Question

Long Word Problem on Newton's Third Law. Help?

Answer 1

(a)
i. Due to push on the boards skater is subjected to acceleration in the opposite direction due to Newton's Third Law.
Acceleration `a` of the ice skater having a mass `m` is found from Newton's Second Law
Force `F=ma` .....(1)
`=>a=F/m`
Inserting given values we get
`a=130.0/54.0=2.4" ms"^-1`

ii. Just after he stops pushing the boards, there is no action. Therefore, no reaction. Force is zero. Implies that acceleration is `0`.

iii. When he digs in his skates, there is action. And from Newton's Third law we know that net force slows the ice skater. Acceleration is calculated from (1)
`-a=38.0/54.0=0.7" ms"^-1`

(b)
i. Once the ice skater stops pushing the boards there is no force to change the velocity. However, the time the skater pushes on the boards has not been given. His velocity can not be calculated due to missing information.

ii. It is given that skater moves with the constant velocity for `4.00s`. His velocity is same as it was in step i above. Value can not be calculated in the absence of time for which the slowing down occurred or distance moved during slowing down.