JEE Advanced 2013 · Paper 1 · Physics Q7

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PhysicsSingle correct+2Question 7 of 60

A particle of mass $m$ is projected from the ground with an initial speed $u_0$ at an angle $\alpha$ with the horizontal. At the highest point of its trajectory, it makes a completely inelastic collision with another identical particle, which was thrown vertically upward from the ground with the same initial speed $u_0$ . The angle that the composite system makes with the horizontal immediately after the collision is

\dfrac\pi4

\dfrac\pi4+\alpha

\dfrac\pi2-\alpha

\dfrac\pi2

Answer and worked solution

Official answer: A

First particle at the top: horizontal velocity $u_0\cos\alpha$ , at height $H=\dfrac{u_0^2\sin^2\alpha}{2g}$ .

Second particle at that height: $v^2=u_0^2-2gH=u_0^2\cos^2\alpha$ , so it moves up at $u_0\cos\alpha$ .

Collision. Equal masses with equal horizontal and vertical momenta: the composite moves at 45°.

(A)

Topics:Elastic and inelastic collisions Projectile motion

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JEE Advanced 2013 · Paper 1, Physics question 7