Answer to Question #171707 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

Draw a diagram:

Apply conservation of momentum for the asteroids before and after the collision:

$p_A +p_B=p'_A+p'_B ​ .$

Assuming asteroids were moving from left to right, wrote the conservation of momentum along each axis:

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

From the second equation, find how velocity of A depends on velocity of B:

$v'_A=v'_B\sqrt2.$

Substitute this in the first equation for Ox and find final velocity of B:

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

Kinetic energy before and after the collision:

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

Dissipated energy:

$ΔK=K_1 −K_2=\frac m2(v_A^2-v'^2_A).\\\space\\ k=\frac{\Delta K}{K_1}=1-\frac{v'^2_A}{v^2_A}=0.46$