Two snowboarders coast down a frictionless incline that is angled at 30 degrees.

Question

Two snowboarders coast down a frictionless incline that is angled at 30 degrees. Snowboarder A (with gear) has a mass of 70kgs, and snowboarder B, 85 kgs. They begin the race at a height of 30 meters above the ground. Immediately before a gun goes off, Snowboarder B pushes snowboarder A backwards for 0.5 seconds with an average force of 50 Newtons, pushing himself (snowboarder B) forward, in front of snowboarder A.

A. When the race starts, what is the relative velocity of snowboarder B with respect to snowboarder A as a result of the push?

B. What is B’s velocity when he arrives at the bottom of the incline?

C. When snowboarder B arrives at the bottom of the incline, how far behind is snowboarder A, in meters?

D. How much time has snowboarder B effectively gained from the push (compared to his time had he not pushed)?

E. The next year, snowboarder B tries the same shenanigan but he has a coefficient of friction between his board and the incline of 0.2. What is his finish time?

Explanation / Answer

Angle = 30o
Ma = 70 Kg
Mb = 85 Kg

Impulse is given by,
F * T = 70* v
50 * 0.5 = 70 * va
va = 0.36 m/s

Using Momentum conservation,
ma * va = mb * vb
70 * 0.36 = 85 * vb
vb = 0.30 m/s

(A)
Relativ velocity, = va + vb = 0.30 + 0.36
Vab = 0.66 m/s

(B)
Using Energy conservation,
Initial Kinetic Energy + Initial Potential Energy = Final Kinetic Energy
1/2 * 0.3^2 + 9.8*30 = 1/2*vf^2
vf = 24.25 m/s
Snowboarder B’s velocity when he arrives at the bottom of the incline, vf = 24.25 m/s


(C)
a = 9.8*sin(30) m/s^2
a = 4.9 m/s^2

vf = vi + a*t
24.25 = 0.3 + 4.9 * t
t = 4.88 s

Now, Distance travelled by Snowboarder a,
S = u*t + 1/2*a*t^2
S = - 0.36 * 4.88 + 1/2 * 4.9 * 4.88^2
S = 56.58 m

30/sin(30) - 56.58 = 3.42 m
Far behind is snowboarder A, in meters, X = 3.42 m

(D)
time has snowboarder B effectively gained from the push,
Using Energy conservation,
Initial Kinetic Energy + Initial Potential Energy = Final Kinetic Energy
1/2 * 0^2 + 9.8*30 = 1/2*vf^2
vf = 24.248 m/s

vf = vi + a*t
24.248 = 4.9 * t
t = 4.95 s

Time gained = 4.95 - 4.88 = 0.07 s

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