Question
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.)
(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.)
Answers
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
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
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)