Question

6. Two balls weighing .200 kg collide head on. Their respective velocities before the collision are 10.0 m/s and -15.0 m/s. a. The collision is completely elastic. What are their respective velocities after the collision. (Hint both conservation laws need to be used) b. If the ba;;s are made of putty and they fuse after the collision what is their speed after the collision.

Answer 1

6: mass of the ball 1, m, = 0.200kg and ball 2, m₂=0.200 1.200 kg velocity of ball 1 , v, = 1000m)s velocity of ball 2, W2 =-15.0m/s a. The collision is completely elastic. using conservation of momentum, miv,+ m2 V₂ = m,u,+ m₂ u₂ where, u, = velocity of ball 1, after collision u, - velocity of balls, after collision .: 0. 200 X 10-0 0-200 X 45.0 = 0.200 (U,+4) 0.200 ( 10.0 -15.0) = 0.200 (4, +4₂) u, +4₂= - 5.0 using conservation of energy £m, N, ² + Ź m2 12²= 2m, u,² + 2 Mau?? mir,afor, V2² = miniat mau2² 0.200 (v +122) 0.200 (4,² + u2²) >> (400) + (-16.0)>= 4, +42?

a u,²+ 4,²= 100+225 325 4,²+ u₂² squaring equation o (4+1)= -50) u " +20,42+42= 25 > u,²+42² +20,422 25 325 & 24, 4z = 25 (using equation ) 24,42= 25 - 325 u,"₂ = - 300 2 u, uz 150 = 375 again from equation (1,-42) ²+ 24, 42 = 325 [as a²76 ola-b) 7206] -> (4,-42) + 2*(-150) → (u, -U2) ? - 300 >> (4,-2) = 325 +300 - (Up-42) = 625 - 325 25 Solving equation ① and @ we have ei = 25-5 -20.0 m/s 2 10.om/s

U2 = -5.0-us 2 - 5.0 10.0 - 15,0 mis .: Velocities of the ball be for after collisari 10.0mis and -15.0 m/s are (b) Let u be the speed of their combination after collision . using conservation of momentum mi vit m2 V₂ = (m, +M2 )V 0-200X (V, +V2) - 2X 0.200 V 0.200* (10-0–15.0 0) = 2x0.200x V 2V 5.0 ó. Va 5.0 -2.5 mis