In: Physics
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.
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