Question

A stone is dropped at t = 0. A second stone, with twice the mass of the first, is dropped from the same point at t = 234 ms.
How far below the release point is the center of mass of the two stones at t = 326 ms? (Neither stone has yet reached the ground.)

Tries 0/10

How fast is the center of mass of the two-stone system moving at that time?

Answer 1

consider the motion of first stone :

m₁ = mass = m

t = time of travel = 326 ms = 0.326 s

v_o = initial velocity = 0 m/s

v = final velocity = ?

x = distance traveled from the starting position

a = acceleration = acceleration due to gravity = 9.8 m/s²

distance traveled by the first stone is given as

x = v_o t + (0.5) a t²

x = (0) (0.326) + (0.5) (9.8) (0.326)²

x = 0.521 m

velocity is given as

v = v_o + a t

v = 0 + (9.8) (0.326)

v = 3.195 m/s

consider the motion of the second stone :

m₂ = mass = 2 m

t = time of travel = 326 ms - 234 ms = 0.326 s - 0.234 s = 0.092 sec

v_o = initial velocity = 0 m/s

v' = final velocity = ?

x' = distance traveled from the starting position

a = acceleration = acceleration due to gravity = 9.8 m/s²

distance traveled by the first stone is given as

x' = v_o t + (0.5) a t²

x' = (0) (0.092) + (0.5) (9.8) (0.092)²

x' = 0.0415 m

velocity is given as

v' = v_o + a t

v' = 0 + (9.8) (0.092)

v' = 0.902 m/s

consider the initial starting position at origin ,

x_cm = center of mass = (m₁ x + m₂ x')/(m₁ + m₂) = ((m)(0.521) + (2m) (0.0415))/(m + 2m) = 0.201 m

velocity of center of mass is given as

v_cm = (m₁ v + m₂ v')/(m₁ + m₂) = ((m)(3.195) + (2m) (0.902))/(m + 2m) = 1.67 m/s