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Center of Mass

Section: Physics

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Question : 2 of 18

Marks: +1, -0

A disc of mass 0.2 kg is kept floating in air without falling by vertically firing bullets each of mass 0.05 kg on the disc at the rate of 10 bullets per every second. If the bullets rebound with the same speed, then the speed of each bullet is (Acceleration due to gravity

=10\ \mathrm{ms}^{-2}

[AP EAPCET 24 May 2025 S1]

$\;2\ \mathrm{m\,s^{-1}}$
$\;10\ \mathrm{m\,s^{-1}}$
$\;20\ \mathrm{m\,s^{-1}}$
$1\ \mathrm{m\,s^{-1}}$

Solution:

As disc is floating in air without falling, force by bullets balances gravitational force.

\Rightarrow \; \left( \frac{N \Delta p}{\Delta t} \right) = m_d g

Here,

N=

number of bullets fired

\; \frac{\Delta p}{\Delta t} =

Rate of change of one bullet

m_d=

mass of disc

Hence, bullets rebound with same speed, momentum change for each bullet

=m(v-(-v))=2 m v

\therefore \; \frac{N}{t} \cdot 2 m_b v = m_d g

Substituting given values,

10 \times 2 \times 0.05 \times v = 0.2 \times 10

\therefore \; v = \frac{0.2 \times 10}{10 \times 2 \times 0.05} = 2\ \text{m/s}

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