Question

As shown below, 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 of length and negligible mass. What is the minimum value of v such that the pendulum bob will barely swing through a complete vertical circle? (Use L for , g for gravity, and M and m as appropriate.)

v =

Answer 1

Concepts and reason

The main concept required to solve this problem are conservation of momentum and conservation of energy.

Firstly, use the conservation of energy and find the expression for the velocity of bob.

Finally, apply the conservation of momentum and substitute the expression for the velocity of bob in it to find the velocity of bullet.

Fundamentals

The momentum of an object is as follows:

$p = mv$

Here, m is the mass of the object and v is the velocity of the object.

The expression for the kinetic energy (KE) of an object is as follows:

$KE = \frac{1}{2}m{v^2}$

Here, m is the mass of the object and v is the velocity of the object.

The potential energy of a body is given as follows:

$PE = mgh$

Here, m is the mass, h is the height of the body, and g is the acceleration due to gravity.

The conservation of momentum of a system states that the momentum ( ${p_{\rm{i}}}$ ) of the system before collision is equal to the momentum ( ${p_{\rm{f}}}$ ) of the system after collision where no external forces are acting on the system.

${p_{\rm{i}}} = {p_{\rm{f}}}$

The conservation of energy of an isolated system states that the total energy of a system remains constant always.

$K{E_{\rm{i}}} + P{E_{\rm{i}}} = K{E_{\rm{f}}} + P{E_{\rm{f}}}$

Consider the bob when it is hit by the bullet. The initial velocity is given to the bob by the bullet collision. The height of the bob from the ground is zero (for reference).

The kinetic energy of the bob initially is as follows:

$K{E_{\rm{i}}} = \frac{1}{2}M{v_{\rm{b}}}^2$

The initial potential energy of the bob is as follows:

$P{E_{\rm{i}}} = Mgh$

Here, M is the mass of the bob, g is the acceleration due to gravity, and h is the initial height from the ground.

Substitute 0 for h in the above expression as the initial height is at 0 m from the ground.

$P{E_{\rm{i}}} = 0$

The final position of the bob is the top of the vertical circle so that the bob is at a diameter away from the ground reference. The bob stops at that point so that the final velocity is zero.

The kinetic energy of the bob finally is as follows:

$K{E_{\rm{f}}} = \frac{1}{2}M{v_{\rm{f}}}^2$

Here, M is the mass of the bob and v_f is the final velocity of the bob.

Substitute 0 for v_f in the above expression.

$K{E_{\rm{f}}} = 0$

The final potential energy of the bob is as follows:

$P{E_{\rm{f}}} = Mgh$

Here, M is the mass of the bob, g is the acceleration due to gravity, and h is the final height from the ground.

The bob is at a diameter away from the ground which is twice the radius of the vertical circle.

Substitute 2L for h in the above expression.

$P{E_{\rm{f}}} = Mg\left( {2L} \right)$

Apply conservation of energy on the system. The conservation of energy of an isolated system states that the total energy of a system remains constant always.

$K{E_{\rm{i}}} + P{E_{\rm{i}}} = K{E_{\rm{f}}} + P{E_{\rm{f}}}$

Substitute $\frac{1}{2}M{v_{\rm{b}}}^2$ for $K{E_{\rm{i}}}$ , 0 for $P{E_{\rm{i}}}$ , 0 for $K{E_{\rm{f}}}$ , and $Mg\left( {2L} \right)$ for $P{E_{\rm{f}}}$ in the above expression.

$\begin{array}{c}\\\frac{1}{2}M{v_{\rm{b}}}^2 + 0 = 0 + Mg\left( {2L} \right)\\\\{v_{\rm{b}}}^2 = 4gL\\\\{v_{\rm{b}}} = 2\sqrt {gL} \\\end{array}$

Consider a system of bullet and bob.

The bullet was fired with some initial velocity. At this point, the velocity of the bob is zero as it is at rest.

The initial momentum of bullet is as follows:

${p_{\rm{i}}} = mv$

Here, m is the mass of the bullet and v is the initial velocity of the bullet.

The initial momentum of the bob is zero as it is at rest.

$p{'_{\rm{i}}} = 0$

The final velocity of the bob is $\frac{v}{2}$ and by hitting the bob, it transferred some momentum into the bob.

The final momentum of the bullet is as follows:

${p_{\rm{f}}} = m\frac{v}{2}$

Here, m is the mass of the bullet and $\frac{v}{2}$ is the final velocity of the bullet.

The final momentum of the bob is as follows:

$p{'_{\rm{f}}} = M{v_{\rm{b}}}$

Here, M is the mass of the bob and ${v_{\rm{b}}}$ is the velocity of the bob.

Apply conservation of momentum.

${p_{\rm{i}}} + p{'_{\rm{i}}} = {p_{\rm{f}}} + p{'_{\rm{f}}}$

Substitute mv for ${p_{\rm{i}}}$ , 0 for $p{'_{\rm{i}}}$ , $m\frac{v}{2}$ for ${p_{\rm{f}}}$ , and $M{v_{\rm{b}}}$ for $p{'_{\rm{f}}}$ in the above equation.

$\begin{array}{c}\\mv + 0 = m\frac{v}{2} + M{v_{\rm{b}}}\\\\v = \frac{{2M{v_{\rm{b}}}}}{m}\\\end{array}$

Substitute $2\sqrt {gL}$ for ${v_{\rm{b}}}$ in the above expression.

$\begin{array}{c}\\v = \frac{{2M\left( {2\sqrt {gL} } \right)}}{m}\\\\ = \frac{{4M}}{m}\sqrt {gL} \\\end{array}$

Ans:

The minimum velocity of bullet is $\frac{{4M}}{m}\sqrt {gL}$ .