JEE Main · 2023 · 30 Jan 2023 · Shift 1 · Conservation of Momentum

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A ball of mass 200 g rests on a vertical post of height 20 m. A bullet of mass 10 g, travelling in horizontal direction, hits the centre of the ball. After collision both travels independently. The ball hits the ground at a distance 30 m and the bullet at a distance of 120 m from the foot of the post. The value of initial velocity of the bullet will be (if

g = 10

m/s²) :

(A) $120$ m/s
(B) $60$ m/s
(C) $400$ m/s
(D) $360$ m/s✓ Correct

Answer: (D)

Solution

Both the ball and the bullet leave the top of the post as horizontal projectiles falling the same height

h = 20 m

, so each stays airborne for

t = \frac{2 h}{g} = \frac{2 \times 20}{10} = 2 s

. Their post-collision horizontal speeds follow from the ranges: the ball has

v_{1} = \frac{30}{2} = 15 m/s

and the bullet

v_{2} = \frac{120}{2} = 60 m/s

. Horizontal momentum is conserved during the strike,

m_{b} u = m_{ball} v_{1} + m_{b} v_{2}

, giving

0.01 u = 0.2 \times 15 + 0.01 \times 60 = 3.6

. The bullet's initial speed is

360 m/s

JEE Main · 2023 · 30 Jan 2023 · Shift 1 · Single correct option

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JEE Main · 2023 · 30 Jan 2023 · Shift 1 · Conservation of Momentum

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