Let the ball strike the block with a speed u1. Since the
initial speed (speed before collision) of the block =u2=0 for the
perfectly elastic collision, the speed of the block is given
as
v2=(2*m1*u1)/(m1+m2)
where u1 can be found by conserving energy of the ball between the
position A and B before collision of m1
⇒1/2*m1u1^2−mgl=0
u1=√(2.10*0.45)=3m/s
From this V2=(2*0.86*3)/(0.86+4.5)=0.963m/s. Is the speed of block
after collision
From conservation of momentum
m1u1+m2u2=m1v1+m2v2
0.86*3=(0.86*V1)+(4.5*0.963)
V1=-2.038m/s
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