Question

Question 2 A 50 kg skateboarder is planning a trick for an upcoming competition. She starts out riding her 7 kg skateboard with initial speed of 2.5 m/s. She and the skateboard then slide down a slope of height 3 meters. At the bottom of the slope, she steps from the first skateboard onto a second stationary 7 kg. skateboard, leaving the first skateboard moving at half the speed it was before the switch. Finally, riding the 2nd skateboard, she goes up a quarter-pipe of height 2 meters. Neglecting friction, how high above the quarter-pipe could she go?

please explain each step

Answer 1

Sol) This question can be done in three steb. (i) When the girl starts and reaches to second skateboard, we can use conservation of energy as (P.E. t K.E.Dinitial = (K-E.AP.E.DE I must migh = ļ mu to as hfzo va rupa agh Vf = [ (2.5)² + 2 (9.8)(3) = 18.065 mis step-2) when she leaves the first skateboard and jumps to second skate board which is at resta Her speed when the jumbs from first skateboard is given by conservation of momentum (myt me ) Vj = mg Vg + mBVB We are given VB = Vy = 4.03 mis putting values we get

(50+7) (8.065) - (50)ug + (7) (4.03) ug = 8.63 misl So now the girl jumbs to skateboard that is at rest, tsing conservation of momentum. we get mg vg + meloJ = (mg + Moe) Var (50) (863) = 7.57 mis) 52 Now again using conservation of energy we can calculate the height reached by her Imun = mghman to hman - is . (7.50 219.80 (hman = 2.924 m

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