91影视

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-

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

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-

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

Substitute this into equation (1) , we get:

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

On further solving,

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

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:

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

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