# objects A,B,C with masses m,2 m,and m are kept on a friction less horizontal surface. The object A move towards B with a speed of 9 m/s and makes an elastic collision with it . B makes completely inelastic collision with C. Then velocity of C is?

Mar 13, 2018

With a completely elastic collision, it can be assumed that all the kinetic energy is transferred from the moving body to the body at rest.

$\frac{1}{2} {m}_{\text{initial"v^2 = 1/2m_"other" v_"final}}^{2}$

$\frac{1}{2} m {\left(9\right)}^{2} = \frac{1}{2} \left(2 m\right) {v}_{\text{final}}^{2}$

$\frac{81}{2} = {v}_{\text{final}}^{2}$

$\frac{\sqrt{81}}{2} = {v}_{\text{final}}$

${v}_{\text{final}} = \frac{9}{\sqrt{2}}$

Now in a completely inelastic collision, all kinetic energy is lost, but momentum is transferred. Therefore

${m}_{\text{initial"v = m_"final"v_"final}}$

$2 m \frac{9}{\sqrt{2}} = m {v}_{\text{final}}$

$2 \left(\frac{9}{\sqrt{2}}\right) = {v}_{\text{final}}$

Thus the final velocity of $C$ is approximately $12.7$ m/s.

Hopefully this helps!

Mar 13, 2018

$4$ [m/s]

#### Explanation:

The collision history can be described as

1) Ellastic collision

{(m v_0 = m v_1 + 2m v_2), (1/2m v_0^2= 1/2 m v_1^2+1/2(2m)v_2^2):}

solving for ${v}_{1} , {v}_{2}$ gives

${v}_{1} = - {v}_{0} / 3 , {v}_{2} = \frac{2}{3} {v}_{0}$

2) Inelastic collision

$2 m {v}_{2} = \left(2 m + m\right) {v}_{3}$

solving for ${v}_{3}$

${v}_{3} = \frac{2}{3} {v}_{2} = {\left(\frac{2}{3}\right)}^{2} {v}_{0} = 4$ [m/s]