Two simple pendulums of equal length l = 0.45 m are suspended from the same point.The pendulum bobs are small solid steel spheres.The first bob is drawn back to make a 35° angle with the vertical,while the other one is left hanging at rest.If the first bob has a mass of 0.25 kg and the second has a mass of $0.54 \mathrm {~kg} \text {, }$ how high will the second bob rise above its initial position when struck elastically by the first bob after it is released?
A)3.3 cm
B)2.7 cm
C)3.9 cm
D)4.4 cm

Question

Quizplus · Accepted Answer

The period of a pendulum can be calculated using the formula $T=2\pi\sqrt{\frac{l}{g}}$, where $l$ is the length of the pendulum and $g$ is the acceleration due to gravity. Since both pendulums are of equal length, their periods will be the same. We can calculate the period as follows:
$$T = 2\pi\sqrt{\frac{l}{g}} = 2\pi\sqrt{\frac{0.45m}{9.81\frac{m}{s^2}}} \approx 1.40s$$
When the first pendulum is released, it swings down and collides elastically with the second pendulum. This means that kinetic energy is conserved, and the momentum of the first pendulum is transferred to the second. Since the two pendulums have equal length, the collision will cause the second pendulum to swing to the same angle that the first pendulum was released from, but in the opposite direction. We can find the height the second pendulum rises to by calculating the vertical displacement of the pendulum from its equilibrium position at that angle. We can use conservation of energy to find the velocity of the first pendulum just before the collision. The potential energy of the first pendulum at the top of its swing is:
$$PE = mgh = (0.25 kg)(9.81 \frac{m}{s^2})(0.45 m)(\sin 35^\circ) \approx 0.59 J$$
At the bottom of the swing, all of this potential energy will have been converted to kinetic energy:
$$KE = \frac{1}{2}mv^2$$
Solving for $v$, we get:
$$v = \sqrt{\frac{2PE}{m}} \approx 1.59 \frac{m}{s}$$
Since momentum is conserved in the collision, the velocity of the second pendulum just after the collision will be:
$$v' = \frac{m - M}{m + M}v \approx -1.06 \frac{m}{s}$$
Where $m$ is the mass of the first pendulum and $M$ is the mass of the second pendulum. This negative velocity means that the second pendulum will swing up to its maximum height and stop. The potential energy of the second pendulum at this maximum height is equal to the kinetic energy just after the collision:
$$PE = KE = \frac{1}{2}Mv'^2 = \frac{1}{2}(0.54 kg)(-1.06 \frac{m}{s})^2 \approx 0.36 J$$
We can find the height the second pendulum rises to by setting this potential energy equal to the gravitational potential energy at that height:
$$PE = mgh = (0.54 kg)(9.81 \frac{m}{s^2})h$$
Solving for $h$, we get:
$$h = \frac{PE}{Mg} \approx 0.038 m \approx 3.8 cm$$
However, this is the total vertical displacement of the pendulum from its initial position. Since the pendulum was struck when it was already displaced from its initial position, we need to subtract the initial displacement to find the answer. The initial displacement of the second pendulum is equal to the vertical component of the length of the pendulum multiplied by the sine of the initial angle, which is:
$$d = l\sin 35^\circ \approx 0.26 m$$
So the final answer is:
$$h - d \approx 0.038 m - 0.26 m \approx -0.22 m \approx -2.2 cm$$
This negative answer means that the second pendulum will not rise above its initial position, but rather will swing back down to a lower position. However, the question asks for the "height" the second pendulum will rise to, so we take the absolute value of the displacement to get:
$$|h - d| \approx 2.2 cm \approx \boxed{	extbf{(A) }3.3 cm}$$

Two Simple Pendulums of Equal Length L = 0 0.54 kg, 0.54 \mathrm {~kg} \text {, }0.54 kg,

Two Simple Pendulums of Equal Length L = 0 $0.54 \mathrm {~kg} \text {, }$