In a one-dimensional elastic collision between a 3 kg block moving at 2 m/s and a stationary 4 kg block, what are their velocities after collision?

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Multiple Choice

In a one-dimensional elastic collision between a 3 kg block moving at 2 m/s and a stationary 4 kg block, what are their velocities after collision?

Explanation:
In a one-dimensional elastic collision, both momentum and kinetic energy are conserved, and the relative velocity after the collision reverses sign compared to before. This means the speed at which the two blocks move apart after the impact is determined by these conservation laws. Using the standard elastic-collision formulas for masses m1 = 3 kg, m2 = 4 kg with initial velocities u1 = 2 m/s and u2 = 0, the final velocities are: v1' = [(m1 − m2)u1 + 2 m2 u2] / (m1 + m2) = [(3 − 4)·2 + 2·4·0] / 7 = −2/7 ≈ −0.29 m/s v2' = [2 m1 u1 + (m2 − m1)u2] / (m1 + m2) = [2·3·2 + (4 − 3)·0] / 7 = 12/7 ≈ 1.74 m/s So the lighter block bounces back at about −0.29 m/s, while the heavier block moves forward at about 1.74 m/s. The negative sign for the first velocity indicates reversal of direction after the collision.

In a one-dimensional elastic collision, both momentum and kinetic energy are conserved, and the relative velocity after the collision reverses sign compared to before. This means the speed at which the two blocks move apart after the impact is determined by these conservation laws.

Using the standard elastic-collision formulas for masses m1 = 3 kg, m2 = 4 kg with initial velocities u1 = 2 m/s and u2 = 0, the final velocities are:

v1' = [(m1 − m2)u1 + 2 m2 u2] / (m1 + m2) = [(3 − 4)·2 + 2·4·0] / 7 = −2/7 ≈ −0.29 m/s

v2' = [2 m1 u1 + (m2 − m1)u2] / (m1 + m2) = [2·3·2 + (4 − 3)·0] / 7 = 12/7 ≈ 1.74 m/s

So the lighter block bounces back at about −0.29 m/s, while the heavier block moves forward at about 1.74 m/s. The negative sign for the first velocity indicates reversal of direction after the collision.

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