Question

a 48 g bullet traveling at 196 m/s buries itself in a 5.56 kg pendulum hanging on a 2 m length of string, which makes the pendulum swing upward in an arc. determine the horizontal component (in meters) of the displacement of the pendulum.
δx = ?
0.74 m
8.59 m
2.73 m
2.93 m

Explanation:

Step1: Apply conservation of momentum

The initial momentum of the bullet is \(p_{i}=m_{bullet}v_{bullet}\), and after the collision, the combined mass \((m_{bullet} + m_{pendulum})\) has a velocity \(v\). By conservation of momentum \(m_{bullet}v_{bullet}=(m_{bullet}+m_{pendulum})v\).
Given \(m_{bullet}=48\space g = 0.048\space kg\), \(v_{bullet}=196\space m/s\), \(m_{pendulum}=5.56\space kg\)

$$v=\frac{m_{bullet}v_{bullet}}{m_{bullet}+m_{pendulum}}=\frac{0.048\times196}{0.048 + 5.56}\space m/s$$

$$v=\frac{9.408}{5.608}\space m/s\approx1.68\space m/s$$

Step2: Apply conservation of mechanical energy

The kinetic energy of the combined mass \(\frac{1}{2}(m_{bullet}+m_{pendulum})v^{2}\) is converted into gravitational potential energy \((m_{bullet}+m_{pendulum})gh\). So \(\frac{1}{2}(m_{bullet}+m_{pendulum})v^{2}=(m_{bullet}+m_{pendulum})gh\), and \(h = \frac{v^{2}}{2g}\) (where \(g = 9.8\space m/s^{2}\))

$$h=\frac{(1.68)^{2}}{2\times9.8}=\frac{2.8224}{19.6}\space m\approx0.144\space m$$

Step3: Use geometric relation

If the length of the string is \(L = 2\space m\), and using the Pythagorean theorem \(L^{2}=(L - h)^{2}+\Delta x^{2}\)

$$L^{2}=L^{2}-2Lh+h^{2}+\Delta x^{2}$$

\(\Delta x=\sqrt{2Lh - h^{2}}\)
Substitute \(L = 2\space m\) and \(h=0.144\space m\)

$$ \Delta x=\sqrt{2\times2\times0.144-(0.144)^{2}}=\sqrt{0.576 - 0.020736}=\sqrt{0.555264}\approx0.74\space m$$

Answer:

\(0.74\space m\)