Q130 / 160 · TS EAMCET 2018 (05 May Shift 1) · Physics

Collision takes place between two solid spheres denoted as 1 and 2 . The initial velocities of the spheres are $u_1=3 \mathrm{~m} / \mathrm{s}$ and $u_2=1.5 \mathrm{~m} / \mathrm{s}$ and the final velocities are $v_1=2.5 \mathrm{~m} / \mathrm{s}$ and $v_2=3.5 \mathrm{~m} / \mathrm{s}$. The coefficient of restitution between the materials of the spheres is nearly

  • Correct answer
    A
    0.67
  • B
    0.78
  • C
    0.83
  • D
    0.96

Solution

Coefficient of restitution, $$ \begin{aligned} & e=\frac{\text { Relative velocity of separation }}{\text { Relative velocity of approach }} \\ & \Rightarrow \quad e=\frac{v_2-v_1}{u_1-u_2}=\frac{3.5-25}{3-1.5}=\frac{1}{1.5}=\frac{2}{3}=0.67 \end{aligned} $$

Hence, the correct answer is option: 'A'.

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