### JEE Mains Previous Years Questions with Solutions

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1

This question has statement ${\rm I}$ and statement ${\rm I}$${\rm I}. Of the four choices given after the statements, choose the one that best describes the two statements. Statement - {\rm I}: A point particle of mass m moving with speed \upsilon collides with stationary point particle of mass M. If the maximum energy loss possible is given as f\left( {{1 \over 2}m{v^2}} \right), then f = \left( {{m \over {M + m}}} \right). Statement - {\rm II}: Maximum energy loss occurs when the particles get stuck together as a result of the collision. A Statement - {\rm I} is true, Statement - {\rm II} is true; Statement - {\rm II} is the correct explanation of Statement - {\rm I}. B Statement - {\rm I} is true, Statement - {\rm II} is true; Statement - {\rm II} is not the correct explanation of Statement - {\rm I}. C Statement - {\rm I} is true, Statement - {\rm II} is false D Statement - {\rm I} is false, Statement - {\rm II} true. ## Explanation Initial energy = {{{P^2}} \over {2m}}, where P is the momentum and m is the mass of the moving particle. Loss of energy is maximum when collision is inelastic means when the particles get stuck together as a result of the collision. So after collision energy = {{{P^2}} \over {2\left( {m + M} \right)}} \therefore Maximum energy loss = {{{P^2}} \over {2m}} - {{{P^2}} \over {2\left( {m + M} \right)}}. \left[ \right. As \left. {K.E. = {{{P^2}} \over {2m}} = {1 \over 2}m{v^2}\,\,} \right] = {{{P^2}} \over {2m}}\left[ {{M \over {\left( {m + M} \right)}}} \right] = {1 \over 2}m{v^2}\left\{ {{M \over {m + M}}} \right\} \therefore f = \left( {{M \over {m + M}}} \right) So statement I is wrong. Statement {\rm I}{\rm I} says "Maximum energy loss occurs when the particles get stuck together as a result of the collision." This is a case of perfectly inelastic collision. Hence statement {\rm I}$${\rm I}$ is correct.

2

### AIEEE 2010

The figure shows the position$-$time $(x-t)$ graph of one-dimensional motion of body of mass $0.4$ $kg.$ The magnitude of each impulse is
A
$0.4$ $Ns$
B
$0.8$ $Ns$
C
$1.6$ $Ns$
D
$0.2$ $Ns$

## Explanation

Motion is uniform as graph is straight line. In x -t graph, velocity is ${x \over t}$ and the slope of the graph tells the direction of the velocity. Here because of impulse the direction of velocity or slope of the graph changes.

During each collision

Initial velocity ${v_1} = {2 \over 2} = 1\,m{s^{ - 1}}$

Final velocity ${v_2} = - {2 \over 2} = - 1\,m{s^{ - 1}}$

Impulse $=$ Change in momentum

$= m\left| {{v_2} - {v_1}} \right| = 0.4 \times 2 = 0.8\,\,Ns$
3

### 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.
4

### 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

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

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

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