Question 127 of 183

A block of mass $m=10kg$ rests on a horizontal table. The coefficient of friction between the block and the table is $0.05$. When hit by a bullet of mass $50g$ moving with speed $v$, that gets embedded in it, the block moves and comes to stop after moving a distance of $2m$ on the table. If a freely falling object were to acquire speed $\dfrac{v}{10}$ after being dropped from height $H$, then neglecting energy losses and taking $g=10m{s}^{-2}$, the value of $H$ is close to

A$0.2km$.B$0.5km$.C$0.4km$.DNone of these.

Show answer and solution▾

Answer: (D) None of these.

By the conservation of linear momentum,

${m}_{1}v=\left(m+{m}_{1}\right){v}_{1}$

$\Rightarrow {v}_{1}=\dfrac{{m}_{1}v}{m+{m}_{1}}...(1)$

After collision by work energy theorem, we have

${W}_{friction}=∆k$

$\Rightarrow -\mu \left(m+{m}_{1}\right)gx=-\dfrac{1}{2}\left(m+{m}_{1}\right){v}_{1}^{2}$

$\Rightarrow \mu gx=\dfrac{1}{2}{v}_{1}^{2}$

$\Rightarrow 0.05\times 10\times 2=\dfrac{1}{2}{\left(\dfrac{{m}_{1}v}{m+{m}_{1}}\right)}^{2}$

$\Rightarrow \dfrac{5}{100}\times 10\times 4={\left(\dfrac{50}{1000}\right)}^{2}\times \dfrac{{v}^{2}}{{\left(10+\dfrac{50}{1000}\right)}^{2}}$

$\Rightarrow 2=\dfrac{1}{400}\dfrac{{v}^{2}\times 400}{{\left(201\right)}^{2}}$
${v}^{2}={\left(201\right)}^{2}2$

$v=\sqrt{2}\times 201$

For a freely falling body,

$v'=\sqrt{2gH}$
$\dfrac{v}{10}=\sqrt{2gH}$
$\dfrac{\sqrt{2}\times 201}{10}=\sqrt{2}\times \sqrt{gH}$

$\dfrac{{\left(201\right)}^{2}}{100}=10H$
$H=\dfrac{{\left(201\right)}^{2}}{1000}=40m$

$=0.04km$