In this question the speed of the bullet AFTER it passes through M is always 100

Question

In this question the speed of the bullet AFTER it passes through M is always 100m/s As shown in the figure below, a bullet of mass m = 0.1kg and speed v passes completely through a pendulum bob of mass M = 2kg. The bullet emerges with a speed of 100m/s. The pendulum bob is suspended by a stiff rod of length l = 50 cm. and negligible mass. a) assuming that the velocity of the bullet was 200 m/s, what is the velocity of the bob just after the bullet passes through it. b) What is the minimum value of v such that the pendulum bob will barely swing through a complete vertical circle? c) Now suppose the original speed of the bullet was 350m/s, what would be the tension in the rod, just as it reaches the highest point?

Explanation / Answer

a) m=0.1kg, M = 2kg, Vmi = 200m/s, VMi = 0,

Conserving the momentum before and after the passing of bullet

m×Vmi + M×VMi = m×Vmf + M×VMf

=> 0.1×200 + 2×0 = 0.1×200/2 + 2×VMf

=> VMf = (20-10)/2 = 2.5 m/s

b) For complete rotation, the minimum kinetic energy at the bottom should be equal to the potential energy at the top. Consider the minimum velocity at the bottom to VM. Conserving energy at the bottom and top

1/2 M VM2 = Mg(2l) => VM = (4 g l) = 4.43 m/s

Considering the bob is at rest initially, to determine the velocity of the bullet, we can conserve the momentum

0.1 × Vm + 2×0 = 0.1×Vm/2 + 2×VM

=>Vm = 177.2 m/s

c) If the velocity of the bullet is 350m/s, the velocity of the bob will be

m×Vmi + M×VMi = m×Vmf + M×VMf

=> 0.1×350 + 2×0 = 0.1×350/2 + 2×VMf

=> VMf = (35-17.5)/2 = 8.75 m/s

Calculating the velocity of bob at the top will be

0.5 × 2 × 8.752 = 2 × 9.8 × (2×0.5) + 0.5 × 2× V2

=>V = 7.55 m/s

The tension in the string will be T = MV2/R - Mg = 2×7.552/0.5 - 2 × 9.8 = 208.41 N (upward)

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