Answer to Question #78273 in Mechanics | Relativity for Assertiveguy

Question

particles of mass m1 and m2 initially sitting at same position start moving simultaneously at t = 0 with velocities v1 and v2 respectively. after a time 't' the angular momentum of the particle of the particle of mass m2 with respect to the particle of mass m1 is
A. |v1-v2|^2*(m1m2/m1+m2)*t
B. |v1-v2|^2*m2*t
C. |v1-v2|^2*(m2^2/m1)*t
D. 0

Answer 1

L=r×p=mr×v
r=(v_2-v_1)t
v=(v_2-v_1)
Thus,
L=m(v_2-v_1 )t×(v_2-v_1 )=mt((v_2-v_1 )×(v_2-v_1 ))
But
(v_2-v_1 )×(v_2-v_1 )=0
Thus,
L=0
L=0
Answer: D. 0.