A 0.400 kg bead slides on a straight frictionless wire and moves with a velocity of 3.50 cm/s to the right, as shown below. The bead collides elastically with a larger 0.600 kg bead that is initially at rest. After the collision, the smaller bead moves to the left with a velocity of 0.70 cm/s.

What is the total kinetic energy of the system of beads after the collision?

A 1.40 * 10-4 J

B 2.45 * 10-4 J

C 4.70 * 10-4 J

D 4.90 * 10-4 J

Question

22 October, 17:02

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Answers (1)

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Mcmillan · Answer 1

B

Explanation:

KE = 1/2 m v^2 = 1/2 (0.400) (3.50^2)

= 2.45 J