Question

Wo particles of masses m1 and m2 separated by a horizontal distance D are let go from the same height h at different times. Particle 1 starts at t=0, and particle 2 is let go at t=T. Find the vertical position of the center of mass at a time before the first particle strikes the ground. Assume no air resistance.

a) 1/2 ghT^2
b) 1/2 gh(T^2 −1)
c) 1/2 gh(T^2−2T)
d) 1/2 ghT(T−1)

Answer 1

The vertical position of the center of mass of two particles can be found by calculating the weighted average of their vertical positions.

Step-by-step explanation:

The vertical position of the center of mass of two particles of masses m1 and m2 separated by a horizontal distance D can be found by calculating the weighted average of their vertical positions. Since both particles start at the same height h, the vertical position of the center of mass at a time before the first particle strikes the ground can be calculated using the following formula:

y_cm = (m1 * h + m2 * (h - 1/2 * g * T^2)) / (m1 + m2)

Where g is the acceleration due to gravity and T is the time difference between the release of the two particles. Therefore, the correct answer is option d) 1/2 ghT(T-1).