A 3 kg block moving at 2 m/s collides elastically with a stationary 1 kg block. After the collision, what are the velocities of the two blocks?

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Multiple Choice

A 3 kg block moving at 2 m/s collides elastically with a stationary 1 kg block. After the collision, what are the velocities of the two blocks?

Explanation:
In a one-dimensional elastic collision, both momentum and kinetic energy are conserved, and the relative speed of the two blocks after impact reverses sign compared to before. With a 3 kg block moving at 2 m/s and a 1 kg block at rest, you can use the elastic-collision formulas or solve the two conservation equations. Using the standard elastic-collision result: v1' = [(m1 − m2)/(m1 + m2)] u1 + [2m2/(m1 + m2)] u2 v2' = [2m1/(m1 + m2)] u1 + [(m2 − m1)/(m1 + m2)] u2 Plugging in m1 = 3, m2 = 1, u1 = 2, u2 = 0: v1' = [(3 − 1)/4]·2 + [2·1/4]·0 = (2/4)·2 = 1 m/s v2' = [2·3/4]·2 + [(1 − 3)/4]·0 = (6/4)·2 = 3 m/s So the heavier block ends up at 1 m/s and the lighter block at 3 m/s, both in the original direction. The other possible outcomes don’t satisfy the velocity-reversal condition required by an elastic collision.

In a one-dimensional elastic collision, both momentum and kinetic energy are conserved, and the relative speed of the two blocks after impact reverses sign compared to before. With a 3 kg block moving at 2 m/s and a 1 kg block at rest, you can use the elastic-collision formulas or solve the two conservation equations.

Using the standard elastic-collision result:

v1' = [(m1 − m2)/(m1 + m2)] u1 + [2m2/(m1 + m2)] u2

v2' = [2m1/(m1 + m2)] u1 + [(m2 − m1)/(m1 + m2)] u2

Plugging in m1 = 3, m2 = 1, u1 = 2, u2 = 0:

v1' = [(3 − 1)/4]·2 + [2·1/4]·0 = (2/4)·2 = 1 m/s

v2' = [2·3/4]·2 + [(1 − 3)/4]·0 = (6/4)·2 = 3 m/s

So the heavier block ends up at 1 m/s and the lighter block at 3 m/s, both in the original direction. The other possible outcomes don’t satisfy the velocity-reversal condition required by an elastic collision.

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