The Ballistic Pendulum Goal Combine the concepts of conservation of energy and conservation of momentum in inelastic collisions. Problem The ballistic pendulum (Fig. 6.10a) is a device used to measure the speed of a fast-moving projectile such as a bullet. The bullet is fired into a large block of

Question

The Ballistic Pendulum
Goal Combine the concepts of conservation of energy and conservation of momentum in inelastic collisions.

Problem The ballistic pendulum (Fig. 6.10a) is a device used to measure the speed of a fast-moving projectile such as a bullet. The bullet is fired into a large block of wood suspended from some light wires. The bullet is stopped by the block, and the entire system swings up to a height h. It is possible to obtain the initial speed of the bullet by measuring h and the two masses. As an example of the technique, assume that the mass of the bullet, [latex] m_{1} [/latex], is 5.00 g, the mass of the pendulum, [latex] m_{2} [/latex], is 1.000 kg, and h is 5.00 cm. Find the initial speed of the bullet, [latex] v_{1 i} [/latex].

Strategy First, use conservation of momentum and the properties of perfectly inelastic collisions to find the initial speed of the bullet, [latex] v_{1 i} [/latex], in terms of the final velocity of the block–bullet system, [latex] v_{f} [/latex]. Second, use conservation of energy and the height reached by the pendulum to find [latex] v_{f} [/latex]. Finally, substitute this value of [latex] v_{f} [/latex] into the previous result to obtain the initial speed of the bullet.

Accepted Answer

Use conservation of momentum, and substitute the known masses. Note that [latex] v_{2 i}=0 ext { and } v_{f} [/latex] is the velocity of the system (block + bullet) just after the collision.

[latex]\begin{aligned}p_{i} &=p_{f} \m_{1} v_{1 i}+m_{2} v_{2 i} &=\left(m_{1}+m_{2} ight) v_{f} \\left(5.00 imes 10^{-3} kg ight) v_{1 i}+0 &=(1.005 kg ) v_{f}\end{aligned}[/latex] (1)

Apply conservation of energy to the block–bullet system after the collision:

[latex] (K E+P E)_{ ext {after collision }}=(K E+P E)_{ ext {top }} [/latex]

Both the potential energy at the bottom and the kinetic energy at the top are zero. Solve for the final velocity of the block–bullet system, [latex] v_{f} [/latex]:

[latex]\begin{aligned}&\frac{1}{2}\left(m_{1}+m_{2} ight)v_{f}^{2}+0=0+\left(m_{1}+m_{2} ight) g h \&v_{f}^{2}=2 g h \&v_{f}=\sqrt{2 g h}=\sqrt{2\left(9.80 m / s ^{2} ight)\left(5.00 imes 10^{-2} m ight)} \&v_{f}=0.990 m / s\end{aligned}[/latex]

Finally, substitute [latex] v_{f} [/latex] into Equation 1 to find [latex] v_{1 i} [/latex], the initial speed of the bullet:

[latex] v_{1 i}=\frac{(1.005 kg )(0.990 m / s )}{5.00 imes 10^{-3} kg }=199 m / s [/latex]

Remarks Because the impact is inelastic, it would be incorrect to equate the initial kinetic energy of the incoming bullet to the final gravitational potential energy associated with the bullet–block combination. The energy isn’t conserved!

Question 6.4: The Ballistic Pendulum Goal Combine the concepts of conserva...

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