Question

Your answer Question cover CLO2: 10 Marks Q 1)-- A 0.2 kg billiard ball moving with a 10 points speed of 10 m/s along the + x-axis, approaches an identical ball at rest, as shown in below. The collision between the balls is elastic. After the collision the incident ball moves with a speed of 6.43 m/s at an angle of 50degree below the + x-axis. What is the angle at which the target ball moves after the collision is? 10 m/s 10 m/s +x 0.2 kg 50 6.43 m/s 0.2 kg Your answer Question cover CLO3: 10 Marks Q2) --The angular position "O" of a 0.36 10 points

Answer 1

Answer
mass 50° Given of each ball= 0.2 kg Initial speed of ball 1= uF lomls Final speed of ball 1 = = 6:43 m/s Anglet below x-axis (8) Let speed (final) of boll 2= v. mis Angle (say) Conser eving momentum ( horizontally muiloso + m₂ V2 Los 6 ofx 10 + 0 = 0.226.43 x los 50+02 xv Los 4.13.+ v cos above horizontal mi ut M2 42= 10 = v loso 5.87 (i)

Conserving momentum (vertically) sind 0 = mi w, sino tama V₂ As directions alle opposite so mik, sino = M2 V₂ Sind OAX 6:43 X Sin 50 = 0% x v sind 4.925 vsind Dividing (1) by (1) (1) tonda 44.92 5.87 tand= 0.838 0 = 39.99