As shown in the figure below, a bullet of mass m = 0.1kg and speed v passes comp

Question

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 t = 50 cm. and negligible mass. 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. What is the minimum value of v such that the pendulum bcto will barely swing through a complete vertical circle? Now suppose the original speed of the bullet was 350m/s. what would be the tension the rod, just as it reaches the highest point?

Explanation / Answer

a)Using the conservation of momentum, mv = Mvb + mv/2
where vp = velocity of bob
0.1x200= 25xvb + 0.1x200/2
=>10=25vb
=>vb=0.4m/s

b)Conservation of energy for the pendulum bob gives:

Ki + Ui = Kf + Uf
(1/2)Mv^2 + 0 = Mg(2l) + 0 <- Pendulum reaches the height of twice its radius at the top

v^2 = 4gl
v = Sqrt(4gl)

mv1 + Mv2 = mv1 +Mv2
mv1 + 0 = m(v/2) + M(Sqrt(4gl))
v = v/2 + M*Sqrt(4gl)/m
v/2 = M*Sqrt(4gl)/m
v = 2*MSqrt(4gl)/m = 4Sqrt(gl)*M/m
=4Sqrt(9.8x0.5)*25/0.1
=2213.59 m/s

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