A body of mass 30 kg moving with a velocity $20 \mathrm{~ms}^{-1}$ undergoes one dimensional elastic…

A body of mass 30 kg moving with a velocity $20 \mathrm{~ms}^{-1}$ undergoes one dimensional elastic collision with another ball of same mass moving in the opposite direction with a velocity of $30 \mathrm{~ms}^{-1}$. After collision the velocities of first and second bodies respectively are
  1. $25 \mathrm{~ms}^{-1}, 30 \mathrm{~ms}^{-1}$
  2. $30 \mathrm{~ms}^{-1}, 30 \mathrm{~ms}^{-1}$
  3. $30 \mathrm{~ms}^{-1}, 20 \mathrm{~ms}^{-1}$
  4. $40 \mathrm{~ms}^{-1}, 15 \mathrm{~ms}^{-1}$

Solution


$m_1=m_2=30 \mathrm{~kg}, u_1=20 \mathrm{~m} / \mathrm{s}, u_2=-30 \mathrm{~m} / \mathrm{s}$ By conservation of momentum, $\begin{aligned} & \mathrm{m}_1 \mathrm{u}_1+\mathrm{m}_2 \mathrm{u}_2=\mathrm{m}_1 \mathrm{v}_1+\mathrm{m}_2 \mathrm{v}_2 \\ & \Rightarrow 20-30=\mathrm{v}_1+\mathrm{v}_2 \end{aligned}$ $\therefore \quad \mathrm{v}_1+\mathrm{v}_2=-10$ $\qquad ...\mathrm{(i)}$ Also, $\mathrm{v}_2-\mathrm{v}_1=e\left(\mathrm{u}_1-\mathrm{u}_2\right)=1 \times(20-(-30))$ $\Rightarrow-v_1+v_2=50$ $\qquad ...\mathrm{(ii)}$ From eq (i) and (ii), we get $\begin{aligned} & 2 v_2=40 \Rightarrow v_2=20 \mathrm{~m} / \mathrm{s} \\ & \therefore \quad v_1=v_2-50=20-50=-30 \mathrm{~m} / \mathrm{s} \end{aligned}$ $\therefore$ Two bodies moves with velocity $30 \mathrm{~m} / \mathrm{s} \& 20 \mathrm{~m} / \mathrm{s}$ in opposite directions.

Asked in: AP EAMCET 2024 (22 May Shift 2)

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