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A triangular wedge of mass M is moving with uniform velocity v_(0) along a smooth horizontal surface in the leftward direction. A particle of mass m falls from rest from height h onto the inclined face, colliding elastically with it. Find velocity of the ball and wedge after the impact taking M = 2m. |
| Answer» <html><body><p></p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/line-1074199" style="font-weight:bold;" target="_blank" title="Click to know more about LINE">LINE</a> of impact is a <a href="https://interviewquestions.tuteehub.com/tag/straight-633156" style="font-weight:bold;" target="_blank" title="Click to know more about STRAIGHT">STRAIGHT</a> line perpendicular to the incline. Normal reaction force between the body and the wedge acts along the impact line. This normal force becomes internal force when we consider (wedge + body) as a total system. But normal reaction is considerable in <a href="https://interviewquestions.tuteehub.com/tag/magnitude-1083080" style="font-weight:bold;" target="_blank" title="Click to know more about MAGNITUDE">MAGNITUDE</a> because the impact force during collision has contribution towards making of the normal force. Momentum of the system is conserved along the line perpendicular to this normal force.....(i)<br/>Momentum of the system is conserved along horizontal.....(ii)<br/> Momentum of the particle is conserved along the common tangent at the point of impact......(iii)<br/> As this is an elastic collision, relative velocity of separation along the impact line = Relative velocity of approach......(<a href="https://interviewquestions.tuteehub.com/tag/iv-501699" style="font-weight:bold;" target="_blank" title="Click to know more about IV">IV</a>)<br/>If the wedge were at rest then the particle would deflect along the horizontal after collision because it is an elastic collision.....(v)<br/>As, in this case, wedge is moving, the <a href="https://interviewquestions.tuteehub.com/tag/object-11416" style="font-weight:bold;" target="_blank" title="Click to know more about OBJECT">OBJECT</a> would not be deflected horizontal, but at an angle `.alpha.` to the line of impact....(vi)<br/> Velocity of particle before impact `u=sqrt(2gh)` along `vec(PO)`<br/> Velocity of particle after collision `= v_(1)` along `vec(OQ)`<br/>Velocity of wedge before collision `= v_(0)` along `vec(RO)`<br/>Velocity of wedge after collision `= v_(2)` along `vec(RO)` <br/>Velocity of approach = velocity of separation<br/>`v_(0)cos 45^(@)-ucos 45^(@)=-v_(1)cos alpha - v_(2)cos 45^(@)`<br/>`rArr v_(1)cos alpha + (v_(2))/(sqrt(2))=(u)/(sqrt(2))-(v_(0))/(sqrt(2))`<br/>`sqrt(2)v_(1)cos alpha + v_(2)=u-v_(0) "" ....(A)`<br/> Conserving momentum along the horizontal<br/>`Mv_(0)=Mv_(2)-mv_(1)cos (45^(@)-alpha)`<br/>`rArr 2v_(0)=2v_(2)-v_(1)cos (45^(@)-alpha)`<br/>`rArr 2v_(2)-v_(1)cos(45^(@)-alpha)=2v_(0) ""`....(B) <br/>Conserving momentum along the common tangent<br/>`mu sin 45^(@)=mv_(1)sin alpha`<br/>`rArr v_(1)sin alpha = u ""`....(C )<br/> On solving (A), (B) and (C )<br/>`v_(1)=sqrt(u^(2)[(u(4-sqrt(2))-6v_(0))/(20)]^(2))` and `v_(2)=(v_(0)+(sqrt(2)+1)u)/(5)`<br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/FIITJEE_PHY_MB_01_C01_E01_018_S01.png" width="80%"/></body></html> | |