Question

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

Answer 1

Answer:

-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

Answer 2

Answer:

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