In a one-dimensional elastic collision, a 5 kg block moving at 2 m/s collides with a stationary 1 kg block. What are their velocities after collision?

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Multiple Choice

In a one-dimensional elastic collision, a 5 kg block moving at 2 m/s collides with a stationary 1 kg block. What are their velocities after collision?

Explanation:
In a one-dimensional elastic collision, momentum and kinetic energy are conserved. With a 5 kg block moving at 2 m/s colliding with a stationary 1 kg block, set up the two conservation equations. Momentum: 5(2) + 1(0) = 5 v1 + 1 v2 → 10 = 5 v1 + v2. Kinetic energy: 0.5(5)(2)^2 + 0.5(1)(0)^2 = 0.5(5) v1^2 + 0.5(1) v2^2 → 10 = 2.5 v1^2 + 0.5 v2^2. Substitute v2 = 10 − 5 v1 into the energy equation and solve: 10 = 2.5 v1^2 + 0.5(10 − 5 v1)^2 gives 3 v1^2 − 10 v1 + 8 = 0, so v1 = 2 or v1 = 4/3. The nontrivial, physically relevant solution for a collision is v1' = 4/3 m/s ≈ 1.33 m/s. Then v2' = 10 − 5 v1' = 10 − 20/3 = 10/3 ≈ 3.33 m/s. Thus the heavier block ends up at about 1.33 m/s and the lighter block at about 3.33 m/s, matching the given result. Momentum and energy checks confirm the values.

In a one-dimensional elastic collision, momentum and kinetic energy are conserved. With a 5 kg block moving at 2 m/s colliding with a stationary 1 kg block, set up the two conservation equations.

Momentum: 5(2) + 1(0) = 5 v1 + 1 v2 → 10 = 5 v1 + v2.

Kinetic energy: 0.5(5)(2)^2 + 0.5(1)(0)^2 = 0.5(5) v1^2 + 0.5(1) v2^2 → 10 = 2.5 v1^2 + 0.5 v2^2.

Substitute v2 = 10 − 5 v1 into the energy equation and solve:

10 = 2.5 v1^2 + 0.5(10 − 5 v1)^2

gives 3 v1^2 − 10 v1 + 8 = 0, so v1 = 2 or v1 = 4/3.

The nontrivial, physically relevant solution for a collision is v1' = 4/3 m/s ≈ 1.33 m/s. Then v2' = 10 − 5 v1' = 10 − 20/3 = 10/3 ≈ 3.33 m/s.

Thus the heavier block ends up at about 1.33 m/s and the lighter block at about 3.33 m/s, matching the given result. Momentum and energy checks confirm the values.

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