1)/(5)`. Find the friction coefficient between wall and the particle form `(X)/(10)` and fill value of X : |
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Answer» Correct Answer - 5 Since `e = (1)/(5)` `:.` Final normal component of velocity `= (v cos 37^(@))/(5)` As the angle of rebound is equal to the angle before impact. Therefore, both normal & tangential components of velocities must change by the same factor. `:.` Tangential velocity after impact becomes `(v sin 37^(@))/(5)` Let the time of impact be `Delta t` `N=(m(v cos 37^(@)+(v cos 37^(@))/(5)))/(Delta t)=(6mv cos 37^(2))/(5 Deltat)` wher N is the normal force imparted on the ball by the wall. Frictional force `= mu N = (6)/(5) (mu mv cos 37^(@))/(Delta t)` Also frictional force `=(m[v sin 37^(2)-(v sin 37^(@))/(5)])/(Delta t)` `rArr(m[v sin 37^(@)-(v sin 37^(@))/(5)])/(Deltat)` `=(6)/(5)(mu mv cos 37^(@))/(Delta t)` `rArr mu=(2)/(3)tan 37^(@)` `rArr mu =(2)/(3).(3)/(4)=(1)/(2)`. |
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