# An elastic collision in two dimensions

Doc Al
Mentor
and How do I get the K? I know its 1/2 mv2, my guess is that

.5(2.00kg)[(13.42m/s)(cos 63.43)]2
Since 1/2mV^2 = 1/2m(Vx^2 + Vy^2), you can find the contributions to the total KE of each component separately.

2.00kg object 1.00kg object total
px 12.00kgm/s 5.00kgm/s 9.00kgm/s 11.00kgm/s 21.00kgm/s 16.00kgm/s
py -24.00kgm/s -26.00kgm/s -9.00kgm/s -19.00kgm/s -33.00kgm/s -45.00kgm/s
K .. .. .. .. .. ..

.5(2.00kg)(13.42)^2(cos 63.43) = 80.55 J
.5(2.00kg)(13.42)^2(sin 63.43) = 161.1 J

.5(1.00kg)(12.73)^2(cos 45) = 57.3 J
.5(1.00kg)(12.73)^2(cos

man im really confused right now, those arent right are they

Doc Al plz come back! I'm so close to finished, i've been on these forums for the past 7 hours trying to finish all this homework

Doc Al
Mentor
Note that when you calculate the KE, you are already given the speeds for the particles before the collision and the speed of the green particle after the collision. So for them there is no need to use components, just use KE = 1/2mV^2, where V is the given speed of the particle.

For the speed of the red particle after the collision, you'll have to figure that out. Using momentum equations, you can solve for Vx & Vy for that particle. Then you can calculate its KE either by first finding the total speed (V^2 = Vx^2 + Vy^2) or just by using the components of its velocity:
KE = 1/2mV^2 = 1/2mVx^2 + 1/2mVy^2.