91Ó°ÊÓ

Initial velocity of asteroid A is$v_{\mathrm{A}}=40\mathrm{m}/\mathrm{s}$.

The total momentum of an isolated system remains conserved during the collision. Thus, the total momentum of all the constituents of the system before the collision and after the collision will be the same.

Conservative of momentum states that initial and final momenta should be same,

$p_A+p_B=p_A^{\text{'}}+p_B^{\text{'}}$

Here, $p_A$and $p_B$are the momentum of the two asteroids before the collision and localid="1665130324336" $p_{{}_A}^{\text{'}}and{p}_B^{\text{'}}$are the momentum of the two asteroids after the collision.

Before collision, asteroid B was at rest and its momentum is 0.

So, along the x -axis, the law of conservation of momentum, gives the equation-

$m{v}_A=m{v}_A^{\text{'}}\cos 30°+m{v}_B^{\text{'}}\cos 45°....................................\left(1\right)$

Here, $v_A$is the velocity of the asteroid A before the collision and ${\mathrm{v}}_{\mathrm{A}}^{\text{'}}\mathrm{and}{\mathrm{v}}_{\mathrm{B}}^{\text{'}}$are the velocities of the two asteroids after the collision.

along the y-axis, the law of conservation of momentum, gives the equation-

$$\begin{gathered} 0=m{v}_A^{\text{'}}\sin 30°-m{v}_B^{\text{'}}\sin 45° \\ m{v}_A^{\text{'}}\sin 30°=m{v}_B^{\text{'}}\sin 45° \\ {v}_A^{\text{'}}=v_B^{\text{'}}\sqrt{2}............................\left(2\right) \end{gathered}$$

Substitute this into equation (1) , we get:

$$\begin{gathered} m{v}_A=m\left(v_B^{\text{'}}\sqrt{2}\right)\cos 30°+m{v}_B^{\text{'}}\cos 45° \\ {v}_A=v_B^{\text{'}}\left(\sqrt{2}\cos 30°+\cos 45°\right) \\ {v}_B^{\text{'}}=\frac{V_A}{\left(\sqrt{2}\cos 30°+\cos 45°\right)} \\ {v}_B^{\text{'}}=\frac{40m/s}{\left(\sqrt{2}\cos 30°+\cos 45°\right)} \end{gathered}$$

On further solving,

$$\begin{gathered} v_B^{\text{'}}=\frac{40\mathrm{m}/\mathrm{s}}{\left(\sqrt{2}\cos 30°+\cos 45°\right)} \\ =20.7\mathrm{m}/\mathrm{s} \end{gathered}$$

Substituting value of $v_B^{\text{'}}$in equation (2)

$$\begin{gathered} v_A^{\text{'}}=v_B^{\text{'}}\sqrt{2} \\ =\left(20.7m/s\right)\sqrt{2} \\ =29.3\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the velocities are $v_A^{\text{'}}=29.3\mathrm{m}/\mathrm{s}$and $v_B^{\text{'}}=20.7\mathrm{m}/\mathrm{s}$.

The kinetic energy of asteroid A before and after collision are given as:

$$\begin{gathered} K_1=\frac{1}{2}m{v}_A^2\left(\mathrm{before}\mathrm{collision}\right) \\ {K}_2=\frac{1}{2}m{v}_A^{\text{'}2}\left(\mathrm{after}\mathrm{collision}\right) \end{gathered}$$

Now, the fraction of dissipated energy is given as:

$$\begin{gathered} f=\frac{∆K}{K_1} \\ =\frac{K_1-K_2}{K_1} \\ =1-\frac{v_A^{\text{'}2}}{v_A^2} \\ =0.46 \end{gathered}$$

Hence, the fraction of the original kinetic energy of asteroid A that dissipates during the collision is 0.46 .