1

### JEE Main 2017 (Online) 9th April Morning Slot

Two particles A and B of equal mass M are moving with the same speed $\upsilon$ as shown in the figure. They collide completely inelastically and move as a single particle C. The angle $\theta$ that the path of C makes with the X-axis is given by : A
tan $\theta$ = ${{\sqrt 3 + \sqrt 2 } \over {1 - \sqrt 2 }}$
B
tan $\theta$ = ${{\sqrt 3 - \sqrt 2 } \over {1 - \sqrt 2 }}$
C
tan $\theta$ = ${{1 - \sqrt 2 } \over {\sqrt 2 \left( {1 + \sqrt 3 } \right)}}$
D
tan $\theta$ = ${{1 - \sqrt 3 } \over {1 + \sqrt 2 }}$

## Explanation Using conservation of linear moments,

Along X-axis,

2MV'cos$\theta$ = MVsin30o $-$ MVsin45o . . . . . (1)

Along Y - axis,

2Mv' sin$\theta$ = Mv cos30o + Mv cos45o . . . . . (2)

Dividing (2) by (1), we get,

${{\sin \theta } \over {\cos \theta }}$ = ${{\cos {{30}^ \circ } + \cos {{45}^ \circ }} \over {\sin {{30}^ \circ } - \sin {{45}^ \circ }}}$

= ${{{{\sqrt 3 } \over 2} + {1 \over {\sqrt 2 }}} \over {{1 \over 2} - {1 \over {\sqrt 2 }}}}$

$\therefore\,\,\,$ tan$\theta$ = ${{\sqrt 3 + \sqrt 2 } \over {1 - \sqrt 2 }}$
2

### JEE Main 2018 (Offline)

In a collinear collision, a particle with an initial speed v0 strikes a stationary particle of the same mass. If the final total kinetic energy is 50% greater than the original kinetic energy, the magnitude of the relative velocity between the two particles, after collision, is :
A
${{{v_0}} \over {\sqrt 2 }}$
B
${{v_0}} \over 4$
C
$\sqrt 2 {v_0}$
D
${{v_0}} \over 2$

## Explanation

From conservation of linear momentum,

mv0 = mv1 + mv2

or v0 = v1 + v2 ........(1)

According to the question,

Kf = ${3 \over 2}$Ki

$\Rightarrow$ ${1 \over 2}mv_1^2 + {1 \over 2}mv_2^2 = {3 \over 2} \times {1 \over 2}mv_0^2$

$\Rightarrow$ $v_1^2 + v_2^2 = {3 \over 2}v_0^2$

Using eq (1) ${\left( {{v_1} + {v_2}} \right)^2} = v_0^2$

$\Rightarrow$ $v_1^2 + v_2^2 + 2{v_1}{v_2}$ = $v_0^2$

$\Rightarrow$ $2{v_1}{v_2}$ = $v_0^2 - {3 \over 2}v_0^2$ = $- {1 \over 2}v_0^2$

Now, ${\left( {{v_1} - {v_2}} \right)^2}$ = ${\left( {{v_1} + {v_2}} \right)^2} - 4{v_1}{v_2}$

= $v_0^2 - \left( { - v_0^2} \right)$ = $2v_0^2$

$\therefore$ ${{v_1} - {v_2}}$ = $\sqrt 2 {v_0}$
3

### JEE Main 2018 (Online) 15th April Evening Slot

A proton of mass m collides elastically with a particle of unknown mass at rest. After the collision, the proton and the unknown particle are seen moving at an angle of 90o with respect to each other. The mass of unknown particle is :
A
${m \over 2}$
B
m
C
${m \over {\sqrt 3 }}$
D
2 m
4

### JEE Main 2018 (Online) 16th April Morning Slot

Two particles of the same mass m are moving in circular orbits because of force, given by $F\left( r \right) = {{ - 16} \over r} - {r^3}$

The first particle is at a distance r = 1, and the second, at r = 4. The best estimate for the ratio of kinetic energies of the first and the second particle is closest to :
A
$6 \times {10^{ - 2}}$
B
$3 \times {10^{ - 3}}$
C
${10^{ - 1}}$
D
$6 \times {10^{ 2}}$

## Explanation

In circular motion the force required

$\left| F \right| = {{m{v^2}} \over r}$

$\therefore\,\,\,$ ${{m{v^2}} \over r} = {{16} \over r} + {r^3}$

$\Rightarrow$ mv2 = 16 + r4

$\therefore\,\,\,$ kinetic energy (K) = ${1 \over 2}$ mv2 = ${1 \over 2}$ [ 16 + r4]

$\therefore\,\,\,$ Kinetic energy of first particle (K1) = ${1 \over 2}$ [16 + 1]

Kinetic energy of second particle (K2) = ${1 \over 2}$ [16 + 44]

$\therefore\,\,\,\,$ ${{{K_1}} \over {{K_2}}}$ = ${{{{16 + 1} \over 2}} \over {{{16 + 256} \over 2}}}$ = ${{17} \over {272}}$

$\Rightarrow$ $\,\,\,$ ${{{K_1}} \over {{K_2}}} = 6 \times {10^{ - 2}}$

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