A skater of mass m standing on ice throws a stone of mass M with speed v in a ho

Question

A skater of mass m standing on ice throws a stone of mass M with speed v in a horizontal direction. Find the distance over which the skater will move in the opposite direction if the coefficient of kinetic friction between the skater and the ice is mu_k. A skater of mass 45.0 kg standing on ice throws a stone of mass 7.65 kg with a speed of 20.9 m/s in a horizontal direction. Find the distance over which the skater will move in the opposite direction if the coefficient of kinetic friction between the skater and the ice is 0.03.

Explanation / Answer

19.

m = mass of skater

vi = initial velocity of skater before throw = 0

M = mass of stone

Vi = initial velocity of stone before throw = 0

Vf = final velocity of stone = v

vf = final velocity of skater = ?

using conservation of momentum

m vi + M Vi = m vf + M Vf

m (0) + M(0) = m v + M Vf

Vf = mv/M

a = acceleration due to friction = - uk g

Vf' = final velocity of skater after stop = 0 m/s

d = stopping distance

using the equation

Vf'2 = Vf2 + 2 a d

02 = (mv/M)2 + 2 (- uk g) d

d = m2 v2/(2uk gM2)

20.

inserting the values in the above expression for "d"

d = (45)2 (20.9)2/(2 (0.03) (9.8) (7.65)2)

d = 25705 m

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