Question

A bullet of mass $m$ moving with a horizontal velocity $v$ strikes a stationary block of mass $M$ suspended by a string of length $L$. The bullet gets embedded in the block. What is the maximum angle made by the string after impact ?

• A. $\theta =co{s}^{-1}\left(\frac{M}{m}\right)$
• B. $\theta =co{s}^{-1}\left(\frac{m}{M}\right)$
• C. $\theta =co{s}^{-1}\left(\frac{mV}{M+m}\right)$
• D. $\theta =co{s}^{-1}(1-\frac{1}{2gL}{\left(\frac{mv}{m+M}\right)}^2)$

Answer 1

The correct option is D $\theta =co{s}^{-1}(1-\frac{1}{2gL}{\left(\frac{mv}{m+M}\right)}^2)$

If $V$ is the velocity of the block- bullet system

just after collision, then by conservation of linear

momentum $mv=(M+m)V$

or $v=\left[\frac{M+m}{m}\right]v.......\left(1\right)$

So the KE of the block- bullet system just after

collision is $\frac{1}{2}(m+M)V^2$ (which is less than

$\frac{1}{2}m{v}^2$ as collision is inelastic).

Now due to this remaining KE if the system rises upto a height $h$, conservation of ME for this part of problem yields,

$\frac{1}{2}(m+M)V^2=(m+M)gh$,

i.e $V=\sqrt{2gh}....\left(2\right)$

Substituting this valueof $V$ from Eqn. (2) in(1),

we get.

$v=\left(\frac{M+m}{m}\right)\sqrt{2gh}.........\left(3\right)$

Since $h=L-Lcos\theta$

$cos\theta =1-\frac{h}{L}$

$cos\theta =1-\frac{1}{2gL}{\left(\frac{mv}{M+m}\right)}^2$