Question

2. In a game of billiards, in order to put the red ball into the corner pocket, it must be hit at a 38° angle away from the line connecting it to the cue ball. The cue ball hits it with a speed of 2.4 m/s, and the cue ball itself heads off at a 46° angle away from the corner pocket. The two balls have the same mass, and the shot is successful. Before After Vaf 33 \, ;= 2, 4 m/s V2; ²0 m/s 46 Vit a) How fast is each ball moving after the collision? b) Calculate the percent kinetic energy lost in the collision, AK/Kx 100%, to determine “how elastic" the collision was (perfectly elastic is 0% lost).

Answer 1

(a) Applying along conservation of linear momentum horizontal direction mui + muzi im Vip Cos 46° + m Vzf Cos 38° Here m= mass of ball Vii + Vz Vif Cos 46° + U2f Cos 38" Given Vii= 2.4 m/s Vi=0ms 2.4 m/s 0.695 Vif Vzf 0 along Vertical direction 0 to --(Vif Sin 46°) + Vzf Sin 38° + 0.788 Vif Sin 46° Vzf Sin 38° Vf 0-719 0.616 Vif = 0.616 V24 0.719 Vif [V2f = 1.17 Vif 2 Put this value in 2.4 m/s 0.695 vit +0.788 ("17 Vif) lismis 1,617 Vif 2.4 m/s If Put this value in € 124 = (1:17)(1.5 m/s) Vaf = 1.8 m/s Thus Vif ت lisms, V2f = 1.8 m/s

amit my zi ki 2 3 (6 Initial Kinetic energy = 1 / m + = 1 m vit + 14m (0)2 ki = 1 m vi Final Kinetic energy of ffm vir Ak=kg - ki = I m vir + 1 mv2 r - 1 m vi? Im vor+ 1 / m Vr - 1/2 mV ² 2 £m vzt AK ki 1 mV,; 2 m In this Voi? + Vi Vi 2 2 Vai 1.8m/s zrumis + 11.5m/s 2.ums DK lehetet) 2-1 2-0047 k: OK Ki Lho.o - 1 0.047*100 DK x 100 = ki = 4.7