for a one dimensional ballistic pendulum. a block of M is hung from the ceiling

Question

for a one dimensional ballistic pendulum. a block of M is hung from the ceiling by a massless rope. a ball m is thrown at the block at (v) the instant before it hits the block.

the coefficient of restitution for the collision is e= 0.75, find the velocities of the ball and the block the instant after the collision

after the collision what is the maximum height the center of the block will rise to before momentarily stopping?

i am having trouble incorporating the coefficient of restitution into my formula
if you could help me with the complete set of formulas that would be great.
thanks

Explanation / Answer

Given Mass of the block is M Mass of the ball is m Initial velocity of the ball is v And the initial velocity of block is vi = 0 m/s (at rest) Let the final velocities of the ball be v' and block be v'' The coefficient of restitution for the collision is e = 0.75 By law of conservation of momentum m v = m v' + M v'' ----1 Since e = v'' - v' / v - vi        0.75 = v'' - v' / v - 0     v'' - v' = 0.75 v       v'' = 0.75 v + v'    ----2 Substitute v'' value in eq 1 m v = m v' + M (0.75 v + v') mv = m v' + 0.75 Mv + Mv' v'(M +m) = (m - 0.75 M) v v' = (m - 0.75 M) v / (M +m)    is the final velocity of the ball substitute v' value in eq 2, we get v'' = 0.75 v +  (m - 0.75 M) v / (M +m) v'' = 1.75 v m / (M + m) is the final velocity of the block ----------------------------------------------------------------- By law of conservationof energy for block ,    1/2 M v''^2 = M gh   [1.75 v m / (M + m)]^2 / 2 g= h     h = 0.15 ( m v / M +m)^2 is the maximum height the block rised

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