ExamBro
ExamBro
JEE Advanced · Physics · 7. COM & Collisions

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. A Total momentum of the system is \(3 \mathrm{~kg}-\mathrm{ms}^{-1}\)
  2. B Momentum of \(5 \mathrm{~kg}\) mass after collision is \(4 \mathrm{~kg}-\mathrm{ms}^{-1}\)
  3. C Kinetic energy of the centre of mass is \(0.75 \mathrm{~J}\)
  4. D Total kinetic energy of the system is \(4 \mathrm{~J}\)
Verified Solution

Answer & Solution

Correct Answer

(C) Kinetic energy of the centre of mass is \(0.75 \mathrm{~J}\)

Step-by-step Solution

Detailed explanation


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