pivots 5) [10 pts] A slightly-modified Newton\'s cradle: two balls are suspended

Question

pivots 5) [10 pts] A slightly-modified Newton's cradle: two balls are suspended from light strings as shown in the figure. The lighter ball (mass 1.50 kg), on the left, is drawn up to a height of 0.300 m above the bottom of its swing and released. At the bottom it hits a more massive ball (mass 3.50 kg) which is initially at rest. Assume the collision is elastic and the strings holding the balls are the same length. a. Find the velocities (speed and direction) of both balls just after the collision. b. Find how high each ball swings after the collision. Phys 131 Final, Dec 19th, 2011 Page 5 21

Explanation / Answer

5. (a) speed of lighter ball at bottom,

v0 = sqrt(2 g h ) { from energy conservation}

v0 = sqrt(2 x 9.8 x 0.300) = 2.42 m/s

for elastic collision,

velocity of approach = velocity of separation

2.42 = v + V

Applying momentum conservation,

1.50 x 2.42 = - 1.50 v + 3.50 V

- 1.50 v + 3.50 V = 3.63

Solving v = 0.968 m/s .....Speed of lighter ball to the left

V = 1.452 m/s .........Speed of heavy ball, to the right

(B)

h = v^2 / 2 g (from energy conservation)

h_lighter = 0.048 m

h_heavry = 0.108 m

(C) T = 2 pi sqrt[ L / g ]

both balls are tied with same length string so time period will be same.

hence will come back at bottom at same time.

Ans: at the bottom

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