2d momentum /physics

Question

A 1210 kg car is driving NE (at

45.0°) at 15.2 m/s when it is

struck by a moving 1540 kg car.

Afterward, they stick together and

move directly east (at 0°) at 23.3

m/s. What was the y-component

of the second car's initial velocity?

Can someone answer this

Thoroughly I would greatly appreciate it !:)

Answer 1

p_afterx = (M+m)*v (x-direction) M = 1540kg, m = 1210kg, v=23.3 m/s

p_afterx = 64075 Ns (x-direction)

p_aftery = (M+m)*v (y-direction) M = 1540kg, m = 1210kg, v=0

p_aftery = 0 (y-direction)

p_beforey = M*v_y + m*v*sin(45°) (y-direction) M = 1540kg, m = 1210kg, v=15.2 m/s

p_{beforey =} p_aftery

M*v_y + m*v*sin(45°) = 0

v_y = (1210*15.2*sin(45°))/1540

v_y = -8.44 m/s

p_beforex = M*v_x + m*v*cos(45°) (x-direction) M = 1540kg, m = 1210kg, v=15.2 m/s

p_{beforey =} p_aftery

M*v_x + m*v*cos(45°) = 64075

1540v_x =64075 - 13005.1

v_x = 33.2 m/s