Question

A 39 g bullet traveling at 186 m/s buries itself in a 5.2 kg pendulum hanging on a 2.8 m length of string, which makes the pendulum swing upward in an arc. Determine the horizontal component (in meter) of the displacement of the pendulum

Answer #1

By momentum conservation,

mu= (M+m)v where m and M are mass of bullet and pendulum, u, v are initial and final velocities of bullet.

0.039*186 = (5.2+0.039)v

v = 0.039*186/ (5.2+0.039) = 1.3846 m/s

Now by energy conservation,

maximum potential energy = initial KE

(m+M)g*L*(1- cos theta) = 0.5*(m+M)v^2 where theta is maximum angle with vertical,

g*L*(1- cos theta) = 0.5*v^2

cos theta = 1 - 0.5*v^2 /gL = 1-0.5*1.3846^2/(9.8*2.8) = 0.965

theta = arc 0.965 = 15.2 degree

horizontal component of displacement = L sin theta

= 2.8*sin 15.2 degree

**= 0.734 m answer**

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