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### JEE Main 2019 (Online) 9th January Morning Slot

Three blocks A, B and C are lying on a smooth horizontal surface, as shown in the figure. A and B have equal masses, m while C has mass M, Block A is given an initial speed $\upsilon$ towards B due to which it collides with B perfectly inelastically. The combined mass collides with C, also perfectly inelastically ${5 \over 6}$th of the initial kinetic energy is lost in whole process. What is value of M/m ?

A
5
B
2
C
4
D
3

## Explanation

As in elastic or in elastic clollision, momentum is conserved.

$\therefore$   Pi = Pf

Pi = Initial momentum

Pf = Final Momentum

mv = (2m + m) Vf

$\Rightarrow$   Vf = ${{mv} \over {2m + M}}$

Here due to collision ${5 \over 6}$th of kinetic energy is lost.

$\therefore$    Remaining kinetic energy,

Kf = ${1 \over 6}$ Ki

$\Rightarrow$    ${1 \over 2}$(2m + M) $\times$ ${{{{\left( {mv} \right)}^2}} \over {{{\left( {2m + M} \right)}^2}}}$ = ${1 \over 6} \times {1 \over 2}m{v^2}$

$\Rightarrow$   ${{{m^2}{v^2}} \over {2m + M}}$ = ${1 \over 6}m{v^2}$

$\Rightarrow$   ${m \over {2m + M}}$ = ${1 \over 6}$

$\Rightarrow$   6m = 2m + M

$\Rightarrow$   M = 4m

$\Rightarrow$    ${M \over m}$ = 4
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### JEE Main 2019 (Online) 9th January Evening Slot

A force acts on a 2 kg object so that its position is given as a function of time as x = 3t2 + 5. What is the work done by this force in first 5 seconds ?
A
850 J
B
950 J
C
875 J
D
900 J

## Explanation

Displacement,

x = 3t2 + 5

$\therefore$  v = ${{dx} \over {dt}} = 6t$

At t = 0,   velocity = 6 $\times$ 0 = 0

at t = 5, velocity = 5 $\times$ 6 = 30 m/s

we know from work energy theorem,

Work (W) = change in kinetic energy ($\Delta$K)

= ${1 \over 2}mv_F^2 - {1 \over 2}mv_i^2$

= ${1 \over 2}$ $\times$ 2 $\times$ (30)2 $-$ 0

= 900 J
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### JEE Main 2019 (Online) 10th January Morning Slot

A block of mass m is kept on a platform which starts from rest with constant acceleration g/2 upward, as shown in figure. Work done by normal reaction on block in time is -

A
${{m{g^2}{t^2}} \over 8}$
B
${{3m{g^2}{t^2}} \over 8}$
C
$-$ ${{m{g^2}{t^2}} \over 8}$
D
0

## Explanation

N $-$ mg = ${{mg} \over 2}$ $\Rightarrow$ N = ${{3mg} \over 2}$

The distance travelled by the system in time t is

S = ut + ${1 \over 2}a{t^2} = 0 + {1 \over 2}\left( {{g \over 2}} \right){t^2} = {1 \over 2}{g \over 2}{t^2}$

Now, work done

W = N.S = $\left( {{3 \over 2}mg} \right)\left( {{1 \over 2}{g \over 2}{t^2}} \right)$

$\Rightarrow$  W = ${{3m{g^2}{t^2}} \over 8}$
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### JEE Main 2019 (Online) 10th January Evening Slot

A particle which is experiencing a force, given by $\overrightarrow F = 3\widehat i - 12\widehat j,$ undergoes a displacement of $\overrightarrow d = 4\overrightarrow i$ particle had a kinetic energy of 3 J at the beginning of the displacement, what is its kinetic energy at the end of the displacement ?
A
9 J
B
10 J
C
12 J
D
15 J

## Explanation

Work done = $\overrightarrow F \cdot \overrightarrow d$

$=$ 12 J

work energy theorem

wnet $=$ $\Delta$K.E.

12 $=$ Kf $-$ 3

Kf = 15 J