A 10.0-g bullet moving with an initial speed of V_i fired into and passes throug

Question

A 10.0-g bullet moving with an initial speed of V_i fired into and passes through a 1.60-kg pendulum bob as shown in the figure below. After impact, the pendulum bob rises up to a maximum height of h = 20 cm above its original position, and the speed of the bullet reduces to one-third of its initial speed. Let V_bob denote the speed of the pendulum bob immediately after impact. Find the speed of the pendulum bob immediately after impact. Apply conservation of momentum for the collision process to obtain an equation involving V_i and V_bob. Find the initial speed of the bullet V_i.

Explanation / Answer

a)We apply law of conservation of energy after impact,

MEi = MEf

KEibob = PEfbob

½*mbob*vbob2 = mbob*g*h

vbob =sqrt(2gh)

vbob = sqrt(2*9.8*0.20)

vbob = 1.98m/s

b)

We apply law of conservation of momentum as below,

Pi = Pf

mbullet*vi =mbob*vbob + mbullet*vf

mbullet*vi =mbob*vbob + mbullet*vi/3

vbob = [mbullet*vi - mbullet*vi/3]/mbob

vbob = [2/3*mbullet*vi ]/mbob

vi = [3/2*mbob*vbob]/mbullet

c)

Plugging values in above equation,

vi = (3/2*1.60*1.98)/0.010

vi = 475.2 m/s

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