During hockey practice, two pucks are sliding across the ice in the same direction. At one instant, a 0.18-kg puck is moving at 16 m/s while the other puck has a mass of 0.14 kg and a speed of 3.8 m/s. What is the velocity of the center of mass of the two pucks?

10.66 m/s

Explanation:

Applying,

The law of conservation of momentum,

Total momentum before collision = Total momentum after collision

mu+m'u' = V(m+m') Equation 1

Where m = mass of the first puck, u = initial velocity of the first puck, m' = mass of the second puck, u' = initial velocity of the second puck, V = Velocity of the center of the two pucks

make V the subject of the equation

V = (mu+m'u')/(m+m') Equation 2

Given: m = 0.18 kg, u = 16 m/s, m' = 0.14 kg, u' = 3.8 m/s

Substitute these values into equation 2

V = [(0.18×16)+(0.14×3.8)]/(0.18+0.14)

V = (2.88+0.532)/(0.32)

V = 3.412/0.32

V = 10.66 m/s

to be honest as much as i would love to answer this. i feel as if something's missing. just saying.

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