4.0 kg object moving with unknown velocity collides with a 6.1 kg stationaty object. after the collision, the 4.0kg object travels with a velocity of 2.8 m/s 22 degress N of W and the 6.1 object travels with a velocity of 1.5m/s 37 degrees S of W. What was the velocity of th 4.0kg object before the collision? was it elastic or inelastic?
west momentum after collision = 4*2.8*cos 22 + 6.1*1.5*cos 37 = 10.4+7.31 = 17.7kg m/s
south momentum after collision = -4*2.8*sin22+ 6.1*1.5*sin37 = - 4.20 + 5.51 = 9.71 kg m/s
same before so if s = speed and A = angle south of west
17.7 = 4 s cos A
9.71 = 4 s sin A
s sin A = 2.43
s cos A = 4.43
so
tan A =2.43/4.43 = 0.549
A = 28.7 deg S of W
s = 2.43 / sin28.7 = 2.43 / 0.481 = 5.05m/s
=================================
m v^2 before = 4 (5.05)^2 = 102
mv^2 after = 4*(2.8)^2 + 6.1*(1.5)^2 = 31.4 + 13.7 = 45.1
half of kinetic energy lost means NOT an elastic collision