Classical Mechanics – Linear Momentum

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As shown in the picture, a ball of mass m, suspended on the end of a wire, is released from height h and collides elastically, when it is at its lowest point, with a block of mass 2m at rest on a frictionless surface. After the collision, the ball rises to a final height equal to

A. 1/9 h
B. 1/8 h
C. 1/3 h
D. 1/2 h
E. 2/3 h

(GR9677 #07)

Solution:

Conservation of momentum of the system:

mavmbvb = mava‘ mbvb‘ 

Given:
mm
m= 2m
vb = 0

mv+ 0 = mva‘ + 2mvb‘  
v = va‘ + 2vb‘    (Eq.1)

Conservation of kinetic energy of the system:

½ mava² + ½ mbvb² = ½ mava² + ½ mbvb²
mva² + 0 = mva² + 2mvb²
va² = va² + 2vb²  (Eq.2)

(Eq.1) → (Eq.2)
(va‘ + 2vb)² = va² + 2vb²
va² + 4vavb‘ + 4vb² = va² + 2vb²
4vavb‘  = 2vb² − 4vb²
2va = − vb‘  
vb‘  = −2va  (Eq.3)

(Eq.3) →  (Eq.1)
v = va‘ + 2vb‘ 
v = va‘ + 2(−2va)
v = − 3va‘  (Eq. 4)

For the pendulum, conservation of energy:

at the moment of the collision
U T → magh = ½ mava² → va = (2gh)½

after the collision
U’  T‘ → magh’ = ½ mava² → va‘ = (2gh’)½

Thus, (Eq. 4):
v = − 3va‘  
(2gh)½  = − 3(2gh’)½ 
[(2gh)½]² = [− 3(2gh’)½

h  = 9h’
h’ = ¹⁄₉h 

Answer: A

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