Momentum, Impulse and Center of Mass Problem 3: A bullet of mass .05 kg, enters

Question

Momentum, Impulse and Center of Mass

Problem 3:

A bullet of mass .05 kg, enters and passes through a grapefruit of mass 0.25 kg hanging from a length of string forming a pendulum with a length of 0.5 meters. The bullet emerges from the grapefruit at a velocity equal to 1/3 of it’s initial velocity before entering the delicious piece of fruit. At what minimum initial velocity for the bullet does the grapefruit make it to the highest point in the pendulum’s possible rotation 1 meter above the starting point?

Explanation / Answer

Apply the momentum conservation to the given situation in the problem.

0.05 * v = 0.05 * v/3 + 0.25* x

After solving, we get

x = 2v/15 m/s

Apply the work-energy thorem, we get

0.5mx^2 = mgh

x = sqrt [2gh] = sqrt [2*9.8*1] = 4.42 m/s

Thus, the requred initial velocity is,

2v/15 = 4.42 m/s

v = 33.2 m/s

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