1

### JEE Main 2017 (Online) 8th April Morning Slot

An object is dropped from a height h from the ground. Every time it hits the ground it looses 50% of its kinetic energy. The total distance covered as t $\to$ $\infty$ is :
A
3h
B
$\infty$
C
${5 \over 3}$h
D
${8 \over 3}$h

## Explanation

Let,

Kinetic energy (k) = ${1 \over 2}$ m $\upsilon$2 before it hit the ground.

After hitting the ground kinetic energy

(k') = ${1 \over 2}$ m $\upsilon $$_1^2 \therefore\,\,\,According to the question, {1 \over 2} m\upsilon$$_1^2$ = ${1 \over 2}$ $\times$ ${1 \over 2}$ m$\upsilon$2

$\Rightarrow$$\,\,\,$ $\upsilon$1 = ${v \over {\sqrt 2 }}$

After hitting the ground the object will bounce

h' = ${{v_1^2} \over {2g}}$ = ${{{v^2}} \over {4g}}$ = ${h \over 2}$ [ as    h = ${{{v^2}} \over {2g}}$ ]

Total distance travelled from the time it first hits the ground to the next time it hits the ground is = ${h \over 2}$ + ${h \over 2}$ = h

So, this will create a infinite geometric progression with the common ration ${1 \over 2}$.

$\therefore\,\,\,$ Total distance covered

= h (distance travelled by the obhect when first dropped, before it hits the ground)
+ (h + ${h \over 2}$ + ${h \over 4}$ + . . . . . . . .$\propto$)

= h + ${h \over {1 - {1 \over 2}}}$

= h + 2h

= 3h
2

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

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

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

$\therefore$ ${{v_1} - {v_2}}$ = $\sqrt 2 {v_0}$
4

### 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

NEET