west momentum = 1500 * 90
north momentum = 1400 * 72
final west momentum = 2900 Vwest
final north momentum = 2900 Vnorth
final = initial
speed = sqrt(Vwest^2 + Vnorth^2)
tan angle west of north = Vwest/Vnorth
A 1500 kg car traveling West at 90.0 km/h collides with a 1400 kg truck traveling North at 72.0 km/h.The two vehicles entangle after the collision and head off as one. What is the velocity of the wreckage immediately after the collision?
2 answers
Car represents the X vector and the truck represent the Y vector.
Vfx = McVc / (Mc +Mt) because the Vx for the truck is zero
Vfy = MtVt / (Mc +Mt) because the Vy for the car is zero
draw the right triangle and then Vfc+t combo = square root of Vfx squared +Vfy squared.
Direction is using the tan function for a right triangle
Answer is 16.1 m/s or 58 km/h @ 37 degrees.
Vfx = McVc / (Mc +Mt) because the Vx for the truck is zero
Vfy = MtVt / (Mc +Mt) because the Vy for the car is zero
draw the right triangle and then Vfc+t combo = square root of Vfx squared +Vfy squared.
Direction is using the tan function for a right triangle
Answer is 16.1 m/s or 58 km/h @ 37 degrees.
2 answers