A block with mass m1 = 0. 500 kg is released from rest on a frictionless track a

Question

A block with mass m1 = 0. 500 kg is released from rest on a frictionless track at a distance h1 = 2. 50 m above the top of the table. It then collides elastically with a object having a mass m2 = 1. 00 kg that is initially at rest on the table as shown in the figure above, (a) Determine the velocities of the two objects just after the collision, (b) How high up the track does the 0. 500 kg object travel after the collision? (c) How far from the bottom of the table does the 1. 00 kg object land, given that the height of the table is h2 = 2. 00 m? (d) How far away from the bottom of the table does the 0. 500 kg object eventually land?

Explanation / Answer

initial energy= 1/2m1g( h1+h2) now, velocity when it reaches the end of table= m1gh1 = 1/2 m1 u1^2 u1= sqrt( 2g h1) = 7m/s it collides elestically suppose its velocity at the end of ramp is u1 and velocity of mass2 = v2 m1u1+ m2u2 = m1v1+m2v2 m1u1= m2v2 + m1 v1 u1 = 2v2 + v1 for an elastic collision, coefficient of restitution is e= 1 ua - ub / vb- va= 1 u1 - 0 / v2- v1 = 1 u1= v2- v1 now, solve for v2 , v1 , u1 v1= -1/2 v2 7= v2 - v1 = v2 + v2/2 v2= 4.66 m/s v1= -2.33 m/s conserving kinetic energy , 1/2 m1 u1^2 = 1/2 m1 v1^2 + 1/2 m2 v2^2 b.now, height the body goes up the ramp mgh = 1/2 m1 v1^2 c. distance landed= v X t v= v2 t= sqrt( 2h2/ g) D= v2 sqrt( 2h2/g) d. the distance landed by m1 , the velocity when the m1 reaches end of table= v1 D= v1 X sqrt( 2h2/g)

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