Question

A simple pendulum of length 1 m has a wooden bob of mass 1 kg. It is struck by a bullet of mass \(10^{-2}kg\) moving with a speed of \(2\times 10^2 \,ms^{-1}.\) The bullet gets embedded into the bob. The height to which the bob rises before swinging back is. (use g = 10 \(m/s^2\))

(1) 0.30 m 

(2) 0.20 m 

(3) 0.35 m 

(4) 0.40 m

Answer 1

(2) 0.20 m 

\(P_i =P_f \Rightarrow 10^{-2}\times 2 \times 10^2 =(1+.01)v\)

\(v = \frac {2}{1.01}= 1.98\,m/sec.\)

\(\frac {1}{2}\times m_T \times v^2 = m_Tgh\)

\(h=\frac {v^2}{2g}=\frac {1.98 \times 1.98}{20}\)

h = 0.20 m