### JEE Mains Previous Years Questions with Solutions

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1

### AIEEE 2010

Statement - 1 : Two particles moving in the same direction do not lose all their energy in a completely inelastic collision.

Statement - 2 : Principle of conservation of momentum holds true for all kinds of collisions.
A
Statement - 1 is true, Statement - 2 is true; Statement - 2 is the correct explanation of Statement - 1
B
Statement - 1 is true, Statement - 2 is true; Statement - 2 is not the correct explanation of Statement - 1
C
Statement - 1 is false, Statement - 2 is true
D
Statement - 1 is true, Statement - 2 is false

## Explanation

In completely inelastic collision,

${m_1}{v_1} + {m_2}{v_2} = {m_1}v + {m_2}v$

after collision both particle have common velocity $v$, so all energy is not lost

$\therefore$ Statement - $1$ is true

The principle of conservation of momentum applicable for all kinds of collisions.

So statement - $2$ is also true.

Statement - $2$ explains statement - $1$ correctly because applying the principle of conservation of momentum, we can get the common velocity and hence the kinetic energy of the combined body.
2

### AIEEE 2006

A bomb of mass $16kg$ at rest explodes into two pieces of masses $4$ $kg$ and $12$ $kg.$ The velocity of the $12$ $kg$ mass is $4\,\,m{s^{ - 1}}.$ The kinetic energy of the other mass is
A
$144$ $J$
B
$288$ $J$
C
$192$ $J$
D
$96$ $J$

## Explanation

Here linear momentum is conserved as no external force is acting on the bomb.

Let the velocity and mass of $4$ $kg$ piece be ${v_1}$ and ${m_1}$ and that of $12$ $kg$ piece be ${v_2}$ and ${m_2}$.

Applying conservation of linear momentum

$0 = {m_2}{v_2} - {m_1}{v_1}$

$\Rightarrow {v_1} = {{12 \times 14} \over 4} = 12\,m{s^{ - 1}}$

$\therefore$ $K.E{_1} = {1 \over 2}{m_1}v_1^2 = {1 \over 2} \times 4 \times 144 = 288\,J$
3

### AIEEE 2005

A mass $'m'$ moves with a velocity $'v'$ and collides inelastically with another identical mass. After collision the ${1^{st}}$ mass moves with velocity ${v \over {\sqrt 3 }}$ in a direction perpendicular to the initial direction of motion. Find the speed of the ${2^{nd}}$ mass after collision.
A
${\sqrt 3 v}$
B
$v$
C
${v \over {\sqrt 3 }}$
D
${2 \over {\sqrt 3 }}v$

## Explanation

Assume speed of second mass = ${v_1}$

As momentum is conserved,

In $x$-direction, $mv = m{v_1}\cos \theta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,....(1)$

In $y$-direction, ${{mv} \over {\sqrt 3 }} = m{v_1}\,\sin \theta \,\,\,\,\,\,\,\,\,\,...(2)$

Squaring and adding eqns.$(1)$ and $(2)$

${\left( {m{v_1}\cos \theta } \right)^2} + {\left( {m{v_1}\sin \theta } \right)^2}$$= {\left( {mv} \right)^2} + {\left( {{{mv} \over {\sqrt 3 }}} \right)^2}$

$\Rightarrow$ $v_1^2 = {v^2} + {{{v^2}} \over {\sqrt 3 }}$

$\Rightarrow {v_1} = {2 \over {\sqrt 3 }}v$

4

### AIEEE 2004

A machine gun fires a bullet of mass $40$ $g$ with a velocity $1200m{s^{ - 1}}.$ The man holding it can exert a maximum force of $144$ $N$ on the gun. How many bullets can he fire per second at the most?
A
Two
B
Four
C
One
D
Three

## Explanation

Assume the man can fire $n$ bullets in one second.

$\therefore$ change in momentum per second $= n \times mv = F$

[ $m=$ mass of bullet, $v=$ velocity, $F$ = force) ]

$\therefore$ $n = {F \over {mv}} = {{144 \times 1000} \over {40 \times 1200}} = 3$