Two ice skaters have masses m1 and m2 and are initially stationary. Their skates

Question

Two ice skaters have masses m1 and m2 and are initially stationary. Their skates are identical. They push against one another, as shown below, and move in opposite directions with different speeds. While they are pushing against each other, any kinetic frictional forces acting on their skates can be ignored. However, once the skaters separate, kinetic frictional forces eventually bring them to a halt. As they glide to a halt, the magnitudes of their accelerations are equal, and skater 1 glides 3 times as far as skater 2. What is the ratio m1/m2 of their masses?

Explanation / Answer

I see you understanding why m1*v1=m2*v2, where m1 is the mass of the boy, v1 is his initial speed, m2 is the mass of the girl, v2 is her initial speed;
? also it says that acceleration of both is a= v1/t1 =v2/t2, where t1 is time of the boy to come to his stop, t2 is time of the girl to come to her stop;
? path of the boy is
p1=0.5*a*t1^2, while path of the girl is
p2=0.5*a*t2^2, and p1=3*p2, therefore we get
0.5*a*t1^2 = 3* 0.5*a*t2^2, hence t1=t2*?2;
from ? above we get: v1/(t2*sqrt(3)) = v2/t2, hence v1 =v2*sqrt(3);
from ? above we get: m1*(v2*sqrt(3)) =m2*v2, hence
m1/m2 =1/sqrt(3);

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