A point mass of $1 \mathrm{~kg}$ collides elastically with a stationary point mass of $5 \mathrm{~kg}$.…

A point mass of $1 \mathrm{~kg}$ collides elastically with a stationary point mass of $5 \mathrm{~kg}$. After their collision, the $1 \mathrm{~kg}$ mass reverses its direction and moves with a speed of $2 \mathrm{~ms}^{-1}$. Which of the following statement(s) is/are correct for the system of these two masses?
  1. Total momentum of the system is $3 \mathrm{~kg}-\mathrm{ms}^{-1}$
  2. Momentum of $5 \mathrm{~kg}$ mass after collision is $4 \mathrm{~kg}-\mathrm{ms}^{-1}$
  3. Kinetic energy of the centre of mass is $0.75 \mathrm{~J}$
  4. Total kinetic energy of the system is $4 \mathrm{~J}$

Solution


$ \begin{aligned} & v_1^{\prime}=\left(\frac{m_1-m_5}{m_1+m_5}\right) v_1+\left(\frac{2 m_5}{m_1+m_5}\right) v_5 \\ & -2=\left(\frac{1-5}{1+5}\right) v_1+0 \quad\left(\text { as } v_5=0\right) \\ & \therefore \quad v_1=3 \mathrm{~ms}^{-1} \\ & v_5^{\prime}=\left(\frac{m_5-m_1}{m_1+m_5}\right) v_5+\left(\frac{2 m_1}{m_1+m_2}\right) v_1 \\ & =0+\left(\frac{2 \times 1}{6}\right)(3)=1 \mathrm{~ms}^{-1} \\ & P_{\mathrm{CM}}=P_i=(1)(3) \\ & =3 \frac{\mathrm{kg}-\mathrm{m}}{\mathrm{s}} \\ & P_5{ }^{\prime}=(5)(1)=5 \frac{\mathrm{kg}-\mathrm{m}}{5} \\ & K_{\mathrm{CM}}=\frac{P_{\mathrm{CM}}^2}{2 M_{\mathrm{CM}}} \\ & =\frac{9}{2 \times 6}=0.75 \mathrm{~J} \\ & K_{\text {total }}=\frac{1}{2} \times 1 \times(3)^2 \\ & =4.5 \mathrm{~J} \\ & \end{aligned} $ $\therefore$ correct options are (a) and (c).

Asked in: JEE Advanced 2010 (Paper 1)

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