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Center of mass of two-stone system moving; speed after impact

A stone is dropped at t = 0. A second stone, with 3 times the mass of the first, is dropped from the same point at t = 190 ms. (a) How far below the release point is the center of mass of the two stones at t = 440 ms? (Neither stone has yet reached the ground.) (b) How fast is the center of mass of the two-stone system moving at that time?

A cart with mass 380 g moving on a frictionless linear air track at an initial speed of 1.2 m/s undergoes an elastic collision with an initially stationary cart of unknown mass. After the collision, the first cart continues in its original direction at 0.49 m/s. (a) What is the mass (in g) of the second cart? (b) What is its speed after impact? (c) What is the speed of the two-cart center of mass?

Solution Preview

Assume the mass of the first stone is m, then the mass of the second stone would be 3m.
At t = 440 ms, the position of the first stone is

h1 = (1/2)*g*t^2 = (1/2)*9.8*0.44^2 = 0.95 (m) (below the release point)

At t = 440 ms, the position of the ...

Solution Summary

The center of mass of a two-stone system moving and the speed after impact is examined.