A simple pendulum of length 1 m has a wooden bob of mass 1 kg . It is struck by a bullet of mass 10 -2 kg moving with a speed of 2 × 10 2 ms -1 . The bullet gets embedded into the bob. The height to which the bob rises before swinging back is. (use g =10 m / s 2 )

Physics · System Of Particles

JEE Main 2024 — 1 February, Shift 1 — Question 43

A simple pendulum of length 1 m has a wooden bob of mass 1 kg . It is struck by a bullet of mass $10^{-2} \mathrm{~kg}$ moving with a speed of $2 \times 10^{2} \mathrm{~ms}^{-1}$ . The bullet gets embedded into the bob. The height to which the bob rises before swinging back is. (use $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}$ )

Option A:
0.30 m
Option B:
0.20 m
Correct
Option C:
0.35 m
Option D:
0.40 m

Answer: B

Step-by-step solution

$\text{Given:} \quad M = 1\,\text{kg}, \quad m = 10^{-2}\,\text{kg}, \quad v = 2 \times 10^{2}\,\text{m/s}, \quad g = 10\,\text{m/s}^2.$

Step 1: Conservation of momentum

m v = (M + m) V

V = \frac{m v}{M + m}

V = \frac{(10^{-2})(200)}{1 + 10^{-2}} = \frac{2}{1.01} \approx 1.98\,\text{m/s}

Step 2: Convert kinetic energy to potential energy

\frac{1}{2} (M + m) V^2 = (M + m) g h

h = \frac{V^2}{2g}

h = \frac{(1.98)^2}{20} \approx 0.196 \approx 0.20\,\text{m}

\boxed{h \approx 0.20\,\text{m}}

Answer key and solution verified before publishing.

Practise System Of Particles

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Main 2024
Paper: 1 February, Shift 1
Subject: Physics
Chapter: System Of Particles
Topic: Collisions in One Dimension

The identical spheres each of mass 2 M are placed at the corners of a right angled triangle with mutually perpendicular sides equal to 4 m…

Answer: B

Step-by-step solution

Practise System Of Particles

More System Of Particles questions from this paper