A block of mass m=10 kg rests… - JEE Main 2015 Physics PYQ

JEE Main 2015 — Physics Mechanics

2015mcqhard

A block of mass $m=10 \mathrm{kg}$ 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 $\frac{v}{10}$ after being dropped from height $H$ , then neglecting energy losses and taking $g=10 m{s}^{-2}$ , the value of $H$ is close to

Official previous-year question

Held on 10 Apr 2015 · Verified 6 Jul 2026.

Options

A
$0.2\mathrm{km}$ .
B
$0.5\mathrm{km}$ .
C
$0.4\mathrm{km}$ .
D
None of these.

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Solution

JEE Main 2015 Physics, Work Power Energy — solution figure

By the conservation of linear momentum,

${m}_{1}v=(m+{m}_{1}){v}_{1}$

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

After collision by work energy theorem, we have

${W}_{\mathrm{friction}}=\Delta k$

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

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

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

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

$\Rightarrow 2=\frac{1}{400}\frac{{v}^{2}\times 400}{{(201)}^{2}}$

${v}^{2}={(201)}^{2} 2$

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

For a freely falling body,

$v'=\sqrt{2gH}$

$\frac{v}{10}=\sqrt{2gH}$

$\frac{\sqrt{2}\times 201}{10}=\sqrt{2}\times \sqrt{gH}$

$\frac{{(201)}^{2}}{100}=10H$

$H=\frac{{(201)}^{2}}{1000}=40 m$

$=0.04 \mathrm{km}$

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Solution

JEE Main 2015 Physics, Work Power Energy — solution figure

By the conservation of linear momentum,

${m}_{1}v=(m+{m}_{1}){v}_{1}$

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

After collision by work energy theorem, we have

${W}_{\mathrm{friction}}=\Delta k$

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

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

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

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

$\Rightarrow 2=\frac{1}{400}\frac{{v}^{2}\times 400}{{(201)}^{2}}$

${v}^{2}={(201)}^{2} 2$

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

For a freely falling body,

$v'=\sqrt{2gH}$

$\frac{v}{10}=\sqrt{2gH}$

$\frac{\sqrt{2}\times 201}{10}=\sqrt{2}\times \sqrt{gH}$

$\frac{{(201)}^{2}}{100}=10H$

$H=\frac{{(201)}^{2}}{1000}=40 m$

$=0.04 \mathrm{km}$

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Solution

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Options

Did you get this right?

Solution

Your note

JEE Main Mechanics in other years

More JEE Main Mechanics questions

Other Physics topics for JEE Main

Electromagnetism

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Mechanics