Question

The skiing duo of Brian (84 kg) and Ashley (42 kg) is always a crowd pleaser. In one routine, Brian, wearing wood skis, starts at the top of a 222-m-long, 23^o slope. Ashley waits for him halfway down (i.e., at 111 meters from the top of the slope). As he skis past, she leaps into his arms and he carries her the rest of the way down the slope. (Note: The coefficient of kinetic friction between wood skis and snow is 0.04, and g = 10 m/s²)



What is Brian's speed after descending 111 meters (i.e. just before Ashley leaps into his arms)? (Hint:Draw the Free Body Diagram to find acceleration and then, use kinematics!)
m/s

A: 11.9

B: 15.8

C: 21.1

D: 28.0

Tries 0/1

What is the final velocity just after Ashley lands in Brian's arms? (Hint: Do a balance of momentum!)
m/s

A: 13.0

B: 14.6

C: 16.5

D: 18.7

Tries 0/1

What is their speed at the bottom of the slope? (Hint: Use kinematics again!)
m/s

A: 28.8

B: 33.7

C: 39.4

D: 46.1

Answer 1

Given,

mB = 84 kg ; mA = 42 kg ; L = 222 m ; theta = 23 deg ; d = 111 m ;uk = 0.04

we know that

Fnet = ma = mg sin(theta) - uk m gcos(theta)

a = g sin(theta) - uk g cos(theta)

a = 10(sin23 - 0.04 x cos23) = 3.54 m/s^2

from eqn of motion

v^2 = u^2 + 2 a S

v = sqrt (0 + 2 x 3.54 x 111) = 28.0 m/s

Hence, (D)28.0 m/s

from conservation of momentum

Pi = Pf

84 x 28 = (84 + 42) vf

vf = 18.7 m/s

Hence, (D)18.7 m/s

again using eqn of motion

v^2 = u^2 + 2 a S

v = sqrt (18.7^2 + 2 x 3.54 x 111) = 33.7 m/s

Hence, (B)33.7 m/s