Question

In: Physics

Starting from rest, two skaters "push off" against each other on smooth level ice, where friction...

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 +2.4 m/s relative to the ice. The mass of the woman is 51 kg, and the mass of the man is 75 kg. Assuming that the speed of light is 4.6 m/s, so that the relativistic momentum must be used, find the recoil velocity of the man relative to the ice.

Solutions

Expert Solution

GIven,

velocity of the woman = v(w) = 2.4 m/s ; her mass =m(w) 51 kg

mass of the man = m(m) = 75 kg ; recoil velocity of man = v(m) = to be determind.

speed of light =c = 4.6 m/s

We know that, the relativistic momentum is given by :

P = m v / sqrt (1 - v2/c2)

In the entire process the momentum of system is conserved. So the final momentum should equal to the intial. The initial mometum of the system is zero.

P(final) = P(m) + P(w) and P(intial) = 0

P(final) =P(intial ) will give us

P(m) = - P(w)

m(m)v(m) / sqrt ( 1 - [v(m)]2 / c2) = m(w)v(w) / sqrt ( 1 - [v(w)]2 / c2)

75 x v(m) / sqrt ( 1 - [v(m)]2 / (4.6)2) = 51 x 2.4 / sqrt ( 1 - (2.4)2 / (4.6)2)

v(m) = 1.92 x sqrt ( 1 - [v(m)]2 / (4.6)2]

squaring both sides we get

[v(m)]2 = (1.92)2 x (1 - [v(m)]2 / 21.16)

[v(m)]2 = 3.69 - 0.174 [v(m)]2

[v(m)]2 + 0.174[v(m)]2 =3.69

1.174[v(m)]2 = 3.69

[v(m)] = 1.77 m/s

hence the recoil velocity of man will be v(m) = 1.77 m/s


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