During the Olympic ice competition, Boris (m = 75 kg) glides at 1.8 m/s to a stationary Juliette (52 kg) and hangs on. How far will the pair slide after the “collision” if the coefficient of kinetic friction between the ice and their skates is .042?
My answer:
Conservation of momentum:
(75)(1.8)+52(0)=v(75+52)
v=1.06299
Force of friction:
Ff=u(Fn)
Ff=(.042)(127)(9.8)
Ff=52.2732
Impulse:
delta P=Ft
135=52.2732t
t=2.582585
d=vt
d=(2.582585)(1.062992126)
d=2.725m
Is this correct. Thanks for your help.
Calculate the velocity of the pair after they "connect", using conservation of momentum. Let the connection velocity be V.
You got that part right, but left out the units. V = 1.063 m/s
After that, use conservation of energy to get the sliding distance, X.
(1/2)MV^2 = MgUX
X = (1/2)*V^2/(Ug) = 1.37 m
Your equation d = v t is not correct since v is decreasing.
U is the friction coefficient,0.042
That isn't very far for a figure skating pair to glide. They must have been deliberately scraping edges to slow down. The speed is also slow, and won't score them many points.
since when did physics problems have to make sense
Your calculations and approach are mostly correct, but there is a small error in your calculation of time (t) for the impulse.
Let's go through the calculations step-by-step:
Step 1: Conservation of momentum:
The initial momentum of Boris is given by (mass of Boris) × (velocity of Boris) = (75 kg) × (1.8 m/s) = 135 kg·m/s.
Since Juliette is initially stationary, her momentum is zero.
The final momentum of the pair after the collision is given by (mass of Boris + mass of Juliette) × (final velocity of the pair).
So, we have: (75 kg + 52 kg) × (final velocity of the pair) = 135 kg·m/s.
Simplifying the equation, we get: (127 kg) × (final velocity of the pair) = 135 kg·m/s.
Therefore, the final velocity of the pair (v) is approximately 1.063 m/s.
Step 2: Force of friction:
The force of friction between the ice and their skates is given by the product of the coefficient of kinetic friction (u) and the normal force (Fn).
The normal force (Fn) is the weight of the pair, which is equal to the sum of their masses multiplied by the acceleration due to gravity (9.8 m/s^2).
So, Fn = (mass of Boris + mass of Juliette) × (acceleration due to gravity) = (75 kg + 52 kg) × (9.8 m/s^2) = 127 kg × 9.8 m/s^2.
Now we can calculate the force of friction:
Force of friction (Ff) = coefficient of kinetic friction (u) × normal force (Fn) = 0.042 × (127 kg × 9.8 m/s^2) = 52.2732 N.
Step 3: Impulse:
The impulse experienced by the pair is equal to the product of the force of friction and the time (t) over which the force is applied.
Impulse = Force × time.
Impulse = 52.2732 N × t.
Since impulse is equal to the change in momentum (135 kg·m/s), we can set up the equation: 52.2732 N × t = 135 kg·m/s.
Now, we can solve for t:
t = (135 kg·m/s) / (52.2732 N) ≈ 2.584 s.
Step 4: Distance (d):
The distance traveled (d) by the pair is equal to the product of the final velocity (v) and the time (t).
d = v × t = 1.063 m/s × 2.584 s ≈ 2.748 m.
So, the correct answer is approximately 2.748 meters.
Note: It's always a good idea to double-check the values and units in each step to ensure accuracy.
According to this reference:
http://hypertextbook.com/facts/2004/GennaAbleman.shtml
the figure skate/ice friction coefficient you were given is about ten times too high. Low coefficients can be achieved by skating on edges of the blades, in the direction the blades are aligned.