Question

In a ballistic pendulum, an object of mass m is fired with an initial speed v₀ at a pendulum bob. The bob has a mass M, which is suspended by a rod of length L and negligible mass. After the collision, the pendulum and object stick together and swing to a maximum angular displacement θ as shown.

A. Find an expression for v₀, the initial speed of the fired object.

Express your answer in terms of some or all of the variables m, M, L, and θ and the acceleration due to gravity, g.

v₀ =

B. An experiment is done to compare the initial speed of bullets fired from different handguns: a 9mm and a .44 caliber. The guns are fired into a 10-kg pendulum bob of length L. Assume that the 9-mm bullet has a mass of 6 g and the .44-caliber bullet has a mass of 12 g . If the 9-mm bullet causes the pendulum to swing to a maximum angular displacement of 4.3° and the .44-caliber bullet causes a displacement of 10.1° , find the ratio of the initial speed of the 9-mm bullet to the speed of the.44-caliber bullet,  (v₀)9/(v₀)44.

Express your answer numerically.

(v₀)9/(v₀)44=

Answer 1

Concepts and reason

The principle of conservation of momentum and the law of conservation of mechanical energy are used to solve the problem.

First, apply the law of conservation of linear momentum to the system of bullet and the block. Then, use the expression for the initial speed of the bullet calculated in part A to calculate the ratio of the speeds of the bullets.

Fundamentals

When no external forces act on a system of two particles, the total momentum of the system remains constant. When the bullet is shot into the block of a ballistic pendulum, the initial momentum is equal to the final momentum of the block and the bullet embedded in the block.

The block along with the bullet swings upwards to a height h, when its velocity becomes zero. If no energy is dissipated in the process, the initial kinetic energy of the system is converted into gravitational potential energy.

By writing equations for the conservation of momentum and for energy, the initial velocity of the bullet can be determined.

(A)

The bullet of mass m and velocity v_o is shot into a block of mass M, which forms the bob of a pendulum of length L, which is at rest. The bullet enters the block and is embedded in the block. The block and the bullet acquire a speed v.

Apply the law of conservation of linear momentum to the system of block and bullet.

$m{v_0} = \left( {m + M} \right)v$

Therefore,

${v_0} = \left( {\frac{{m + M}}{m}} \right)v$ …….(1)

The pendulum swings upwards to reach a maximum height h.

Draw a diagram representing the motion of the bob.

The pendulum bob with the bullet embedded in it has a speed v at point A. It swings upwards and reaches point B, where its speed becomes equal to zero. It is therefore displaced upwards and reaches a height h above the point A. The string makes an angle $\theta$ with the vertical.

Take the zero of potential energy along the horizontal line passing through the point A. The block of mass $\left( {m + M} \right)$ has no potential energy at A, while it has kinetic energy due to its speed v. At B, its speed becomes equal to zero, therefore, it has no kinetic energy at B, but since it has made a vertical displacement h, it has gained gravitational potential energy.

Apply the law of conservation of energy to the motion of the pendulum.

$\frac{1}{2}\left( {m + M} \right){v^2} = \left( {m + M} \right)gh$

Therefore,

$v = \sqrt {2gh}$ ……(2)

In $\Delta OCB$ ,

$\cos \theta = \frac{{L - h}}{L}$

Hence,

$h = L\left( {1 - \cos \theta } \right)$ …….(3)

Substitute equation (3) in equation (2).

$\begin{array}{c}\\v = \sqrt {2gh} \\\\ = \sqrt {2gL\left( {1 - \cos \theta } \right)} \\\end{array}$ …….(4)

Substitute equation (4) in equation (1)

$\begin{array}{c}\\{v_0} = \left( {\frac{{m + M}}{m}} \right)v\\\\ = \left( {1 + \frac{M}{m}} \right)\sqrt {2gL\left( {1 - \cos \theta } \right)} \\\end{array}$

(B)

The bullet of mass ${m_9}$ is shot from the 9-mm gun with a speed ${\left( {{v_0}} \right)_9}$ . The ballistic pendulum swings upwards and makes an angle ${\theta _9}$ with the vertical.

Write the expression for the initial speed ${\left( {{v_0}} \right)_9}$ of the 9 mm bullet.

${\left( {{v_0}} \right)_9} = \left( {1 + \frac{M}{{{m_9}}}} \right)\sqrt {2gL\left( {1 - \cos {\theta _9}} \right)}$

The bullet of mass ${m_{44}}$ is shot from a .44 caliber gun with a speed ${\left( {{v_0}} \right)_{44}}$ . The pendulum makes an angle ${\theta _{44}}$ with the vertical when it reaches the maximum height.

Write the expression for the initial speed ${\left( {{v_0}} \right)_{44}}$ of the .44 caliber bullet.

${\left( {{v_0}} \right)_{44}} = \left( {1 + \frac{M}{{{m_{44}}}}} \right)\sqrt {2gL\left( {1 - \cos {\theta _{44}}} \right)}$

Find the ratio $\frac{{{{\left( {{v_0}} \right)}_9}}}{{{{\left( {{v_0}} \right)}_{44}}}}$ .

$\begin{array}{c}\\\frac{{{{\left( {{v_0}} \right)}_9}}}{{{{\left( {{v_0}} \right)}_{44}}}} = \frac{{\left( {1 + \frac{M}{{{m_9}}}} \right)\sqrt {2gL\left( {1 - \cos {\theta _9}} \right)} }}{{\left( {1 + \frac{M}{{{m_{44}}}}} \right)\sqrt {2gL\left( {1 - \cos {\theta _{44}}} \right)} }}\\\\ = \frac{{\left( {1 + \frac{M}{{{m_9}}}} \right)}}{{\left( {1 + \frac{M}{{{m_{44}}}}} \right)}}\sqrt {\frac{{\left( {1 - \cos {\theta _9}} \right)}}{{\left( {1 - \cos {\theta _{44}}} \right)}}} \\\end{array}$

Express the masses ${m_9}$ and ${m_{44}}$ in kg.

$\begin{array}{c}\\{m_9} = \left( {6{\rm{ g}}} \right)\left( {\frac{{{{10}^{ - 3}}{\rm{ kg}}}}{{1{\rm{ g}}}}} \right)\\\\ = 6 \times {10^{ - 3}}{\rm{ kg}}\\\\{m_{44}} = \left( {12{\rm{ g}}} \right)\left( {\frac{{{{10}^{ - 3}}{\rm{ kg}}}}{{1{\rm{ g}}}}} \right)\\\\ = 1.2 \times {10^{ - 2}}{\rm{ kg}}\\\end{array}$

Substitute $6 \times {10^{ - 3}}{\rm{ kg}}$ for ${m_9}$ , $1.2 \times {10^{ - 2}}{\rm{ kg}}$ for ${m_{44}}$ , 10 kg for M, 4.3^o for ${\theta _9}$ and 10.1^o for ${\theta _{44}}$ in the expression,

$\begin{array}{c}\\\frac{{{{\left( {{v_0}} \right)}_9}}}{{{{\left( {{v_0}} \right)}_{44}}}} = \frac{{\left( {1 + \frac{M}{{{m_9}}}} \right)}}{{\left( {1 + \frac{M}{{{m_{44}}}}} \right)}}\sqrt {\frac{{\left( {1 - \cos {\theta _9}} \right)}}{{\left( {1 - \cos {\theta _{44}}} \right)}}} \\\\ = \left[ {\frac{{1 + \left( {\frac{{10{\rm{ kg}}}}{{6 \times {{10}^{ - 3}}{\rm{ kg}}}}} \right)}}{{1 + \left( {\frac{{10{\rm{ kg}}}}{{1.2 \times {{10}^{ - 2}}{\rm{ kg}}}}} \right)}}} \right]\sqrt {\frac{{1 - \cos 4.3^\circ }}{{1 - \cos 10.1^\circ }}} \\\\ = 0.85\\\end{array}$

Ans: Part A

The initial speed of the bullet is given by the expression ${v_0} = \left( {1 + \frac{M}{m}} \right)\sqrt {2gL\left( {1 - \cos \theta } \right)}$