Doubt from Centre of mass

elastic-collision
moderate
centre-of-mass
finding-height
unsolved
pp
#1


Please write linear momentum conservation and mechanical energy conservation seperately

1 Like
#2

@Gaurav_2020_3

See, here let the final velocities of the heavy bob and light bob be v_1 and v_2 respectively.

From energy conservation initially, we can see that the initial velocities of each bob will be \sqrt{2gH} and let we call it as v_0.

Now, applying momentum conservation along the direction of approach and taking rightward velocities as positive, we get :

P_i=P_f \implies 2mv_0-mv_0=mv_2-2mv_1 \implies v_2-2v_1=v_0 \space\space\space\space(1)

From the equation for coefficient of restitution, we get :

v_{sep}=ev_{app} \implies v_2+v_1=e(v_0+v_0)=2ev_0 \implies v_2+v_1=2v_0 \space\space(as\space e=1)\space\space\space (2)

From (1) and (2) : v_1=\frac{1}{3}v_0 \space \& \space v_2=\frac{5}{3}v_0

Now, applying energy conservation for the larger bob : \frac{1}{2}(2m)v_1^2=(2m)gh_1 \implies h_1=\frac{1}{9}H

Similarly, for the other one, we will get that : \frac{1}{2}mv_2^2=mgh_2 \implies h_2=\frac{25}{9}H

Do let me know if you have any doubts.