A simple pendulum of length 1m has a wooden bob of mass 1kg . It is struck by a bullet of mass 10 ^ - 2 kg moving with a speed of 2 × 10 ^ 2 m s ^ - 1 . The bullet gets embedded into the bob. The height to which the bob

Correct answer: B. 0 . 20m Linear momentum will be conserved, therefore mu = (M + m)V ⇒ 10 ^ - 2 × 2 × 10 ^ 2 ≅ 1 × V ⇒ V ≅ 2m s ^ - 1 Again from energy conservation, we get height achieved, h = V ^ 2 2g = 0 .2 m .

JEE Main 20241 FebruaryMorning Shift

Exam: JEE Main
Year: 2024
Date: 1 February
Shift: Morning Shift
Subject: Physics
Chapter: Center of Mass, Momentum and Collision

JEE Main 2024 · 1 February · Morning Shift · Physics · Center of Mass, Momentum and Collision

A simple pendulum of length 1m has a wooden bob of mass 1kg . It is...

A simple pendulum of length

1m

has a wooden bob of mass

1kg

. It is struck by a bullet of mass

{10}^{- 2}kg

moving with a speed of

2 \times {10}^{2}m{s}^{- 1}

. The bullet gets embedded into the bob. The height to which the bob rises before swinging back is. (use

g = 10m{s}^{- 2}

)

A simple pendulum of length 1m has a wooden bob of mass 1kg . It is struck by a bullet of mass 10 ^ - 2 kg moving with

Options

A $0 . 30m$
B $0 . 20m$
C $0 . 35m$
D $0 . 40m$

Correct answer

B. $0 . 20m$

Step-by-step solution

A simple pendulum of length 1m has a wooden bob of mass 1kg . It is struck by a bullet of mass 10 ^ - 2 kg moving with

Linear momentum will be conserved, therefore

$mu = \left(M + m\right)V$

$⇒ {10}^{- 2} \times 2 \times {10}^{2} ≅ 1 \times V$

$⇒ V ≅ 2m{s}^{- 1}$

Again from energy conservation, we get height achieved, $h = \dfrac{{V}^{2}}{2g} = 0 .2 m$ .

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