p0520

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Unformatted text preview: 258 CHAPTER 5. IMPULSE AND MOMENTUM 5.20 Chapter 5, Problem 20 Problem: A ball of mass m and coefficient of restitution e is dropped from a height H above a fixed incline of angle β to the horizontal as shown. The height of the point of impact relative to the ground is h. (a) Determine the ball’s velocity, v = vx i + vz k, immediately after bouncing from the incline. Express your answer in terms of e, H , β and gravitational acceleration, g . 5 (b) Now, assume the impact is perfectly elastic, β = 30o and h = 16 H . Compute the time at which the ball first strikes the ground. Express your answer in terms of g and H . . ........ ....... . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ... . . .... . .. . . . . . . . . . . . .. .. . .. . .. . .. .. . .. .. . . .. .. .. .. . .. .. . . . .. . .. . .. . . .. . .. . .. . . . .. . .. . . .. .. . . . . .. . .. . . . .. .. . .. . .. .. . . . .. . . . . .. . .. .. . .. . .. ........ .. . .. ..... ......... . .. .. .. ............. . ....... . ..... ................ .... . .. . .. .. ..... . .. .. ....................... . ....... . . ......................... . . ................ . .... . . . ................................ . . .. .. .... . ..................................... ...... .. . . .. . .. . .. ......................................... .. ............................................. ................................................ .. .. . .. . .. ..................................................... .. ... . ........................................................... ......................................................... .. . .. . .. .. ................................................................. .. ..... .. ................................................................. ............................................................................................................................................................................................................................................... ... .. ....... .. . .. ................................................................................................................................................................................................................................................... ......... . .. .. ........................................................................................................................................................................................................................................................... ................................................................................................................................................................................................................................................... . . ................................................................................................................................................................................................................................................... ................................................................................................................................................................................................................................................... f v H z h β x Solution: To solve this problem, we first use energy conservation to determine the velocity of the ball just prior to its impact with the fixed incline. Next, we determine the velocity of the ball just after the impact. This is the initial velocity for the ball’s subsequent motion. Finally, using Newton’s Second Law for the ball’s flight after bouncing from the incline, we compute the time at which the ball first strikes the ground. (a) The line of impact is normal to the inclined surface. Before proceeding, we must establish the unit vectors along and normal to the line of impact. This will facilitate determining the horizontal and vertical velocity components for computation of the ball’s trajectory after it bounces from the incline. Referring to the figure below, we find n = i sin β + k cos β and t = i cos β − k sin β To transform from nt coordinates to xz coordinates, we can use the fact that i = n sin β + t cos β and n k = n cos β − t sin β k ............................................................................................ . .............................................................................................. ...................................................................................................... ........................................................................................................ ............................................................................................................ .... . ................................................................................................................. . ......................................................................................................................... . ............................................................................................................................. ............................................................................................... ............... ............................................................................................ ........ .................................................................................................................................. ....................................................................................................................................... ................................................................................................................................................... .................................................................................................................................................. ......... ......... ....................................................................................................................................................... ............................................................................................................................................................. ................................................................................................................................................................ ..................................................................................................................................................................... ......................................................................................................................................................................... ............................................................................................................................................................................. .................................................................................................................................................................................. ...................................................................................................................................................................................... . β As he ba moves from s n a pos on o he ns an us before energy s conserved e T1 + V1 = T2 + V2 mpac s he nc ne o a mechan ca 5.20. CHAPTER 5, PROBLEM 20 259 Since the ball starts from rest, we know that T1 = 0. Also, setting the origin of our coordinate system at the ground, we have 1 0 + mg (H + h) = mV 2 + mgh 2 Solving for V gives V = 2gH Since the initial velocity of the ball is v = −V k, there follows vt vn = t · v = V sin β = n · v = −V cos β Turning to the impact between ball and the incline, the tangential velocity component is unchanged, which tells us that vt = V sin β where v = vn n + vt t is the ball’s velocity just after the impact. The impact relation tells us that since the coefficient of restitution is e, and since the incline is immovable, 0 − vn = e (vn − 0) Hence, just after the impact, the ball’s velocity is v = eV cos β n + V sin β t Substituting for n and t from above, we have v = eV cos β (i sin β + k cos β ) + V sin β (i cos β − k sin β ) = (eV sin β cos β + V sin β cos β ) i + eV cos2 β − V sin2 β k 1 sin 2β , 2 1 (1 − cos 2β ), 2 1 (1 + cos 2β ) 2 =⇒ vn = −evn = eV cos β Making use of the following trigonometric identities sin β cos β = sin2 β = cos2 β = a simple algebraic exercise yields 1 1 1−e (1 + e)V sin 2β and vz = (1 + e)V cos 2β − 2 2 1+e √ Finally, since we found above that V = 2gH , we have vx = vx = 1 (1 + e) 2gH sin 2β 2 and vz = 1 1−e (1 + e) 2gH cos 2β − 2 1+e (b) In the special case of a perfectly-elastic impact for which e = 1, the ball’s velocity components just after the impact simplify to vx = 2gH sin 2β and vz = 2gH cos 2β So, after the ball bounces, the equation governing its vertical motion is m d2 z = −mg, dt2 z (0) = h, z (0) = ˙ 2gH cos 2β 260 Solving this initial-value problem, we find z =h+ CHAPTER 5. IMPULSE AND MOMENTUM 1 2gH t cos 2β − gt2 2 The ball strikes the ground for the first time when z = 0. Thus, h+ 1 2gH t cos 2β − gt2 = 0 2 =⇒ t2 − 2 2H 2h t cos 2β − =0 g g Solving the quadratic and rejecting the physically irrelevant root with the minus sign before the radical, we have h 2H t= cos 2β + cos2 2β + g H Now, we are given β = 30o and h = t= 2H g 1 + 2 5 16 H . Substituting into the solution for t yields 2H g 1 + 2 4 5 + = 16 16 2H g 13 + 24 1 5 + = 4 16 Therefore, the ball strikes the ground for the first time at t= 5 4 2H g ...
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