(a) [tex]\frac{4}{3}v[/tex]
Let's write the law of conservation of momentum for the collision:
[tex]m_A u_A + m_B u_B = m_A v_A + m_B v_B[/tex] (1)
where u refers to the initial velocities and v refers to the velocity after the collision.
The problem gives us the following information:
- car B is one-half as massive as car A, so we can write:
[tex]m_A = 2m\\m_B = m[/tex]
- car B is initiall stationary:
[tex]u_B=0[/tex]
- After the collision, car A moves in the same direction as before with a speed v/3:
[tex]u_A = v\\v_A = \frac{1}{3}u_A=\frac{1}{3}v[/tex]
So we can rewrite (1) as
[tex](2m) v = (2m) \frac{v}{3}+mv_B[/tex]
and solving for [tex]v_B[/tex], we find the final speed of car B:
[tex]v_B = \frac{6v-2v}{3}=\frac{4}{3}v[/tex]
(b) Elastic
To find if the collision is elastic or inelastic, we have to check if the total kinetic energy has been conserved or not.
The total kinetic energy before the collision is:
[tex]K_i = \frac{1}{2}m_A u_A^2 = \frac{1}{2}(2m)(v)^2=mv^2[/tex]
The total kinetic energy after the collision is:
[tex]K_f = \frac{1}{2}m_A v_A^2 + \frac{1}{2}m_B v_B^2=\frac{1}{2}(2m)(\frac{v}{3})^2+\frac{1}{2}m(\frac{4}{3}v)^2=\frac{1}{9}mv^2+\frac{8}{9}mv^2=mv^2[/tex]
The total kinetic energy has been conserved: so, the collision is elastic.
