Question

Two billiard balls of mass m1 = 0.5 kg and m2 = 1.5 are hit at a given moment traveling m1 to the right at 1.0 m / s and m2 traveling to the left at -0.5 m / s. For this case, the magnitude of the velocities just after the shock ends by suppressing a) A perfectly elastic collision e = 1.0, b) Perfectly inelastic collision e = 0.0, c) e = 0.35.

Answer 1

(a)

A perfectly elastic collision e = 1.0,

v1 = ( m1 + m2 / m1 - m2 )*u1 + ( 2m2 / m1 + m2 )*u2

v1 = (0.5 + 1.5 / 0.5 - 1.5)* 1 + (2*1.5 / 0.5 + 1.5) * -0.5

v1 = -2.75 m/s

v2 =  (2m1 / m1 + m2 )*u1 + (m1 + m2 / m1 - m2 )*u2

v2 = (2*0.5 / 2)*1 + (2 / -1)*-0.5

v2 = 1.5 m/s

(b)

Perfectly inelastic collision e = 0.0,

m1u1 + m2u2 = (m1 + m2)*v

v = 0.5*1 + 1.5*-0.5 / (0.5 + 1.5)

v = -0.125 m/s

(c)

inelastic collision e = 0.35

From newton's law,

v2 - v1 / u2 - u1 = -e

v2 = v1 - e(u2 - u1)

From conservation of momentum,

m1u1 + m2u2 = m1v1 + m2v2

v1 = u1(m1 - em2) / m1+m2 + u2*(m2 (1+e)) / m1+m2

v1 = (0.1*(0.5 - 0.35*1.5) / 2) + (-0.5*(1.5*1.35) / 2)

v1 = -0.5075 m/s

v2 = v1 - e*(u2 - u1)

v2 = -0.5075 - 0.35*(-0.5 - 1)

v2 = 0.0175 m/s