Question

1.Playing a game of pool, you hit the 8 ball into the stationer 8 ball so that the 8 ball bounces of at an angle of 25 degrees if the initial velocity of the 8 ball is 3.7m/s. What is the velocity of the 8 ball after the collision?.

2. A 50 gram steel ball hangs from a 15 cm long string. The ball is brought back to an angle of 15 degrees from vertical at the bottom of the swing it contacts and bounces off the 25 gram ball on a 15cm string. Find the angle of both balls.

Answer 1

When two idon such that one identical object collida with each oth I one is motion and another is in ter collision two balts, bounces peope objeet than 3 in rest peopendicular to Ab each each other of as shown figure muisings" in figure mrvi mbias mui cases Rest efore 650 collision Collision » mv2c08650 musinoso Before collision mono ·lersion momentum of Est Ball = (3. 7m Before After collision m - Collision momentum of 2nd Balso collision mucosas - Collision momentum of Ist Ball Afur Colliston momentum î tmvisinasi of and Ball = mv, cosos i - musings? By conservation of momentum momentum Before a momentum After Collision (system) Collision (System) m (3.7) it o] = prv, casas êt mv, sinas d ruglesosî-shirsinos] (~, cos 25+ Va Cosos) î = 3.7 ê - Luisings - V2 SIG65) 9 = o f - from Eph ② V, 512 25 = Uz sings 12 = Visines & V₂ = V1 X 0.42 Put in & 10 0.90 be sin 65

ogo cosest V2 Cesos 3.7 where v2 = u(0.42) (0.90) + V110.421 (0-42) = 3.7 W(60.90)2 + (0:42)%) - 3.7 0.90 v (0.81 +0.18 ) - 3.33 vi (0.99) = 3.33 4= 3.33 14,= 3.36 mis while = v (0.42) (0.90) = 3.36X0.42 0.90 Iv=1.57 mil