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A ball with a velocity of 5 ms^(-1) impinges at angle of 60^(@) with the vertical on a smooth horizontal plane. If the coefficient of restitution is 0.5, find the velocity and direction after the impact. |
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Answer» Solution :Given : velocity of ball: `5ms^(-1)` Angle of INCLINATION with vertical : `60^(@)` Cofficient of restitution =0.5 Notes:Let the angle reflection is and the speed after collision is `v^(.)` the floor exerts a FORCE on the ball along the normal during the collision. There is no force parallel to the surface. Thus, the parallel component of the velocity of the hall remains unchanged . Tihis gives `v^(.)SINTHETA^(.)=vsintheta` ...(1) Vertical component with respect to floor `=v^(.)costheta^(.)` (velocity of separation) `velocity of approach=vcostheta` `"coefficient of restituting e" =("velocity of separation")/("velocity of approach")` `e=(v^(.)costheta^(.))/(v COS theta) therefore v^(.)costheta^(.)=evcostheta` ...(2) from (1) and (2) (ii)`v^(.)sqrt((1-sin^(.2)theta))=evcostheta` `v^(2)(1-sin^(.2)theta)=e^(2)v^(2)cos^(2)theta` `v^(.2)-v^(.2)sin^(.2)theta=e^(2)v^(2)cos^(2)theta` `v^(.2)-v^(.2)sin^(2)theta=e^(2)v^(2)cos^(2)theta [v^(.)sintheta^(.)=vsintheta]` `v^(.)=sqrt(v^(2)sin^(2)theta+e^(2)v^(2)cos^(2)theta)` `v^(.)=vsqrt(sin^(2)theta+e^(2)cos^(2)theta)` The speed after collision `v(.)sqrt(sin^(2)theta+e^(2)cos^(2)theta)` `v^(.)=5sqrt(sin^(2)(60^(@))+(0.5)^(2)cos^(2)60^(@))=5sqrt(3/4+0.25xx1/4)` `=5/2-sqrt(3.25)=2.5xx1.8=4.5ms^(-1)` `"Angle of reflection" theta^(.)=TAN^(-1)((tantheta)/e)=tan^(-1)((tan60^(@))/0.5)=tan^(-1)(sqrt3/0.5)` `=tan^(-1)(3.464)=73.9^(@)` 
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