Answer to Question #171710 in Physics for RHYLYN

Question

Two asteroids of equal mass in the asteroid belt between Mars and Jupiter collide with a glancing blow. Asteroid A, which was initially traveling at 40.0 m/s is deflected 30.0° from its original direction, while asteroid B, which was initially at rest, travels at 45.0° to the original direction of A (see figure below).

a) Find the speed of each asteroid after the collision.

b) What fraction of the original kinetic energy of asteroid A dissipates during this collision?

Answer 1

a) Conservation of momentum states that initial and final momenta conserve:

$$p_A +p_B=p'_A+p'_B.$$

Asteroid B was at rest, so, its momentum is 0. Thus, conservation along the x-axis:

$$mv_a=mv'_A\text{ cos}30°+mv'_B\text{cos}45°$$

Along the y-axis:

$$0=mv'_A\text{ sin}30°-mv'_B\text{ sin}45°,\\\space\\ v'_A=v'_B\sqrt2.$$

Substitute this into equation for x-axis and find velocities of B and A:

$$v'_B=20.7\text{ m/s},\\ v'_A=29.3\text{ m/s}.$$

b) Initial and final kinetic energy of A:

$$K_1=\frac m2v^2_A,\\\space\\ K_2=\frac m2 v'^2_A.$$

The fraction of dissipated energy is

$$f=\frac{\Delta K}{K_1}=\frac{K_1-K_2}{K_1}=1-\frac{v'^2_A}{v^2_A}=0.46.$$