Question

Two balls, ball -1 and ball – 2 are about to collide hands on a frictionless linear track. Ball – 1is traveling east with a speed 8 m/s toward ball-2. Ball-2 has a speed -2 m/s. if the mass of ball -2 is 6 times the mass of ball-1, what is the speed of ball -2 after the collision? We assume that collision is perfectly elastic.

Answer 1

Here, mass of ball -1 = m1 = m

mass of ball-2 = m2 = 6m

first apply conservation of momentum -

m1u1 + m2u2 = m1v1 + m2v2

=> m*8 + 6m*(-2) = mv1 + 6mv2

=> -4 = v1 + 6v2

=> v1 = -(6v2+4)-------------------------------------------------------(i)

again, collision is perfectly elastic, so kinetic energy shall be conserved.

so,

(1/2)m*8^2 + (1/2)*6m*(-2)^2 = (1/2)mv1^2 + (1/2)*6m*v2^2

=> 64 + 24 = v1^2 + 6v2^2

=>  v1^2 + 6v2^2 = 88

putting the value of v1 from (i) -

(6v2+4)^2 + 6v2^2 = 88

=> 36v2^2 + 48v2 + 16 - 88+6v2^2 = 0

=> 42v2^2 + 48v2 -72 = 0

=> 0.6v2^2 + 0.7v2 -1 = 0

so, v2 = [-0.7+sqrt(0.49 + 2.4)]/1.2 = [-0.7 + 1.7]/1.2 = 0.83 m/s

the other value of v2 = [-0.7 - sqrt(0.49 + 2.4)]/1.2 = [-0.7 - 1.7]/1.2 = -2.0 m/s

the second value of m2 is the same as before.

so discarding this value, v2 = 0.83 m/s