Starting with an initial speed of 5.00 m/s at a height of 0.335 m, a 1.65-kg bal

Question

Starting with an initial speed of 5.00 m/s at a height of 0.335 m, a 1.65-kg ball swings downward and strikes a 4.55-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.
1)   m/s

(b) Assuming that the collision is elastic, find the velocities (magnitude and direction) of both balls just after the collision.
1)  m/s (1.65 kg-ball)
2) m/s (4.55 kg-ball)

(c) How high does each ball swing after the collision, ignoring air resistance?
1) m (1.65 kg-ball)

2)   m (4.55 kg-ball)

Explanation / Answer

a.) Applying principle of energy conservation for 1.65 Kg ball,

0.5mvf^2 + mghf = 0.5mv0^2 + mgh0

hf =0

So vf = sqrt(v0^2 + 2gh0 ) = sqrt( 5^2 + (2*9.8*0.335) )= 5.62 m/s

b.) If we assume, the collision is elastic, then

vf1 = (m1-m2)*v01/(m1+m2)

vf2 = 2m1*v01/(m1+m2)

m1 = 1.65

m2 = 4.55

v01 = 5.62

So vf1 = - 2.63 m/s (1.65 Kg ball)

vf2 = 2.99 m/s (4.55 Kg ball)

c.) 0.5mvf^2 + mghf = 0.5mv0^2 + mgh0

h0 = 0

vf = 0

So hf = v0^2/2g

So for 1.65 Kg ball

hf = (2.63^2)/(2*9.8) = 0.353 m/s

So for 4.55 Kg ball

hf = (2.99^2)/(2*9.8) = 0.456 m/s

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