A particle of mass m moving with a speed v hits elastically another stationary particlebof mass 2m on a smooth horizontal circular tube of radius r. What is the time in which the next collision will take place?
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The velocity of the particle of mass m after the collision can be called v1, and the velocity of the particle of mass 2m v2.
Thus, mv=mv1+2mv2
Eventually giving v=v1+2v2
12mv2=12mv21+122mv22
Eventually, it is
v2=v21+2v21
Dividing both sides of the second equation by v−v1 we get
v+v1=2v222v2=v2
Thus, the relative speed of separation of the two objects is v2−v1=v
When the two bodies collide again, the distances travelled by them must differ by 2πr.
The time taken for this must be t=2πrv2−v1=2πrv
Thus, mv=mv1+2mv2
Eventually giving v=v1+2v2
12mv2=12mv21+122mv22
Eventually, it is
v2=v21+2v21
Dividing both sides of the second equation by v−v1 we get
v+v1=2v222v2=v2
Thus, the relative speed of separation of the two objects is v2−v1=v
When the two bodies collide again, the distances travelled by them must differ by 2πr.
The time taken for this must be t=2πrv2−v1=2πrv
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