Starting with an initial speed of 5.00 m/s at a height of 0.255 m, a 1.65-kg bal

Question

Starting with an initial speed of 5.00 m/s at a height of 0.255 m, a 1.65-kg ball swings downward and strikes a 4.70-kg ball that is at rest, as the drawing shows.

(a) Using the principle of conservation of mechanical energy, find the speed of the 1.65-kg ball just before impact.
----------------------- m/s

(b) Assuming that the collision is elastic, find the velocities (magnitude and direction) of both balls just after the collision.

--------------m/s (1.65 kg-ball)

--------------m/s (4.70 kg-ball)

(c) How high does each ball swing after the collision, ignoring air resistance?

------------------m (1.65 kg-ball)
------------------m (4.70 kg-ball)

Explanation / Answer

Part A:.
Speed of the 1.65-kg ball just before impact:
Applying conservation of energy, we have
m1gh +0.5m1vi2 = 0.5m1vf2
1.65x9.81x0.255 + 0.5x1.65x52 = 0.5x1.65xvf2
vf2 = 30.0031
vf = 5.48 m/s

Part B:
Since the collision is elastic, therefore

Velocity of ball 1 after collision: v1=  [(m1 - m2)/(m1 + m2)]x vf = (1.65-4.7)/(4.7+1.65)] x 5.48 = -2.63m/s
Negaitive sign shows that the ball 1 moves in the opposite direction.

Velocity of ball 2 : v2 = 2x m1/(m1 + m2) x vf = [2x1.65/(1.65+4.7)]x5.48 = 2.85m/s

Part C:

For ball 1:
0.5m1v12 = m1gh1
0.5x1.65x2.632 = 1.65 x 9.81xh1
h1 = 0.353 m

For ball 2:
0.5m2v22 = m2gh2
0.5x2.852 = 9.81xh2
h2 = 0.414 m

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