Answer to Question #125632 in Physics for Mwansa Kunda

Question

Two asteroids of equal mass in the asteroid belt between mars and Jupiter collide with a glancing blow. Asteroid A, which was initially travelling at 40m/s is deflected 30 degrees from its original position while Asteroid B travels at 45 degrees to the original direction of A. Find the speed of each Asteroid after collision and what fraction of the original kinetic energy of Asteroid A dissipates during this collision

Answer 1

According to the law of conservation of momentum, for the initial and final momenta we can write

$$\vec{p}_A+\vec{p}_b=\vec{p'}_A+\vec{p'}_B.$$

If the asteroid B was initially at rest, its momentum is zero, and the equation becomes

$$\vec{p}_A=\vec{p'}_A+\vec{p'}_B.$$

Conservation of momentum along x- and y-axes:

$$Ox: mv_A=mv'_A\text{ cos}30°+mv_B'\text{ cos}45°,\\ Oy: 0=mv_A'\text{ sin}30°-mv_B'\text{ sin}45°.$$

The equation for y-axis gives us a relation between final speeds of A and B:

$$v'_A\text{ sin}30°=v'_B\text{ sin}45°,\\ v_A'=v'_B\sqrt2.$$

Therefore, if we substitute this into the equation for x-axis, we will get the final speed of A:

$$v_A=v'_B\sqrt2\text{ cos}30°+v'_B\text{ cos}45°,\\ v'_B=\frac{v_A}{\sqrt2\text{ cos}30°+\text{ cos}45°},\\\space\\ v'_A=\frac{40}{\text{ cos}30°+\sqrt2\text{ cos}45°}=20.7\text{ m/s}.\\\space\\ v'_A=v'_B\sqrt2=29.3\text{ m/s}.$$

To find what fraction of the original kinetic energy of asteroid A dissipates during this collision, determine the initial kinetic energy of A:

$$K_i=\frac{mv_A^2}{2}.$$

The final energy of A was

$$K_f=\frac{m{v'_A}^2}{2}.$$

Energy dissipated during the collision is

$$\Delta K=K_i-K_f.$$

It corresponds to the following fraction of the original kinetic energy:

$$\epsilon=\frac{\Delta K}{K_i}=1-\frac{v_A'^2}{v_A^2}=0.46.$$

About 46% of original energy dissipated.