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Part A) only A light inextensible string passes over a smooth fixed pulley. One end of...

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roshan-rox | (Level 1) Valedictorian

Posted September 9, 2013 at 11:11 AM via web

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Part A) only

A light inextensible string passes over a smooth fixed pulley. One end of the string carries a bucket of mass M and the other end carries a counterpoise of equal mass. A small ball ball of mass m is dropped vertically so as to strike the horizontal bottom of the bucket with velocity u. If e is the coefficient of restitution,

A)

show that the bucket begins to move with velocity `(m(1+e)u)/(2M+m)`

B)

find the impulse in the string.

C)

Find , also, the time between the first and second impacts of the ball and the bucket.

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted September 9, 2013 at 1:27 PM (Answer #1)

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A light inextensible string passes over a smooth fixed pulley. One end of the string carries a bucket of mass M and the other end carries a counterpoise of equal mass. A small ball of mass m is dropped vertically so as to strike the horizontal bottom of the bucket with velocity u; e is the coefficient of restitution.

The coefficient of restitution is the ratio of relative speed after the collision and the relative speed before collision.

Let the velocity with which the bucket moves be V. u' is the velocity of the ball after the collision.

`(V - u')/u = e`

V = u' + u*e

=> u' = V - u*e

The initial momentum of the system is u*m and the final momentum is m*u' + 2*M*V

`u*m = m*u' + 2*M*V`

=> `u*m = m*(V - u*e) + 2*M*V`

=> `u*m = m*V - m*u*e + 2*M*V`

=> `u*m + m*u*e = V(m + 2*M)`

=> `u*m*(1 + e) = V(m + 2*M)`

=> `V = (u*m*(1 + e))/(m + 2*M)`

The velocity of the bucket after collision is `(u*m*(1 + e))/(m + 2*M)`

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