A 250 g steel ball and a 500 g steel ball each hang from 3.5-m-long strings. At

Question

A 250 g steel ball and a 500 g steel ball each hang from 3.5-m-long strings. At rest, the balls hang side by side, barely touching. The 250 g ball is pulled to the left until the angle between its string and vertical is 16 . The 500 g ball is pulled to a 16 angle on the right. The balls are released so as to collide at the very bottom of their swings. Part A: To what angle does 250 g ball rebound? Express your answers using two significant figures. Part B: To what angle does 500 g ball rebound? Express your answers using two significant figures.

Explanation / Answer

In order to find the speed of the balls before collision, law of conservation of energy is applied

i.e., m g h = 1/2 m v² so that v = (2 g h)

By trigonometric relation,
h = (1 - cos(16º)) L

So h = (1- 0.96) * 3.5
where L = 3.5 m

Hence, h = 0.04*3.5 = 0.14 m

Taking g = 9.8 m/s²

v= sqrt (2*9.8*0.14)

v = 1.656 m/s

Before collision:
m1 = 0.25 kg
m2 = 0.5 kg
v = 1.656 m/s

Conservation of energy and momentum gives...
1/2 m1 v² + 1/2 m2 v² = 1/2 m1 v1² + 1/2 m2 v2² (since collision is elastic)
m1 v - m2 v = m2 v2 + m1 v1
where v1 and v2 are defined to be to the right

By simplifying the above equations, we get

v1 = v (m1 - 3 m2)/(m1+m2) = -1.666 m/s
v2 = v (3m1 - m2)/(m1+m2) =   0.333 m/s

You can then use
h = v² / (2 g) = (1 - cos()) L
to get
= arccos( (-v²+2 L g)/(2 L g)
to get

­1 = arccos( (-v1²+2 L g)/(2 L g)
­2 = arccos( (-v2²+2 L g)/(2 L g)

v1 = v (m1 - 3 m2)/(m1+m2) = -1.666 m/s
v2 = v (3m1 - m2)/(m1+m2) =   0.333 m/s

1 = 0.95954 = 16.35º
2 = 0.99838 = 3.25º

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