Starting with an initial speed of 5.00 m/s at a height of 0.335 m, a 1.45-kg bal

Question

Starting with an initial speed of 5.00 m/s at a height of 0.335 m, a 1.45-kg ball swings downward and strikes a 4.40-kg ball that is at rest, as the drawing shows.

(a) Using the principle of conservation of mechanical energy, find the speed of the 1.45-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 (1.45 kg-ball)
m/s (4.40 kg-ball)

(c) How high does each ball swing after the collision, ignoring air resistance?
m (1.45 kg-ball)
m (4.40 kg-ball)

Explanation / Answer

(a) according to law of conservation of energy,

the mecanical energy of 1.45 kg block:

1/2mv^2 + mgh = 1/2 mV1^2

1/2*25 + 9.8*0.335 = 1/2 V1^2

V1 = 15.78 m/s

he speed of the 1.45-kg ball just before impact is 15.78 m/s.

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(b) By applying law of conservation of momentum

after the collision, let V2 be the speed of the 1.45kg ball and V3 be the speed of the 4.40 kg ball

1.45*(V1) + 0 = 1.45*(V2) + 4.40*(V3)

22.885 = 1.45*V2 + 4.40 *V3-------------- (1)

the spapration of the speeds before collision equals to the after collision

V1 = V3-V2

22.885 = V3 - V2 --------------------- (2)

solving (1) & (2) we get

V2 = -13.29 m/s

V3 = 9.584 m/s

the speed of the 1.45 kg block is 13.19 m/s (opposite direction to initial ) and

the speed of the 4.40 kg ball is 9.584 kg.

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(c) the each ball rises height is 'h' is given by

       h = v^2/2g

    for 1.45 kg ball , h1 = (13.29)^2/2*9.8

                                = 9.01 m

for 4.40 kg ball , h2 = (9.584)^2/19.6

                                  = 4.68 m

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