a.) Is momentum conservedin the x direction? Why or why not? If so, set up the x
ID: 1751708 • Letter: A
Question
a.) Is momentum conservedin the x direction? Why or why not? If so, set up the x momentumconservation equation.
b.) Is momentum conservedin the y direction? Why or why not? If so, set up the y momentumconservation equation.
c.) Is energyconserved before the collision? Why or why not? If so, set up theenergy conservation equation (write an equation for the speed justbefore collision in terms of h).
d.) Is energyconserved after the collision? Why or why not? If so, set up theenergy conservation equation (write an equation for the speed justafter collision in terms of hF).
e.) Is energyconserved during the collision? Why or why not? If so, set up theenergy conservation equation. If not, calculate the energy loss.(Hint: use the answers from parts c and d).
f.) List the completeset of equations derived from these principles. Explain how tosolve for qF in terms of q0 and h.
Explanation / Answer
a) the horizantal component of the velocity of the ballbefore hitting the ground is Vi = Vcos and horizantal velocity of the ball afterhitting the ground Vf = V cos since horizantal velocity remains the same thruout the motion mometum alongX- direction remains constant mvi = mvf b) velocity of the ball in negative y- directionjust before hitting the ground Viy = - [Vosin + (2gh)] h = height of fall Vo = initialvelocity velocity of the ball after hitting the ground is alongpositive direction Vfy = (2g[0.5 * h]) momentum before hitting the ground = mViy isnot equal to momentum after hitting the ground = mVfy moemtum in ydirection is not conserved. c) Energy of the ball just beforecollision KEi= {m[Vosin()]^2 / 2} + {m[(2gh)]^2} / 2 d) energy just after collision KEf= mVf^2 / 2 Vf = {[Vosin()]^2+(gh)} . e) the energy just before collision and energyafter collision are not conserved. energy lossE = KEi- KEf from the given figure we assumed that the ball fallsfrom aheight h with intial velocity Vo and after stricking the surface the ball reached amaximum altitude of h/2 . moemtum in ydirection is not conserved. c) Energy of the ball just beforecollision KEi= {m[Vosin()]^2 / 2} + {m[(2gh)]^2} / 2 d) energy just after collision KEf= mVf^2 / 2 Vf = {[Vosin()]^2+(gh)} . e) the energy just before collision and energyafter collision are not conserved. energy lossE = KEi- KEf from the given figure we assumed that the ball fallsfrom aheight h with intial velocity Vo and after stricking the surface the ball reached amaximum altitude of h/2 .Related Questions
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