Question

1. A ping pong ball (ball #1, mass of 2g) is held just above a tennis ball (ball #2, mass of 57g) with their centers vertically aligned. When the balls are released, they fall through a vertical distance of 1.27 meters. a) Prove that each ball is traveling at 5.00 m/s just before it hits the ground. The tennis ball (#2) hits the ground and bounces elastically off of it before the ping pong ball (#1) actually hits #2. Since the ground is very large, it doesn't move, which means that the velocity of #2 just reverses; it begins moving upward with velocity +5 m/s after it hits the ground. However, #1 has just fallen its own 1.27 meters, so it is still moving downward with velocity - 5 m/s. Thus Vii = -5 m/s and vzi = +5 m/s b) Determine the velocities of each ball just after they collide elastically. c) How high will each ball rise after this collision? (Use energy or kinematics, whichever you prefer.) d) Explain briefly why the ping pong ball is able to go higher than its original 1.27 meters. e) Confirm that energy is conserved from just after the drop until they rise to the highest point on their bounces. Hint: at both the beginning and end, each has only pure PE.

Answer 1

solution : Ball 1, m,= = 2g Ball 2 ) m2=57g yta 127 m y=0 (a) using equation of kinematics uz v² = galy-yo) -2011.27) 1 final initial Velocity velocity a = 09.8 mis? 0o=0 mis (for both balls tin-y-direction しっ 2x9.8x1.27 smis v=smis for they Both Balls Balls before hits the ground. linear momentum collision (6) In. elastic and K.E. of the system remains conserved. m, w, it malze = m. uif + MzUzf 2 m, un + Ź mauli Import + 1 masa CS Scanned with CamScanner

- m2 Vif vi mi m, tmz D + 2m2 tri m, tmz c 2-57 (-5) + 2157) (5) 59 2+57 -0.93% (-5) +11.93x5) 14.31 mist velocity of Ball 1 Uff after collision Uzt 2m, vi i t' m2 m, uzi m, tm2 m, tmz 2(2)(-5) + = 55 (5) 59 59 -0.34 + 4.66 Uzf 4.32 mis Velocity of Ball 2 after collision (C) v2. Yo=0 v² 2aly-yo) (fon Both Balls as initially they y=0 are at v=0 ( То To find put max. height we have to Vzo at that height 0.2 = zay = +0²=+2 gy CS scanned with CamScanner

for Ball 1 » Vo Uif 14.31 mis Y,= (14.312 = 10.45 m 2x9.8 y, 10.45m for Ball 2 → vo= Uzf 4.32 mis Y2= (4.32² -0.95m 2x9.8 Y2=0.95m ( d ) to 1.27 m because Ball 1 lie. Ping Pong ball) is able go higher than its original during collision gets from Ball energy Ball has higher 1 (because mz>m,). it 2 . As 2 higher energy than Ball CS Scanned with CamScanner

(e) Initially both balle are Yo=1.27mto Same height ie. 127 m. (P.E.), = Potential energy of ball 1. (P.E)2 1) 1) )) .2 اره را P.E.-0 y=0 2 Before Collision At point (K.E), =(K.E.),=0 P.e.), mig (1.27) = 0.002x9.8x1.27 0.025J Mag(127) P.E.), = 0 (P.E.)2 → (P.E.)2 0.71J At point 2 9 As we know, velocities at point Vii=-5 mis Uzi=5mis (k.E), = {m,via 0.25J (K.E.), -0.25 J = (P.E.), 1 At point Hence, During this motion energy is conserved for Ball ② At point After collision At point - (P.E), (P. E)zi = 0 li (K.E.) li 4 m, vt 0.20J (k.€ ) 2i = 1 m₂ 5244 = 0.53J 2 CS Scanned with CamScanner

At point (K.E.),4 = (K.E.) 2870 (P.E.) migy, If 0.002*9.8 x 10.45 = 0.203 (P-E) 24 = m29y2 = 0.057x9.8*0.95 = 0.53 J. Applying conservation of energy at point ① and 2 → For Ball 1 → (K. E),4t (p.5) (K.Elit P. E.)ji of = At point at Point 0 + 0.20 0.20 to LHS RHS - Hence conserved for Ball 1. energy is For Ball 2 s (K.E.) aft + P.E.) 24 = (K.E.) zi + (P.E. At point At point 0 + 0.53 0.53+ LHS =RHS Hence, Energy for Ball is conserved 2. CS Scanned with CamScanner