Question

A 8 kg ball is moving with a speed of 10 m/s directly toward a 4 kg ball

which is at rest. The two balls collide and stick together. What is their
velocity immediately after the collision?

Answer 1

v₃ = 6.67 [m/s]

Step-by-step explanation:

To solve this problem we must use the linear momentum conservation theorem, which tells us that momentum is preserved before and after the collision.

Let's take the ball movement of 8 [kg] as positive.

Therefore we can built the following equation:

`(m_(1)*v_(1))+(m_(2)*v_(2))=(m_(1)+m_(2))*v_(3)`

where:

m₁ = mass of the 8 [kg] ball

m₂ = mass of the 4 [kg ] ball

v₁ = velocity of the 8 [kg} ball before the colllision = 10 [m/s]

v₂ = velocity of the 4 [kg] ball before the colllision = 0 [m/s] (at rest)

v₃ = velocity of the two balls after the collision [m/s]

`(8*10)+(4*0)=(8+4)*v_(3)\\80 = 12*v_(3)\\v_(3)=6.67 [m/s]`