Question

An apple with mass 0.4 kg is hanging at rest from the lower end of a light vertical rope. A dart of mass 0.1 kg is shot vertically upward, strikes the bottom of the apple, and remains embedded in it.

a) If the speed of the dart is 5 m/s just before it strikes the apple, what is the speed of the apple just after the collision?

b) How much energy was dissipated during the strike?

c) How high does the apple move upward because of its collision with the dart?

Answer 1

m - Mass of the dart, m = 0.1 kg
M -Mass of the apple, M = 0.4 kg
v - Velocity of the dart, v = 5 m/s
Consider V as the velocity of the dart+apple system.
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a)
Initial momentum = mv
Final momentum = (m +M)V
Using conservation of momentum,
mv = (m+M)V
V = mv / (m+M)
= (0.1 * 5) / (0.1 + 0.4)
= 1 m/s
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b)
Initial kinetic energy, KEi = 1/2 mv²
= 0.5 * 0.1 * 5²
= 1.25 J

Final kinetic energy, KEf = 1/2 (m+M)V²
= 0.5 * (0.4 + 0.1) * 1²
= 0.25 J

Energy dissipated = KEi - KEf
= 1.25 - 0.25
= 1 J
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b)
Initial velocity of the system, V = 1 m/s
Final velocity of the system, Vf = 0
Acceleration, a = - 9.8 m/s²
Consider h as the upward distance moved

Using the formula, Vf² - V² = 2ah,
h = (Vf² - V²) / 2a
= (0² - 1²) / (2 * 9.8)
= 0.05 m