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Q30P

Expert-verifiedFound in: Page 247

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,$\mathrm{v}=\frac{4\mathrm{M}}{\mathrm{m}}\sqrt{\mathrm{gl}}$.

**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)**

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, V
_{1}=V - The final velocity of the bullet is, V
_{f}=$\frac{\mathrm{V}}{2}$ . - The length of the rod is,I.

Initially, the bob has a velocity V_{b}, 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 m_{1 }is the mass of the bullet, m_{2 }is the mass of the bob, V_{1 }and V_{3 }are the initial and final velocity of the bullet, V_{2 }and V_{4 }are the initial and final velocity of the bob.

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

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

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

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