Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

MCQ (Single Correct Answer)

A particle of mass $$m$$ moving in the $$x$$ direction with speed $$2v$$ is hit by another particle of mass $$2m$$ moving in the $$y$$ direction with speed $$v.$$ If the collision is perfectly inelastic, the percentage loss in the energy during the collision is close to:

A

$$56\% $$

B

$$62\% $$

C

$$44\% $$

D

$$50\% $$

Applying conservation of linear momentum in x direction

$$2mv = \left( {2m + m} \right){V_x}$$

$$ \Rightarrow {V_x} = {2 \over 3}v$$

Applying conservation of linear momentum in y direction

$$2mv = \left( {2m + m} \right){V_y}$$

$$ \Rightarrow {V_y} = {2 \over 3}v$$

Final speed of 3m mass, $${V_f}$$ = $$\sqrt {V_x^2 + V_y^2} $$

= $$\sqrt {{{4{v^2}} \over 9} + {{4{v^2}} \over 9}} $$

= $$\sqrt {{{8{v^2}} \over 9}} $$

Initial kinetic energy

$${E_i} = {1 \over 2}m{\left( {2v} \right)^2} + {1 \over 2}\left( {2m} \right){\left( v \right)^2}$$

$$ = 2m{v^2} + m{v^2}$$

= $$3m{v^2}$$

Final kinetic energy,

$${E_f} = {1 \over 2}\left( {3m} \right)$$$${V_f^2}$$

$$ = {{3m} \over 2}\left[ {{{8{v^2}} \over 9}} \right]$$

$$ = {{4m{v^2}} \over 3}$$

Energy loss = $${E_i} - {E_f}$$

= $$3m{v^2} - {{4m{v^2}} \over 3}$$

= $${{5m{v^2}} \over 3}$$

Percentage loss in the energy during the collision

= $${{{E_i} - {E_f}} \over {{E_i}}}$$$$ \times 100$$

= $${{{{5m{v^2}} \over 3}} \over {3m{v^2}}}$$$$ \times 100$$

= $${5 \over 9} \times 100$$

$$ \simeq 56\% $$

2

MCQ (Single Correct Answer)

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.

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.

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.

3

MCQ (Single Correct Answer)

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

$$0.2$$ $$Ns$$

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

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

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

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

4

MCQ (Single Correct Answer)

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

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.

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

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.

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