JEE Advanced · 2013 · 02 Jun 2013 · Paper 1 · Conservation of Momentum

A particle of mass

m

is projected from the ground with an initial speed

u_{0}

at an angle

α

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

(A) $\frac{π}{4}$ ✓ Correct
(B) $\frac{π}{4} + α$
(C) $\frac{π}{2} - α$
(D) $\frac{π}{2}$

Answer: (A)

Solution

At the highest point of the projectile, the first particle has velocity

u_{0} cos α

horizontally.

The maximum height reached is

H = \frac{u _{0}^{2} s i n ^{2} α}{2 g}

.

For the second particle thrown vertically upward, at height

H

v_{y}^{2} = u_{0}^{2} - 2 g H = u_{0}^{2} - u_{0}^{2} sin^{2} α = u_{0}^{2} cos^{2} α

v_{y} = u_{0} cos α

At collision, the first particle has velocity

(u_{0} cos α) \hat{i}

and the second has velocity

(u_{0} cos α) \hat{j}

.

After the completely inelastic collision, by conservation of momentum:

v_{f ina l} = \frac{m ( u _{0} c o s α i ^ ) + m ( u _{0} c o s α j ^ )}{2 m} = \frac{u _{0} c o s α}{2} (\hat{i} + \hat{j})

The angle with horizontal is:

tan θ = \frac{u _{0} c o s α /2}{u _{0} c o s α /2} = 1

θ = 45° = \frac{π}{4}

Hence the answer is (A).

JEE Advanced · 2013 · 02 Jun 2013 · Paper 1 · Single correct option

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JEE Advanced · 2013 · 02 Jun 2013 · Paper 1 · Conservation of Momentum

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