A 20.0 g marble moving to the right at 28.0 cm/s overtakes and collides with a 19.0 g marble moving in the same direction at 16.0 cm/s. After the collision, the 19.0 g marble moves to the right at 24.3 cm/s. Find the velocity of the 20.0 g marble after the collision.

Question

Hudson Mcclain

Physics

22 November, 18:55

A 20.0 g marble moving to the right at 28.0 cm/s overtakes and collides with a 19.0 g marble moving in the same direction at 16.0 cm/s. After the collision, the 19.0 g marble moves to the right at 24.3 cm/s. Find the velocity of the 20.0 g marble after the collision.

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Answers (1)

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Anaya Gould · Answer 1

20.4 cm/s

Explanation:

From the law of conservation of momentum,

mu+m'u' = mv+m'v' ... Equation 1

Where m = mass of the first marble, m' = mass of the second marble, u = Initial velocity of the first marble, u' = initial velocity of the second marble, v = final velocity of the first marble, v' = Final velocity of the second marble.

make v the subject of the formula

v = (mu+m'u'-m'v') / m ... Equation 2

Given: m = 20 g, u = 28 cm/s, m' = 19 g, u' = 16 cm/s, v' = 24.3 cm/s.

Substitute into equation 2

v = [ (20*28) + (19*16) - (19*24.3) ]/20

v = (560+304-456) / 20

v = 20.4 cm/s.

Hence the final velocity of the 20 g marble = 20.4 cm/s