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
1
Expert's answer
2018-06-18T10:02:08-0400
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.
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