Skip to main content

Physics Library

Free Physics Study Material for CSIR NET, IIT JAM, GATE, JEST & TIFR

Physics Library
by Pravegaa Education
Visit Pravegaa.com
Login / Register
Menu
Menu

Chapter 3 Momentum, Collisions and Variable Mass Previous Year Questions with Solutions

Pravegaa PYQ Series · Volume 2: Classical Mechanics & Special Relativity

CM03: Momentum, Collisions and Variable Mass

30 previous-year questions with complete solutions, from IIT JAM, GATE, CSIR NET-JRF, JEST and TIFR, arranged from IIT JAM and GATE through to TIFR and from the most recent paper backwards. Click Show solution under any question.

  • IIT JAM: 14
  • CSIR NET: 3
  • JEST: 7
  • TIFR GS: 6
CM03.1IIT JAM 2025, Q2MCQ | 1 mark

Two point-particles having masses \(m_1\) and \(m_2\) approach each other in perpendicular directions with speeds \(v_1\) and \(v_2\), respectively, as shown in the figure below. After an elastic collision, they move away from each other in perpendicular directions with speeds \(v_1'\) and \(v_2'\), respectively.FigureThe ratio \(\dfrac{v_2'}{v_1'}\) is

  1. \(\dfrac{m_1^2v_1}{m_2^2v_2}\)

  2. \(\dfrac{m_1v_1}{m_2v_2}\)

  3. \(\dfrac{m_1^2v_2}{m_2^2v_1}\)

  4. \(\dfrac{m_1v_2}{m_2v_1}\)

Show solution

Before the collision \(m_1\) moves along \(+x\) and \(m_2\) along \(-y\); after it, \(m_1\) moves along \(-y\) and \(m_2\) along \(+x\). Conserving momentum separately along \(x\) and \(y\): \[x:\; m_1v_1=m_2v_2',\qquad y:\; m_2v_2=m_1v_1'.\] Dividing, \[\frac{v_2'}{v_1'}=\frac{m_1v_1/m_2}{m_2v_2/m_1}=\boxed{\dfrac{m_1^2v_1}{m_2^2v_2}}\]

Answer: (A) \(\dfrac{m_1^2v_1}{m_2^2v_2}\)

CM03.2IIT JAM 2025, Q17MCQ | 2 marks

Two masses, \(M_1\) and \(M_2\), are connected through a massless spring of spring constant \(k\), as shown in the figure below. The mass \(M_1\) is at rest against a rigid wall. Both \(M_1\) and \(M_2\) are on a frictionless surface. The mass \(M_2\) is pushed towards \(M_1\) by a distance \(x\) from its equilibrium position and then released. After \(M_1\) leaves the wall, the speed of the center of mass of the composite system isFigure

  1. \(\sqrt{\dfrac{k}{M_2}}\,x\)

  2. \(\sqrt{\dfrac{k}{M_1+M_2}}\,x\)

  3. \(\dfrac{\sqrt{kM_2}}{M_1+M_2}\,x\)

  4. \(\dfrac{\sqrt{kM_1}}{M_1+M_2}\,x\)

Show solution

While the spring is compressed, the wall holds \(M_1\) in place. When the spring reaches its natural length, all the stored energy is with \(M_2\): \[\frac12M_2v_2^2=\frac12kx^2\;\Rightarrow\;v_2=\sqrt{\frac{k}{M_2}}\,x .\] After this the spring starts to stretch and pulls \(M_1\) off the wall. From then on there is no external horizontal force, so the total momentum \(M_2v_2\) is conserved: \[v_{\rm cm}=\frac{M_2v_2}{M_1+M_2}=\frac{M_2}{M_1+M_2}\sqrt{\frac{k}{M_2}}\,x=\boxed{\dfrac{\sqrt{kM_2}}{M_1+M_2}\,x}\]

Answer: (C) \(\dfrac{\sqrt{kM_2}}{M_1+M_2}\,x\)

CM03.3IIT JAM 2025, Q18MCQ | 2 marks

One end of a long chain is lifted vertically from flat ground to a height \(H\) with constant speed \(v\) by a force of magnitude \(F\). Assume that the length of the chain is greater than \(H\) and that it has a uniform mass per unit length \(\rho\). The magnitude of the force \(F\) at height \(H\) is (\(g\) is the acceleration due to gravity)

  1. \(\rho(gH+v^2)\)

  2. \(\rho(gH+2v^2)\)

  3. \(\rho(2gH+v^2)\)

  4. \(\dfrac{\rho}{2}(gH+v^2)\)

Show solution

Take the lifted part (length \(y\), mass \(\rho y\), moving with speed \(v\)) as the system. The ground supports the part of the chain lying on it, so the net external force on the lifted part is \(F-\rho yg\): \[F-\rho yg=\frac{d}{dt}(\rho y\,v)=\rho v\frac{dy}{dt}=\rho v^2 .\] At \(y=H\), \[\boxed{F=\rho\left(gH+v^2\right)}\]

Answer: (A) \(\rho(gH+v^2)\)

CM03.4IIT JAM 2025, Q28MCQ | 2 marks

Two non-relativistic particles with masses \(m_1\) and \(m_2\) move with momenta \(\vec p_1\) and \(\vec p_2\), respectively, in an inertial frame \(S\). In another inertial frame \(S'\), moving with a constant speed with respect to \(S\), the same particles are observed to have momenta \(\vec p_1^{\,\prime}\) and \(\vec p_2^{\,\prime}\), respectively. Galilean invariance implies that

  1. \(m_2\vec p_1^{\,\prime}-m_1\vec p_2^{\,\prime}=m_2\vec p_1-m_1\vec p_2\)

  2. \(m_2\vec p_1^{\,\prime}+m_1\vec p_2^{\,\prime}=m_2\vec p_1+m_1\vec p_2\)

  3. \(m_1\vec p_1^{\,\prime}-m_2\vec p_2^{\,\prime}=m_1\vec p_1-m_2\vec p_2\)

  4. \(m_1\vec p_1^{\,\prime}+m_2\vec p_2^{\,\prime}=m_1\vec p_1+m_2\vec p_2\)

Show solution

Let \(\vec u\) be the velocity of \(S'\) relative to \(S\). Velocities transform as \(\vec v_i^{\,\prime}=\vec v_i-\vec u\), so \[\vec p_1^{\,\prime}=\vec p_1-m_1\vec u\quad(1),\qquad \vec p_2^{\,\prime}=\vec p_2-m_2\vec u\quad(2).\] To eliminate \(\vec u\), form \(m_2\times(1)-m_1\times(2)\): \[\boxed{m_2\vec p_1^{\,\prime}-m_1\vec p_2^{\,\prime}=m_2\vec p_1-m_1\vec p_2}\] This is \(m_1m_2(\vec v_1-\vec v_2)\), i.e. the relative velocity, which is Galilean invariant.

Answer: (A)

CM03.5IIT JAM 2024, Q26MCQ | 2 marks

A ball is dropped from a height \(h\) to the ground. If the coefficient of restitution is \(e\), the time required for the ball to stop bouncing is proportional to:

  1. \(\dfrac{2+e}{1-e}\)

  2. \(\dfrac{1+e}{1-e}\)

  3. \(\dfrac{1-e}{1+e}\)

  4. \(\dfrac{2-e}{1+e}\)

Show solution

Time of the first fall: \(t_0=\sqrt{2h/g}\), with impact speed \(u=\sqrt{2gh}\). After the \(k\)-th impact the rebound speed is \(e^ku\), and the time of flight up and down is \(2e^ku/g=2e^kt_0\). Total time: \[ T=t_0+2t_0(e+e^2+e^3+\cdots)=t_0\left(1+\frac{2e}{1-e}\right)=\sqrt{\frac{2h}{g}}\;\frac{1+e}{1-e} . \] \[ T\propto\boxed{\frac{1+e}{1-e}} \] (Checks: \(e=0\) gives \(T=t_0\); \(e\to1\) gives \(T\to\infty\).)

Answer: (B) \(\frac{1+e}{1-e}\)

Continue reading — it's free

Register with your mobile number to unlock the full chapter and all 100+ chapters of the Physics Library.

Register / Login with OTP

Already registered? Use the same button — just verify your number.