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NEET Exam Model Paper 6

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Question : 4 of 180

Marks: +1, -0

Ball 1 is released from the top of a smooth inclined plane, of height ‘h’ and at the same instant ball 2 is projected from the foot of the plane with such a velocity that they meet halfway up the incline. Determine

The velocity with which balls are projected

$\sqrt{2 g h}$
$\sqrt{g h}$
$\sqrt{\frac{g h}{2}}$
$2 \sqrt{g h}$

Solution:

Accelerations of both the balls are

a_1 = g \sin \theta

and

a_2 = -g \sin \theta

down the incline

Ball 1 :

\frac{l}{2} = (0)t + \frac{1}{2} g \sin \theta t^2

.......(1)

Ball 2 :

\frac{l}{2} = v_i t + \frac{1}{2} (-g \sin \theta) t^2

...(2)

Adding eqn. (1) and (2), we get

l = v_i t

or

t = \frac{l}{v_i}

Substituting it in eqn. (1), we get

\frac{l}{2} = \frac{1}{2} g \sin \theta \left( \frac{1}{v_i} \right)^2

or

v_i = \sqrt{g l \sin \theta} = \sqrt{g h}

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