Question

A 24.0 g marble moving to the right at 23.0 cm/s overtakes and collides elastically with a 10.0 g marble moving in the same direction at 20.0 cm/s. After the collision, the 10.0 g marble moves to the right at 22.5 cm/s. Find the velocity of the 24.0 g marble after the collision. cm/s to the right?

Answer 1

The final speed of the 24 g marble after the collision is 0.21 m/s.

Step-by-step explanation:

Given that,

Mass of the marble 1, m = 24 g = 0.024 kg

Initial speed of the marble 1, u = 23 cm/s = 0.23 m/s

Mass of the marble 2, m' = 10 g = 0.01 kg

Initial speed of the marble 2, u' = 20 cm/s = 0.2 m/s

After the collision,

Final speed of marble 2, v' = 22.5 cm/s = 0.225 m/s

We need to find the velocity of the 24.0 g marble after the collision. It is a case of elastic collision. Using the conservation of momentum as :

`mu+m'u'=mv+m'v'`

v is the speed of 24.0 g marble after the collision.

`mv=mu+m'u'-m'v'\\\\mv=0.024* 0.23+0.01* 0.2-0.01* 0.225 \\\\v=(0.00527)/(0.024)\\\\v=0.21\ m/s`

So, the final speed of the 24 g marble after the collision is 0.21 m/s.