Question

Starting from rest, two skaters "push off" against each other on smooth level ice, where friction is negligible. One is a woman and one is a man. The woman moves away with a velocity of +1.6 m/s relative to the ice. The mass of the woman is 59 kg, and the mass of the man is 88 kg. Assuming that the speed of light is 2.0 m/s, so that the relativistic momentum must be used, find the recoil velocity of the man relative to the ice. (Hint: This problem is similar to Example 6 in Chapter7.)

Answer 1

Using momentum conservation in Relativistic motion

Pi = Pf

Pi = 0, Since initially both are at rest

So

Pf = 0

Pm + Pw = 0

Pm = -Pw

Mm*Vm/sqrt (1 - (Vm/c)^2) = -Mw*Vw/sqrt (1 - (Vw/c)^2)

Using given values:

Mm = 88 kg, Mw = 59 kg

Vw = +1.6 m/sec

c = 2 m/sec

88*Vm/sqrt (1 - Vm^2/4) = -59*1.6/sqrt (1 - (1.6/2)^2)

88*Vm/sqrt (1 - Vm^2/4) = -157.33

1 - Vm^2/4 = (88/157.33)^2*Vm^2

Vm^2*(1/4 + (88/157.33)^2) = 1

Vm^2 = 1/[(1/4 + (88/157.33)^2)]

Vm^2 = 1.7766

Vm = sqrt (1.7766)

Vm = +1.33 m/sec or -1.33 m/sec

Since given that woman is in +ve direction, so man will be move in -ve direction

Vm = -1.33 m/sec

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