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A(n) 12 g object moving to the right at 29 cm/s overtakes and collides elastically with a 24 g object moving in the same direction at 14 cm/s. Find the velocity of the slower object after the collision. Answer in units of cm/s.

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Answer: 16.9cm/s

Step-by-step explanation:

According to the principle of conservation of linear momentum which states that the sum of momentum of bodies before collision is equal to the sum of their momentum after collision.

Momentum = mass × velocity of the body

Let m1 be mass of the first body= 12g

m2 be mass of the second body= 29cm/s

v1 be velocity of the first body= 24g

v2 be velocity of the second body= 14cm/s.

Note that both objects will move with a common velocity (v) after collision. Using the formula

m1v1 + m2v2 = m1v +m2v

12(24) + 29(14) = (12+29)v

288+406 = 41v

694 = 41v

v = 694/41

v = 16.9cm/s

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