Question

Question 1 of 13 > 2M 3M An object of mass 3M, moving in the +x direction at speed Vo, breaks into two pieces of mass M and 2M as shown in the figure If 0, = 70.0' and 02 = 21.0", determine the final velocities 01 and v2 of the resulting pieces in terms of vo. M =? Vi = UO U2 = DO

Answer 1

Given:

$\theta_1=70.0^0\\ \theta_2=21.0^0$

Initial momentum is,

pi= (3Mvo).

Final velocities of the pieces are:

$\vec{v_1}=v_1\cos\theta_1\hat x-v_1\sin\theta_1\hat y\\ \vec {v_2}=v_2\cos\theta_2\hat x+v_2\sin\theta_2\hat y$

Then, total final momentum is,

$\vec{p_f}=M\vec{v_1}+2M\vec{v_2}\\ =Mv_1\cos\theta_1\hat x-Mv_1\sin\theta_1\hat y+2Mv_2\cos\theta_2\hat x+2Mv_2\sin\theta_2\hat y$

According to the law of conservation of momentum, the initial and final velocities are same. then we get

$\vec{p_i}=\vec{p_f}\\ 3Mv_0\hat x=Mv_1\cos\theta_1\hat x-Mv_1\sin\theta_1\hat y+2Mv_2\cos\theta_2\hat x+2Mv_2\sin\theta_2\hat y$

From this we get

$3v_0=v_1\cos\theta_1+2v_2\cos\theta_2\\ 0.114v_1+0.622v_2=v_0..........(1)\\ 0=2v_2\sin\theta_2-v_1\sin\theta_1\\ 0.717v_2-0.940v_1=0..........(2)$

On solving these linear equations we get