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.

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

what is the speed of a softball that travels 20m in 2secs

speed = 20/2 = 10 m/s