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potential energy of the pendulum at the top of its swings
E  ( M  m) g h 
P
Substitute for the height  Lh 1 (  cos )

E P  (M  m )  Lg 1 (  cos )

Here L is the distance from the pivot point to the centre of mass of the
pendulum-ball system. The potential energy is equal to the kinetic
energy of the pendulum immediately after collision

(m  M ) v 2 m M
E K  .
2
The momentum of the pendulum after collision is

p m M  ( m  M v  ) m M ,

which we substitute into the previous equation to get

2
P m M
E K 
( 2 m  M )
Solving this equation for the pendulum momentum gives

p m M  2 ( m  M ) E .
K
This momentum is equal to the momentum of the ball before the
collision

p  m v  m .
m

Setting these two equations equal to each other and replacing E with
K
known potential energy gives us

p m M  2 ( m  M ) E .
K

Solve this for the ball speed and simplify to get

M  m
v m  2 g  L 1 (  cos )
m

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