# A 29-g rifle bullet traveling 200 m/s embeds itself in a 3.4-kg pendulum hanging on a 3.0-m-long...

## Question:

A 29-g rifle bullet traveling at 200 m/s embeds itself in a 3.4-kg pendulum hanging on a 3.0-m-long string, which makes the pendulum swing upward in an arc.

Part A

Determine the vertical component of the pendulum's maximum displacement. Express your answer in two significant figures and include the appropriate units.

Part B

Determine the horizontal component of the pendulum's maximum displacement. Express your answer in two significant figures and include the appropriate units.

## Energy Conservation in Simple Harmonic Motion:

During a simple harmonic motion (SHM), there is a continued change of energy from potential energy at the ends where amplitude is maximum, and back to kinetic energy which is maximum at the equilibrium point of the oscillation. If no external forces act on the motion, the sum of the potential energy and the kinetic energy is the same at any point during the motion.

Part(a)

When the bullet hits the pendulum, we have a collision, and in any collision, momentum is conserved. Let the initial momentum of the pendulum and initial momentum of the bullet be {eq}p_{1} {/eq} and {eq}p_{2} {/eq}, respectively. We notice that after the collision, the pendulum and the bullet move as one body with the same velocity, v. Let the momentum of the pendulum plus the bullet after the collision be {eq}p_{3} {/eq}.

{eq}\begin{align*} initial \ momentum \ &= \ final \ momentum\\ p_{1} \ + \ p_{2} \ &= \ p_{3}\\ (3.4 \ kg) \ \times \ (0 \ m/s) \ + \ ((2.90 \ \times \ 10^{-2} \ kg) \ \times \ (200 \ m/s)) \ &= \ (3.429 \ kg) \ \times \ v\\ 5.8 \ kg \cdot m/s \ &= \ (3.429 \ kg) \ \times \ v\\ \therefore \ v \ &= \ 1.69 \ m/s \end{align*} {/eq}.

Therefore, the velocity of the pendulum plus the bullet after the collision is v = 1.69 m/s. Now we use the conservation of energy expression. The kinetic energy of the pendulum plus the bullet will equal the potential energy at the maximum height.

{eq}\begin{align*} mgh \ &= \ \frac{1}{2} \ m \ v^2\\ \therefore h \ &= \ \frac{\ v^2}{2g} \\ &= \ \frac{(1.69 \ m/s)^2}{2 \ (9.81) \ m/s^2} \\ &= \ 0.144821652 \ m\\ \therefore \ h \ &= \ 0.14 \ m \end{align*} {/eq}

Part(b)

From the Pythagorean theorem, by considering that the perpendicular distance from the hinge at the top of the string to the pendulum at its maximum height {eq}= l - h = 3 - 0.14 = 2.84m, {/eq} where height h = 0.14m, the horizontal displacement can be calculated as,

{eq}\begin{align*} x \ &= \sqrt {2.84^2 + 0.14^2 } = 2.9m \end{align*} {/eq}, correct to two significant figures.