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### Physics For Scientists & Engineers

Book edition 9th Edition
Author(s) Raymond A. Serway, John W. Jewett
Pages 1624 pages
ISBN 9781133947271

# As shown in Figure P9.30, a bullet of mass m and speed v passes completely through a pendulum bob of mass M. The bullet emerges with a speed of V/2. The pendulum bob is suspended by a stiff rod (not a string) of length l and negligible mass. What is the minimum value of v such that the pendulum bob will barely swing through a complete vertical circle?

The minimum value of v such that the pendulum bob will barely swing through a complete vertical circle is,$\mathrm{v}=\frac{4\mathrm{M}}{\mathrm{m}}\sqrt{\mathrm{gl}}$.

See the step by step solution

## Step 1: Concept

Energy is conserved for the bob-earth system as this is an isolated system, and that can be expressed mathematically as,

${{\mathbf{K}}}_{{\mathbf{i}}}{\mathbf{+}}{{\mathbf{U}}}_{{\mathbf{i}}}{\mathbf{=}}{{\mathbf{K}}}_{{\mathbf{f}}}{\mathbf{+}}{{\mathbf{U}}}_{{\mathbf{f}}}$ (1)

## Step 2: Given data

Given data can be listed below as,

• Mass of the bullet is, m.
• Mass of the pendulum bob, M .
• The initial velocity of the bullet is, V1=V
• The final velocity of the bullet is, Vf=$\frac{\mathrm{V}}{2}$ .
• The length of the rod is,I.

## Step 3: Determination of minimum values of v

Initially, the bob has a velocity Vb, and the height will be zero.

After the bullet passes through the bob, the bob will be at a height 2l with a velocity of 0. The momentum of the bob-bullet system is conserved in the collision, and the mathematical expression can be written as,

${\mathrm{m}}_{1}{\mathrm{v}}_{1}+{\mathrm{m}}_{2}{\mathrm{v}}_{2}={\mathrm{m}}_{1}{\mathrm{v}}_{3}+{\mathrm{m}}_{2}{\mathrm{v}}_{4}$ (2)

Here m1 is the mass of the bullet, m2 is the mass of the bob, V1 and V3 are the initial and final velocity of the bullet, V2 and V4 are the initial and final velocity of the bob.

Substitute the values in the equation (2), and we get,

$\mathrm{mv}+\mathrm{M}\left(0\right)=\mathrm{m}\frac{\mathrm{v}}{2}+{\mathrm{Mv}}_{\mathrm{b}}\phantom{\rule{0ex}{0ex}}\mathrm{mv}=\mathrm{m}\frac{\mathrm{v}}{2}+\mathrm{M}×2\sqrt{\mathrm{gl}}\phantom{\rule{0ex}{0ex}}\mathrm{v}=\frac{4\mathrm{M}}{\mathrm{m}}\sqrt{\mathrm{gl}}\phantom{\rule{0ex}{0ex}}$

Thus, the minimum value of v that the pendulum bob will barely swing through a complete vertical circle is, $\mathrm{v}=\frac{4\mathrm{M}}{\mathrm{m}}\sqrt{\mathrm{gl}}$.