JEE Advanced · 2014 · 25 May 2014 · Paper 1 · Conservation of Momentum

A horizontal circular platform of radius

0.5 m

and mass

0.45 k g

is free to rotate about its axis. Two massless spring toy-guns, each carrying a steel ball of mass 0.05 kg are attached to the platform at a distance

0.25 m

from the centre on its either sides along its diameter (see figure). Each gun simultaneously fires the balls horizontally and perpendicular to the diameter in opposite directions. After leaving the platform, the balls have horizontal speed of

9 m s^{- 1}

with respect to the ground. The rotational speed of the platform inrads

^{- 1}

after the balls leave the platform is

Answer: 4

Solution

By conservation of angular momentum (no external torque):

Initial angular momentum:

L_{i} = 0

(system at rest).

Final angular momentum:

L_{f} = L_{ba l l s} + L_{pl a t f or m}

.

Each ball:

L_{ba l l} = m v r = 0.05 \times 9 \times 0.25

. Two balls in opposite directions contribute

2 \times 0.05 \times 9 \times 0.25 = 0.225

kg·m²/s.

Platform:

L_{pl a t f or m} = I ω = \frac{1}{2} M R^{2} ω = \frac{1}{2} \times 0.45 \times 0. 5^{2} \times ω = 0.05625 ω

.

Conservation:

0 = 0.225 - 0.05625 ω

(balls' angular momentum is opposite to platform's rotation).

Solving:

ω = \frac{0.225}{0.05625} = 4

rad/s.

JEE Advanced · 2014 · 25 May 2014 · Paper 1 · Integer answer

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JEE Advanced · 2014 · 25 May 2014 · Paper 1 · Conservation of Momentum

Solution

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