12. Two ice skaters are initially at rest. The 78.2 kg male ice skater pushes his 48.5 kg female partner forward and away from his body with a velocity of 8.46 m/s. What is the male skater's velocity as a result of the push

-5.247 m/s

Explanation:

From the law of conservation of momentum,

Total momentum before collision = Total momentum after collision

Note: Since both ice skaters were initially at rest, therefore the total momentum before collision = 0

0 = mv+m'v'

-mv = m'v' ... Equation 1

Where m = mass of the male skater, m' = mass of the female skater, v = Final velocity of the male skater, v' = final velocity of the female skater

make v the subject of the equation

v = - (m'v'/m) ... Equation 2

Given: m = 78.2 kg, m' = 48.5 kg, v' = 8.46 m/s

Substitute into equation 2

v = - (48.5*8.46/78.2)

v = - 5.247 m/s.

Hence the velocity of the male skater = - 5.247 m/s.

Note: The negative sign means that the direction of the velocity of the male skater is in opposite direction to that of the velocity of the female skater

The male skater's velocity is 13.71 m/s.

Explanation:

From the law of conservation of momentum:

m1u1 = (m1 + m2) u2

m1 is the mass of the male skater = 78.2 kg

m2 is the mass of the female skater = 48.5 kg

u1 is the velocity of the male skater as a result of the push

u2 is the velocity with which the male skater pushed away the female skater = 8.46 m/s

u1 = [ (78.2+48.5) 8.46] : 78.2 = 1071.882 : 78.2 = 13.71 m/s