(2) A bullet of mass m and speed v passes completely through a pendulum bob of m

Question

(2) A bullet of mass m and speed v passes completely through a pendulum bob of mass M. The bullet emerges with a speed of v{2. The pendulum bob is suspended by a stiff rod of length l and negligible mass. (a) What is the minimum value of v such that the pendulum bob will barely swing through a complete vertical circle? In other words, enough to cause the pendulum bob to reach the top of its circular path. (b) Assuming the rod has length l “ 30cm and the masses are M “ 3kg and m “ 500g, what is the speed v?

Explanation / Answer

(2) Applying momentum conservation,

m v + 0 = M v' + (m v / 2)

v' = m v / 2 M

Now applying energy conservation,

M v'^2 /2 + 0 = 0 + M g (2l)

M (m^2 v^2 ) / (8 M^2) = 2 M g l

m^2 v^2 = 16 M^2 g l

v = 4 M sqrt(gl) / m

(b) putting values,

v = (4 x 3 / 0.500) sqrt(9.81 x 0.30)

v =41.2 m/s

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.