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?

Answer 1

Step-by-step explanation:

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

pafterx = 64075 Ns (x-direction)

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

paftery = 0 (y-direction)

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

pbeforey = paftery

M*vy + m*v*sin(45°) = 0

vy = (1210*15.2*sin(45°))/1540

vy = -8.44 m/s

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

pbeforey = paftery

M*vx + m*v*cos(45°) = 64075

1540vx =64075 - 13005.1

vx = 33.2 m/s