I assume you mean approaching FROM south for example, so moving north
momentum 1500 * 9 N = 13500 N
momentum 700 * 15 E = 10500 E
final mass = 1500 + 700 = 2200
2200 v cos (heading ) = 13500
2200 v sin (heading)= 10500
tan heading = 10500/13500
heading = 37.9 degrees east of north
2200 v cos (37.9) = 13500
so
v = 7.77 m/s
initial ke = (1/2) 1500 (81) + (1/2) 700 (225)
final ke = (1/2) 2200 (7.77)^2
subtract
Two cars collide at an icy intersection and stick together afterward. The first car has a mass of 1500 kg and was approaching at 9.00 m/s due south. The second car has a mass of 700 kg and was approaching at 15.0 m/s due west.
(a) Calculate the final velocity of the cars.
(b) How much kinetic energy is lost in the collision? (This energy goes into deformation of the cars.)
2 answers
No Damon, he meant moving due south. negative on y axis.
You can't use conservation of momentum since it's 2d from what MY book told me.
Use the components of Velocity vectors.
m1v1+m2v2=Vx (m1+m2)
m1v1+m2v2=Vy (m1 +m2)
Vf= sq.rt.(Vx^2 + Vy^2)
theta= arctan(Vy/Vx)
You can't use conservation of momentum since it's 2d from what MY book told me.
Use the components of Velocity vectors.
m1v1+m2v2=Vx (m1+m2)
m1v1+m2v2=Vy (m1 +m2)
Vf= sq.rt.(Vx^2 + Vy^2)
theta= arctan(Vy/Vx)
2 answers