Question

two balls, ball 1 and ball 2are about collide heads on a frictionless linear track. Ball 1 is traveling east with a speed 8m/s toward ball 2. Ball 2 has a speed of 4 m/s. If the mass of the ball 2 is 8 times the mass of ball 1 what is the speed of ball 2 after the collision? We asume that the collision id perfectly elastic

Answer 1

a)

m₁ = mass of ball 1 = m

m₂ = mass of ball 2 = 8 m

v_1i = initial velocity of ball 1 before collision = 8 m/s

v_2i = initial velocity of ball 2 before collision = - 4 m/s

v_1f = final velocity of ball 1 after collision

v_2f = fina velocity of ball 2 after collision

using conservation of momentum

m₁ v_1i + m₂ v_2i = m₁ v_1f + m₂ v_2f

m (8) + (8m) (-4) = m v1f + (8m) v2f

- 24 = v_1f + 8 v_2f

v_1f = - 24 - 8 v_2f                                                Eq-1

Using conservation of kinetic energy

m₁ v²_1i + m₂ v²_2i = m₁ v²_1f + m₂ v²_2f

m (8)² + (8m) (-4)²= m v²_1f + (8m) v²_2f

(8)² + (8) (-4)²= v²_1f + (8) v²_2f

using eq-1

(8)² + (8) (-4)²= (- 24 - 8 v_2f )² + (8) v²_2f

v_2f = - 1.33 m/s

speed = 1.33 m/s