Two equal spheres A and B lie on a smooth horizontal circular groove at opposite ends of a
diameter. At time t =0, A is projected along the groove and it first impinges on B at time t =T1
and again at timet = T2. Ife is the coefficient of restitution the ratio T2/T1 is
Answers
Answer:
The ratio T2/T1 is 2+e/e
Explanation:
According to the problem at time t = 0 sphere A is projected and its impinges at t = T1
Let the velocity of A is u before collision,
and the velocity of B is zero before collision.
As A and B collide two times,
Let the velocities of A and B after collision is v1 and v2
Therefore, π= u /r x T1 or, T1 = πR/u
As given the coefficient of restitution is e
Now we know that,
v1 = (m1 -em2) u1 + (1+e)m2u2/m1+m2
v1 = (m -em) u + (1+e)m x 0 /m+m = (1-e)u/2
v2 = (m2 -em1) u2 + (1+e)m1u1/m1+m2
= (m -em) u2 + (1+e)mu/m+m = (1+e)u/2
Now for the second time t1 ,
T2 = t1+T1
Let A completes θ and B completes (θ +π) to hit A once again.
Therefore θ= v1/r x t1 = (1-e)u/2r t1
again (θ +π) = v2/rx t1 = (1+e)u/2r x t1
By solving these two we can get t1 = 2πr/eu
Therefore,
t1/T1 = 2/e
Therefore,
T2/T1 = T1+ t1 /T1 = 1 + t1 /T1
or T2/T1 = 2+e/e
Answer:
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