Starting with an initial speed of 3.89 m/s at a height of 0.159 m, a 1.45-kg bal

Question

Starting with an initial speed of 3.89 m/s at a height of 0.159 m, a 1.45-kg ball swings downward and strikes a 4.78-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. (b) Assuming that the collision is elastic, find the velocity (magnitude and direction) of the 1.45-kg ball just after the collision. (c) Assuming that the collision is elastic, find the velocity (magnitude and direction) of the 4.78-kg ball just after the collision. (d) How high does the 1.45-kg ball swing after the collision, ignoring air resistance? (e) How high does the 4.78-kg ball swing after the collision, ignoring air resistance?

Explanation / Answer

Given. M1=1.45 kg. M2=4.78 kg. h= 0.159 m

a). Use conservation of mechanical energy to calculate the speed of ball

Potential Energy = Kinetic Energy

M1gh = 1/2 M1V1

V1=2gh

= 2×9.8×0.159

= 1.77 m/s

Since Collision is elastic we know that

b) velocity of the 1.45 kg ball just after the collision

Vs1 = (M1-M2)/(M1+M2)×V1

= (1.45 - 4.78)/(1.45 - 4.78)×1.77

= - 0.95 m/ss

c). Velocity of the ball 4.78 just after the collision

Vs2 = 2 M1/(M1 + M2)× V1

= 2 × 1.45/ (1.45 + 4.78) × 1.77

= 0.82 m/s

We are using conservation of energy to find h1&h2

d). Height of ball 1.45 kg swing after the collision ignoring the air resistance

   Mgh1= 1/2 MV2

  h1 = Vs12/2g

= (0.95)2/ 2×9.8

= 0.046 m

h2 for the ball of mass 4.78 kg

h2 = (Vs2)2/2g

= (0.82)2/ 2×9.8

= 0.034m

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