 JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2002

Speeds of two identical cars are $u$ and $4$$u$ at the specific instant. The ratio of the respective distances in which the two cars are stopped from that instant is
A
$1:1$
B
$1:4$
C
$1:8$
D
$1:16$

Explanation

Given that initial speed of two cars $u$ and $4u$ and final speed $v$ is 0 for both car. Both car is gradually slowing down so acceleration = $(-a)$

So formula becomes, 0 = ${u^2}$ - 2$a$s.

$\Rightarrow {u^2} = 2as$

For first car ${u^2} = 2a{s_1}$ ..........(i)

For second car ${\left( {4u} \right)^2} = 2a{s_2}$ ..........(ii)

Dividing $(i)$ and $(ii),$

${{{u^2}} \over {16{u^2}}} = {{2a{s_1}} \over {2a{s_2}}}$

$\Rightarrow {1 \over {16}} = {{{s_1}} \over {{s_2}}}$
2

AIEEE 2002

From a building two balls A and B are thrown such that A is thrown upwards and B downwards ( both vertically with the same speed ). If vA and vB are their respective velocities on reaching the ground, then
A
${v_B} > {v_A}$
B
${v_A} = {v_B}$
C
${v_A} > {v_B}$
D
their velocities depend on their masses.

Explanation Assume the initial velocity of each particle is = u

And height of building = h

If final velocity of A is vA then vA2 = u2 + 2(-g)(-h) = u2 + 2gh

If final velocity of B is vB then vB2 = u2 + 2gh

$\therefore$ vA = vB

Sign Rule : Take the direction of initial velocity positive opposite direction as negative.

Here for ball A initial velocity u is upward so upward is positive and downward is negative. That is why gravity is = - g and height = - h

And for ball B initial velocity u is downward so downward is positive and upward is negative. That is why gravity is = + g and height = + h
3

AIEEE 2002

A ball whose kinetic energy E, is projected at an angle of $45^\circ$ to the horizontal. The kinetic energy of the ball at the highest point of its height will be
A
E
B
${E \over {\sqrt 2 }}$
C
${E \over 2}$
D
zero

Explanation

Assume the ball of mass m is projected with a speed u. Then the kinetic energy(E) at the point of projection = ${1 \over 2}m{u^2}$

At highest point of flight only horizontal component of velocity $u\cos \theta$ present as at highest point vertical component of velocity is = 0.

Note : The horizontal component of velocity does not change in entire projectile motion.

At highest point the velocity is = $u\cos \theta$ = $u\cos 45^\circ$ = ${u \over {\sqrt 2 }}$

$\therefore$ The kinetic energy at the height point = ${1 \over 2}m{\left( {{u \over {\sqrt 2 }}} \right)^2}$

= ${1 \over 2}m{u^2} \times {1 \over 2}$ = ${E \over 2}$