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 va1 = 40.0 m/s with respect to an inertial frame in which asteroid b was at rest, is deflected 30.0 ∘ from its original direction, while asteroid b travels at 45.0 ∘ to the original direction of a, as shown in

Answer: We have to use conservation of linear momentum; momentum (p) = mass x velocity so the original momentum is p=40m where m is any arbitrary mass Linear momentum must be conserved after the collision so the combined momentum of the two rocks after should equal this amount. x direction 40m=mv (a) cos30+mv (b) cos (-45) 40=v (a) cos30+v (b) cos (-45) this is the total momentum in the x-direction. we need another equation so lets take the y components it initially has no momentum in the y direction so 0=v (a) sin30+v (b) sin (-45) so system of equations looks like this. 5v (a) -.7071v (b) = 0.8660v (a) +.7071v (b) = 40 we can add these two equations to solve for v (a) and you get 1.366v (a) = 40 so v (a) = 29.3 m/s plug this in to either equation to get v (b) = 20.7 m/s for the third part of the question find the kinetic energy of the original asteroid and the kinetic energy of the two final asteroids added together. so initial kinetic energy is 1/2mv^2 =.5m (40) ^2 and final kinetic is. 5m (29.3) ^2+.5m (20.7) ^2 so you get 800m for initial and 429.245+214.245 = 643.49 for final we need to find the fraction lost so subtract and get 800-643.49 = 156.51 find out what percent of the original this amount is 156.51/800=.196 or 19.6%