Question: Explain the correct answer for each part to guarantee points Q3 (7 points, suggested time 13 minutes) VBX = UB 46 A toy car coasts
Explain the correct answer for each part to guarantee points

Q3 (7 points, suggested time 13 minutes) VBX = UB 46 A toy car coasts along the curved track shown above. The car has an initial speed VA when it is at point A at the top of the track, and the car leaves the track at point B with speed v: at an angle O above the horizontal. Assume that the energy loss due to friction is negligible. (a) Suppose the toy car is released from rest at point A (VA = 0). i. After the car leaves the track and reaches the highest point in its trajectory it will be at a different height than it was at point A. Briefly explain why this is so. The car will be at a different height because it is launched at an angle. Although Uqis converted to kinte, the angle would produce a different height than at point A. 1/2 ii. Determine the speed of the car when it is at the highest point in its trajectory after leaving the track, in terms of VB and 0 . Briefly explain how you arrived at your answer. Projectile motion Speed before launch; Agh = Love , 12 gh = VB) The speed of the car is the same Using conservation of energy the speed is equal to vagh. The velocity tays constant ony as no forces actonit ( VB = tcost (b) Suppose the toy car is given an initial push so that it has nonzero speed at point A. Determine the speed VA of the car at point A such that the highest point in its trajectory after leaving the track is the same as its height at point A. Express your answer in terms of VB and 0 . Explain how you arrived at your
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