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A smooth ball of mass 1 kg is projected with velocity 7 m//s horizontal from a tower of height 3.5 m. It collides elastically with a wedge of mass 3 kg and inclination of 45^(@) kept on ground. The ball collides with the wedge at a height of 1 m above the ground. Find the velocity of the wedge and the ball after collision. (Neglect friction at any contact.)

Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :`v_(x)=7m//s` velocity along `y`-<a href="https://interviewquestions.tuteehub.com/tag/axis-889931" style="font-weight:bold;" target="_blank" title="Click to know more about AXIS">AXIS</a> of the <a href="https://interviewquestions.tuteehub.com/tag/ball-891849" style="font-weight:bold;" target="_blank" title="Click to know more about BALL">BALL</a> just before collision. <br/> `v_(y)=sqrt(2xx9.8x2.5)=7m//s` <br/> As `v_(x)=v_(y)` so it strikes the plane of incline perpendicularly. Let the ball rebound with velocity `V` and `v_(1)` be the velocity of the wedge <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/BMS_VOL2_C01_E01_075_S01.png" width="80%"/> <br/> Applying the principle <a href="https://interviewquestions.tuteehub.com/tag/conservation-929810" style="font-weight:bold;" target="_blank" title="Click to know more about CONSERVATION">CONSERVATION</a> of <a href="https://interviewquestions.tuteehub.com/tag/momentum-1100588" style="font-weight:bold;" target="_blank" title="Click to know more about MOMENTUM">MOMENTUM</a> in horizonal direction, we get <br/> `1xx7=1xxv/(sqrt(2))+3v_(1)` <br/> `7sqrt(2)=3sqrt(2)v_(1)-v`...........i <br/> Applying the equation for coefficient of restitution, we get `e=1=(v_(1)//sqrt(2)+v)/(7sqrt(2)),7sqrt(2)=v_(1)/sqrt(2)+v`.........ii <br/> Solving eqn i and ii `v_(1)=4m//s` and `v=5sqrt(2)m//s`</body></html>


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