A point mass of 1 ~kg collides elastically with a stationary point mass of 5 ~kg . After their collision, the 1 ~kg mass reverses its direction and moves with a speed of 2 ~ms^ -1 . Which of the following statement(s)

Correct answer: A. Total momentum of the system is 3 ~kg-ms^ -1 aligned & v_1^ =((m_1-m_5)/(m_1+m_5))v_1+((2 m_5)/(m_1+m_5))v_5 & -2=((1-5)/(1+5))v_1+0 ( as )v_5=0) & v_1=3 ~ms^ -1 & v_5^ =((m_5-m_1)/(m_1+m_5))v_5+((2 m_1)/(m_1+m_2))v_1 & =0+((2 × 1)/(6))(3)=1 ~ms^ -1 & P_ CM =P_i=(1)(3) & =3 kg-m s & P_5 ^ =(5)(1)=5 kg-m 5 & K_ CM = P_ CM ^2 2 M_ CM & =(9)/(2 × 6)=0.75 ~J & K_ total =(1)/(2) ×1 ×(3)^2 & =4.5 ~J & aligned correct options are (a) and (c).

JEE Advanced 2010Paper 1

Exam: JEE Advanced
Year: 2010
Paper: Paper 1
Subject: Physics
Chapter: Center of Mass, Momentum and Collision

JEE Advanced 2010 · Paper 1 · Physics · Center of Mass, Momentum and Collision

A point mass of 1 ~kg collides elastically with a stationary point...

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? A point mass of 1 ~kg collides elastically with a stationary point mass of 5 ~kg . After their collision, the 1 ~kg

A point mass of 1 ~kg collides elastically with a stationary point mass of 5 ~kg . After their collision, the 1 ~kg

Options

ATotal momentum of the system is $3 \mathrm{~kg}-\mathrm{ms}^{-1}$
BMomentum of $5 \mathrm{~kg}$ mass after collision is $4 \mathrm{~kg}-\mathrm{ms}^{-1}$
CKinetic energy of the centre of mass is $0.75 \mathrm{~J}$
DTotal kinetic energy of the system is $4 \mathrm{~J}$

Correct answer

A. Total momentum of the system is $3 \mathrm{~kg}-\mathrm{ms}^{-1}$

Step-by-step 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} \times1 \times(3)^2 \\ & =4.5 \mathrm{~J}\\ & \end{aligned}

\therefore$ correct options are (a) and (c).

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