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 4 times as far as skater 2. What is the ratio m1/m2 of their masses?

Explanation / Answer

v^2 =u^2 +2as
v =0 as they stop
hecnce
U1=(2aS1) ....here a is deceleration

U2=(2aS2)

hence

U1/U2 =(S1/S2) =4 =2 .... as skater one glides 4 time sas skater 1

U1/U2= -2 as direction is opposite

also

when they push each other momentum is conserved as we have neglect friction when they are pushing each other

center of mass doesnt move

hence M1U1 +M2U2 =0

hence M1/M2 = -U2/U1 = 2

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