Answer
(a) Asteroid A will move with a speed of 29.3 m/s.
Asteroid B will move with a speed of 20.7 m/s.
(b) The fraction of kinetic energy that dissipates during the collision is 0.196.
Work Step by Step
(a) Let's assume that asteroid A was originally moving in the +x direction. After the collision, the magnitude of the vertical components of momentum will be equal.
$m~v_A~sin(30.0^{\circ}) = m~v_B~sin(45.0^{\circ})$
$v_A = \frac{v_B~sin(45.0^{\circ})}{sin(30.0^{\circ})}$
We can set up an equation for the horizontal component of momentum.
$m~v_A~cos(30.0^{\circ})+m~v_B~cos(45.0^{\circ}) = m~v_0$
$(\frac{v_B~sin(45.0^{\circ})}{sin(30.0^{\circ})})~cos(30.0^{\circ})+v_B~cos(45.0^{\circ}) = v_0$
$v_B = \frac{v_0}{sin(45.0^{\circ})~cot(30.0^{\circ})+cos(45.0^{\circ})}$
$v_B = \frac{40.0~m/s}{sin(45.0^{\circ})~cot(30.0^{\circ})+cos(45.0^{\circ})}$
$v_B = 20.7~m/s$
$v_A = \frac{v_B~sin(45.0^{\circ})}{sin(30.0^{\circ})}$
$v_A = \frac{(20.7~m/s)~sin(45.0^{\circ})}{sin(30.0^{\circ})}$
$v_A = 29.3~m/s$
Asteroid A will move with a speed of 29.3 m/s.
Asteroid B will move with a speed of 20.7 m/s.
(b) $K_1 = \frac{1}{2}m(40.0~m/s)^2$
$K_1 = (800~m)~J$
$K_2 = \frac{1}{2}m(29.3.~m/s)^2+\frac{1}{2}m(20.7.~m/s)^2$
$K_2 = (643~m)~J$
We can find the fraction of kinetic energy remaining after the collision.
$\frac{K_2}{K_1}= \frac{(643~m)~J}{(800~m)~J} = 0.804$
The fraction of kinetic energy that dissipates during the collision is 1 - 0.804, which is 0.196
