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?

Question

Quizplus · Accepted Answer

Explanation:Momentum is conserved in the collision, so the total momentum before the collision is equal to the total momentum after the collision. Since the second bob is initially at rest, the total momentum before the collision is equal to the momentum of the first bob. After the collision, the first bob has zero velocity, so the total momentum is equal to the momentum of the second bob. Therefore, the momentum of the second bob after the collision is equal to the momentum of the first bob before the collision.The momentum of an object is equal to its mass times its velocity. Therefore, the momentum of the first bob before the collision is equal to 0.25 kg * v1, where v1 is the velocity of the first bob before the collision. The momentum of the second bob after the collision is equal to 0.54 kg * v2, where v2 is the velocity of the second bob after the collision.Since the momentum of the second bob after the collision is equal to the momentum of the first bob before the collision, we have:0.25 kg * v1 = 0.54 kg * v2Solving for v2, we get:v2 = 0.463 v1The kinetic energy of an object is equal to 1/2 * m * v^2, where m is the mass of the object and v is its velocity. Therefore, the kinetic energy of the first bob before the collision is equal to 1/2 * 0.25 kg * v1^2. The kinetic energy of the second bob after the collision is equal to 1/2 * 0.54 kg * v2^2.Since the collision is elastic, the total kinetic energy before the collision is equal to the total kinetic energy after the collision. Therefore, we have:1/2 * 0.25 kg * v1^2 = 1/2 * 0.54 kg * v2^2Substituting v2 = 0.463 v1, we get:1/2 * 0.25 kg * v1^2 = 1/2 * 0.54 kg * (0.463 v1)^2Solving for v1, we get:v1 = 1.41 m/sThe height to which the second bob rises above its initial position is equal to the maximum height reached by the first bob. The maximum height reached by the first bob is equal to the height from which it was released, which is 0.45 m * sin(35°) = 0.26 m.Therefore, the second bob will rise to a height of 0.26 m above its initial position.

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 $0.54 \mathrm {~kg} \text {, }$