In Fig. 9-64, block A (mass 1.6 kg)slides into block B (mass 2.4 kg), along a frictionless surface. The directions of three velocities before (i) and after (f) the collision are indicated; the corresponding speeds are ${\mathit{v}}_{\mathbf{A}\mathbf{i}}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{5}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}$, ${\mathit{v}}_{\mathbf{B}\mathbf{i}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{5}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}$, and ${\mathit{v}}_{\mathbf{B}\mathbf{f}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{9}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}$. What are the (a) speed and (b) direction (left or right) of velocity ${\overrightarrow{\mathbf{v}}}_{\mathbf{A}\mathbf{F}}$? (c) Is the collision elastic?

$$\begin{gathered} v_{Bf}=4.9m/s \\ {v}_{Bi}=2.5m/s \\ {v}_{Ai}=5.5m/s \\ {m}_A(massoftheblockA)=1.6kg \\ {m}_B(massoftheblockB)=2.4kg \end{gathered}$$

Let m_A be the massof the block A on the left, v_Ai be its initial velocity and v_Afbe its final velocity. Let m_B be the mass of the block B on the right, v_Bi be its initial velocity and v_Bfbe its final velocity. The momentum of the two-block system is conserved.

Formula:

Initial momentum= Final momentum.

${\mathit{m}}_{\mathbf{A}}{\mathit{v}}_{\mathbf{A}\mathbf{i}}\mathbf{+}{\mathit{m}}_{\mathbf{B}}{\mathit{v}}_{\mathbf{B}\mathbf{i}}\mathbf{=}{\mathit{m}}_{\mathbf{A}}{\mathit{v}}_{\mathbf{A}\mathbf{f}}\mathbf{+}{\mathit{m}}_{\mathbf{B}}{\mathit{v}}_{\mathbf{B}\mathbf{f}}$

(a)

$$\begin{gathered} v_{Af}=\frac{m_A{v}_{Ai}+m_B{v}_{Bi}-m_B{v}_{Bf}}{m_A} \\ =\frac{\left(1.6kg\right)(5.5m/s)+\left(2.4kg\right)(2.5m/s)-\left(2.4kg\right)(4.9m/s)}{1.6kg} \\ =1.9m/s \end{gathered}$$

(b) The block continues going to the right after the collision.

(c)

To see whether the collision is elastic, we compare the total kinetic energy before the collision with the total kinetic energy after the collision. The total kinetic energy before is

$$\begin{gathered} K_i=\frac{1}{2}{m}_A{\left(v_{Ai}\right)}^2+\frac{1}{2}{m}_B{\left(v_{Bi}\right)}^2 \\ =\frac{1}{2}\left(1.6kg\right){\left(5.5m/s\right)}^2+\frac{1}{2}\left(2.4kg\right){\left(2.5m/s\right)}^2 \\ =31.7J \end{gathered}$$

The total kinetic energy after is

$$\begin{gathered} K_f=\frac{1}{2}{m}_A{\left(v_{Af}\right)}^2+\frac{1}{2}{m}_B{\left(v_{Bf}\right)}^2 \\ =\frac{1}{2}\left(1.6kg\right){\left(1.9m/s\right)}^2+\frac{1}{2}\left(2.4kg\right){\left(4.9m/s\right)}^2 \\ =31.7J \end{gathered}$$

Since$K_i$ and$K_f$ are equal, the collision is found to be elastic.