Get Answers to all your Questions

header-bg qa

Q6 sir A simple pendulum of length 1m has a wooden bob of mass 1kg. It is struck by a bullet of mass 10^-2 kg moving with a speed of 2oo m/s. The bullet gets embedded into the bob. Obtain the height to which the bob rises before swinging back. Take g=10 m/s^2

Answers (1)

best_answer

@ gulamjilani shaikh

Given m=10^{-2}kg

         M=1 kg

         u=2*10^2 ms^{-1}

Apply Momentum conservatiom 

We get mu=(m+M)V \Rightarrow V= \frac{10^{-2}*2*10^2}{1.01}\Rightarrow 1.98 ms^{-1}

Now after bullet is embedded in the wooden block

Apply enegy conversation

change in K.E= change in P.E

\frac{1}{2}(m+M)V^2=(m+M)gh \Rightarrow V^2=2gh \Rightarrow h= V^2/2g \Rightarrow \frac{(1.98)^2 }{2*9.8}\Rightarrow h=0.20 m

Posted by

avinash.dongre

View full answer

NEET 2024 Most scoring concepts

    Just Study 32% of the NEET syllabus and Score up to 100% marks