A bullet of mass 0.20 kg and speed v passes completely through a pendulum bob of

Question

A bullet of mass 0.20 kg and speed v passes completely through a pendulum bob of mass 2 kg. The bullet emerges with half of its initial speed v/2. The pendulum bob is suspended by a stiff rod of length 0.15 m with a negligible mass.

What is the minimum value of v such that the bob will barely swing through a complete vertical circle:

A bullet of mass 0.20 kg and speed v passes completely through a pendulum bob of mass 2 kg. The bullet emerges with half of its initial speed v/2. The pendulum bob is suspended by a stiff rod of length 0.15 m with a negligible mass. What is the minimum value of v such that the bob will barely swing through a complete vertical circle:

Explanation / Answer

mass of bullet = m speed of bullet = v mass of the pendulum bob = M length of the rod = L emerging speed of the bullet = v/2 according to law of conservation of energy , Ki + Ui = Kf + Uf (1/2)MV2 + 0 = 0 + Mg(2L) (1/2)MV2 = 2MgL V2 = 4 gL speed V = 2 ?gL the speed of the bob after the bullet passes through it completely is V = 2 ?gL ............. (1) ....................................................... according to law of conservation of momentum , mv + M(0) = MV + m(v/2) mv = MV + m(v/2) ............... (2) substitute the eq (1) in eq (2) , we get mv = M(2 ?gL) + m(v/2) mv - m(v/2) = M(2 ?gL) m(v/2) = M(2 ?gL) v = (2M/m)(2 ?gL) = (4M/m)(?gL) minimum velocity of v such that the pendulum bob will barely swing through a complete vertical circle is v = (4M/m)(?gL)

= 4*2/0.2* sq(9.81*0.15)

=48.52215988597375
(1/2)MV2 + 0 = 0 + Mg(2L) (1/2)MV2 = 2MgL V2 = 4 gL speed V = 2 ?gL the speed of the bob after the bullet passes through it completely is V = 2 ?gL ............. (1) ....................................................... according to law of conservation of momentum , mv + M(0) = MV + m(v/2) mv = MV + m(v/2) ............... (2) substitute the eq (1) in eq (2) , we get mv = M(2 ?gL) + m(v/2) mv - m(v/2) = M(2 ?gL) m(v/2) = M(2 ?gL) v = (2M/m)(2 ?gL) = (4M/m)(?gL) minimum velocity of v such that the pendulum bob will barely swing through a complete vertical circle is v = (4M/m)(?gL)

= 4*2/0.2* sq(9.81*0.15)

=48.52215988597375

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