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.
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.
3 answers
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.
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.
since when did physics problems have to make sense
1 answer