Answer:

4.8 m/s

Explanation:

This problem is an example of conservation of momentum.The formula for conservation of momentum is at follows:

Sum of Initial Momenta=Sum of Final Momenta

M1V1+M2V1=M1V2+M2V2;

Where m is mass and v is velocity.

Let us first list what we know:

Before Collision:

M1=8 kg

V1=0 m/s(At rest)

M2=5 kg

V1=?

After Collision:

M1=8 kg

V2=3 m/s

M2=5 kg

V2=0 m/s(Stops)

Let us plug them in our conservation of momentum equation:

M1V1+M2V1=M1V2+M2V2

(8)(0)+(5)V1=(8)(3)+(5)(0)

5V1=24

V1=4.8 m/s

Therefore,the velocity of the lighter ball before collision is 4.8 m/s.