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

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

CM01: Kinematics and Newton's Laws

51 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: 16
  • GATE: 1
  • CSIR NET: 9
  • JEST: 8
  • TIFR GS: 17
CM01.1IIT JAM 2026, Q22MCQ | 2 marks

A particle is rotating along a circular path with uniform speed \(v\), as shown in the figure. While moving from the point \(P\) to \(Q\) subtending an angle \(\theta\), the magnitude of the change in its velocity is:Figure

  1. Zero

  2. \(v\cos\theta\)

  3. \(2v\sin\dfrac{\theta}{2}\)

  4. \(v\cos\dfrac{\theta}{2}\)

Show solution

The velocities at \(P\) and \(Q\) both have magnitude \(v\), and the angle between them equals the angle \(\theta\) turned by the radius. Hence \[ |\Delta\vec{v}|^2=v^2+v^2-2v^2\cos\theta=2v^2(1-\cos\theta)=4v^2\sin^2\frac{\theta}{2} \] \[ \boxed{|\Delta\vec{v}|=2v\sin(\theta/2)} \]

Answer: (C) \(2v\sin\dfrac{\theta}{2}\)

CM01.2IIT JAM 2025, Q55NAT | 2 marks

A particle is moving with a constant angular velocity 2 rad/s in an orbit on a plane. The radial distance of the particle from the origin at time \(t\) is given by \(r=r_0e^{2\beta t}\) where \(r_0\) and \(\beta\) are positive constants. The radial component of the acceleration vanishes for \(\beta=\) rad/s. (in integer)

Show solution

In plane polar coordinates \(a_r=\ddot r-r\dot\theta^2\). With \(\dot\theta=2\) rad/s and \(r=r_0e^{2\beta t}\), \(\ddot r=4\beta^2r\): \[a_r=4\beta^2r-4r=4r\left(\beta^2-1\right)=0\;\Rightarrow\;\boxed{\beta=1\ \text{rad/s}}\]

Answer: 1

CM01.3IIT JAM 2024, Q10MCQ | 1 mark

Which of the following types of motion may be represented by the trajectory, \(y(x)=ax^2+bx+c\)?

(Here \(a\), \(b\), and \(c\) are constants; \(x\), \(y\) are the position coordinates)

  1. projectile motion in a uniform gravitational field

  2. simple harmonic motion

  3. uniform circular motion

  4. motion on an inclined plane in a uniform gravitational field

Show solution

Completing the square, \[ \left(x+\frac{b}{2a}\right)^2=\frac{1}{a}\left(y-c+\frac{b^2}{4a}\right), \] which is a parabola. A projectile in a uniform field has \(x=u_xt\), \(y=u_yt-\tfrac12gt^2\), i.e. \(y\) quadratic in \(x\). SHM is along a straight line, circular motion is a circle, and motion on an inclined plane is a straight line. Hence \(\boxed{\text{projectile motion}}\).

Answer: (A) projectile motion

CM01.4IIT JAM 2023, Q46NAT | 1 mark

The sum of the \(x\)-components of unit vectors \(\dot{\hat r}\) and \(\dot{\hat\theta}\) for a particle moving with angular speed 2 rad s\(^{-1}\) at angle \(\theta=215^\circ\) is (rounded off to two decimal places)

Show solution

With \(\hat r=\cos\theta\,\hat i+\sin\theta\,\hat j\) and \(\hat\theta=-\sin\theta\,\hat i+\cos\theta\,\hat j\), \[ \dot{\hat r}=\dot\theta\,\hat\theta=\dot\theta(-\sin\theta\,\hat i+\cos\theta\,\hat j),\qquad \dot{\hat\theta}=-\dot\theta\,\hat r=-\dot\theta(\cos\theta\,\hat i+\sin\theta\,\hat j). \] The sum of the \(x\)-components is \[ -\dot\theta\left(\sin\theta+\cos\theta\right)=-2\left(\sin215^\circ+\cos215^\circ\right)=-2(-0.5736-0.8192)=2.786\approx2.79 . \] \[\boxed{2.79}\]

Answer: 2.79 (range 2.70 to 2.85)

CM01.5IIT JAM 2022, Q17MCQ | 2 marks

Consider a particle of mass \(m\) moving in a plane with a constant radial speed \(\dot r\) and a constant angular speed \(\dot\theta\). The acceleration of the particle in \((r,\theta)\) coordinates is

  1. \(2r\dot\theta^2\,\hat r-\dot r\dot\theta\,\hat\theta\)

  2. \(-r\dot\theta^2\,\hat r+2\dot r\dot\theta\,\hat\theta\)

  3. \(\ddot r\,\hat r+r\ddot\theta\,\hat\theta\)

  4. \(\ddot r\theta\,\hat r+r\ddot\theta\,\hat\theta\)

Show solution

In plane polar coordinates \[ \vec a=(\ddot r-r\dot\theta^2)\,\hat r+(r\ddot\theta+2\dot r\dot\theta)\,\hat\theta . \] With \(\dot r\) and \(\dot\theta\) constant, \(\ddot r=0\) and \(\ddot\theta=0\): \[\boxed{\vec a=-r\dot\theta^2\,\hat r+2\dot r\dot\theta\,\hat\theta}\]

Answer: (B)

CM01.6IIT JAM 2022, Q36MSQ | 2 marks

Let \((r,\theta)\) denote the polar coordinates of a particle moving in a plane. If \(\hat r\) and \(\hat\theta\) represent the corresponding unit vectors, then

  1. \(\dfrac{d\hat r}{d\theta}=\hat\theta\)

  2. \(\dfrac{d\hat r}{dr}=-\hat\theta\)

  3. \(\dfrac{d\hat\theta}{d\theta}=-\hat r\)

  4. \(\dfrac{d\hat\theta}{dr}=\hat r\)

Show solution

\[ \hat r=\cos\theta\,\hat x+\sin\theta\,\hat y,\qquad \hat\theta=-\sin\theta\,\hat x+\cos\theta\,\hat y . \] Both depend only on \(\theta\), so \(\dfrac{d\hat r}{dr}=\dfrac{d\hat\theta}{dr}=0\): (B) and (D) are false. Also \[ \frac{d\hat r}{d\theta}=-\sin\theta\,\hat x+\cos\theta\,\hat y=\hat\theta,\qquad \frac{d\hat\theta}{d\theta}=-\cos\theta\,\hat x-\sin\theta\,\hat y=-\hat r . \] \[\boxed{\text{(A), (C)}}\]

Answer: (A), (C)

CM01.7IIT JAM 2021, Q30MCQ | 2 marks

The radial component of acceleration in plane polar coordinates is given by

  1. \(\dfrac{d^2r}{dt^2}\)

  2. \(\dfrac{d^2r}{dt^2}-r\left(\dfrac{d\theta}{dt}\right)^2\)

  3. \(\dfrac{d^2r}{dt^2}+r\left(\dfrac{d\theta}{dt}\right)^2\)

  4. \(2\dfrac{dr}{dt}\dfrac{d\theta}{dt}+r\dfrac{d^2\theta}{dt^2}\)

Show solution

With \(\vec r=r\hat r\), \(\dot{\hat r}=\dot\theta\hat\theta\) and \(\dot{\hat\theta}=-\dot\theta\hat r\): \[ \vec v=\dot r\hat r+r\dot\theta\hat\theta,\qquad \vec a=\left(\ddot r-r\dot\theta^2\right)\hat r+\left(r\ddot\theta+2\dot r\dot\theta\right)\hat\theta . \] \[ \boxed{a_r=\frac{d^2r}{dt^2}-r\left(\frac{d\theta}{dt}\right)^2} \]

Answer: (B)

CM01.8IIT JAM 2021, Q50NAT | 1 mark

A particle with positive charge \(10^{-3}\,\text{C}\) and mass \(0.2\,\text{kg}\) is thrown upwards from the ground at an angle \(45^\circ\) with the horizontal with a speed of \(5\,\text{m/s}\). The projectile moves through a horizontal electric field of \(10\,\text{V/m}\), which is in the same direction as the horizontal component of the initial velocity of the particle. The acceleration due to gravity is \(10\,\text{m/s}^2\). The range is m. (Round off to three decimal places).

Show solution

\(u_x=u_y=5/\sqrt2\,\)m/s. The field gives a horizontal acceleration \(a_x=qE/m=10^{-3}\times10/0.2=0.05\,\text{m/s}^2\). The vertical motion is unaffected, so the time of flight is \[ T=\frac{2u_y}{g}=\frac{2\times5/\sqrt2}{10}=\frac{1}{\sqrt2}\,\text{s}. \] \[ R=u_xT+\frac12a_xT^2=\frac{5}{\sqrt2}\cdot\frac{1}{\sqrt2}+\frac12(0.05)\left(\frac12\right)=2.5+0.0125=2.5125 \] \[ \boxed{R\approx2.513\,\text{m}} \]

Answer: 2.513 (range 2.510 to 2.515)

CM01.9IIT JAM 2019, Q29MCQ | 2 marks

If the motion of a particle is described by \(x=5\cos(8\pi t)\), \(y=5\sin(8\pi t)\) and \(z=5t\), then the trajectory of the particle is

  1. Circular

  2. Elliptical

  3. Helical

  4. Spiral

Show solution

\[ x^2+y^2=25\cos^2(8\pi t)+25\sin^2(8\pi t)=5^2,\qquad z=5t. \] The projection on the \(xy\)-plane is a circle of radius \(5\), while \(z\) increases uniformly with time. The particle moves on a circular cylinder with constant pitch. \[ \boxed{\text{Helical}} \]

Answer: (C) Helical

CM01.10IIT JAM 2019, Q30MCQ | 2 marks

A ball of mass \(m\) is falling freely under gravity through a viscous medium in which the drag force is proportional to the instantaneous velocity \(v\) of the ball. Neglecting the buoyancy force of the medium, which one of the following figures best describes the variation of \(v\) as a function of time \(t\)?Figure

Show solution

With drag force \(kv\) opposing the motion, \[ m\frac{dv}{dt}=mg-kv\;\Rightarrow\;\int_0^v\frac{dv}{g-\frac{k}{m}v}=\int_0^t dt \;\Rightarrow\;v(t)=\frac{mg}{k}\left(1-e^{-kt/m}\right). \] The velocity starts from zero with initial slope \(g\) and rises exponentially to the terminal velocity \(v_t=mg/k\), with the slope decreasing continuously. This is curve (D). \[ \boxed{v(t)=\frac{mg}{k}\left(1-e^{-kt/m}\right)} \]

Answer: (D)

CM01.11IIT JAM 2018, Q30MCQ | 2 marks

A particle \(P\) of mass \(m\) is constrained to move on the surface of cylinder under a force \(-k\vec r\) as shown in figure (\(k\) is the positive constant). Which of the following statements is correct? (Neglect friction.)

  1. Total energy of the particle is not conserved.

  2. The motion along \(z\) direction is simple harmonic.

  3. Angular momentum of the particle about \(O\) increases with time.

  4. Linear momentum of the particle is conserved.

Figure

Show solution

In cylindrical coordinates the position vector is \(\vec r=\rho\hat\rho+z\hat z\), so \[ \vec F=-k\vec r=-k\left(\rho\hat\rho+z\hat z\right):\qquad F_\rho=-k\rho,\quad F_\phi=0,\quad F_z=-kz . \] The radial component is balanced by the normal reaction of the cylinder. Along the axis, \[ m\ddot z=-kz\;\Rightarrow\;\boxed{\text{simple harmonic motion along }z\ \left(\omega=\sqrt{k/m}\right)} \] The force is conservative and the constraint does no work, so energy is conserved; the force is not zero, so linear momentum is not conserved; the angular momentum about \(O\) does not grow secularly.

Answer: (B)

CM01.12IIT JAM 2018, Q56NAT | 2 marks

A particle of mass \(m\) is moving in a circular orbit given by \(x=R\cos\omega t\); \(y=R\sin(\omega t)\), as observed in an inertial frame \(S_1\). Another inertial frame \(S_2\) moves with uniform velocity \(\vec v=\omega R\hat i\) with respect to \(S_1\). \(S_2\) and \(S_1\) are related by Galilean transformation, such that the origins coincide at \(t=0\). The magnitude of the angular momentum of the particle at \(t=\dfrac{2\pi}{\omega}\), as observed in \(S_2\) about its origin is expressed as \(\left(mR^2\omega\right)x\). Then \(x\) is .

(Specify your answer upto two digits after the decimal point)

Show solution

In \(S_2\) (with \(v=\omega R\)): \begin{align*} \vec r\,'&=(x-vt)\hat i+y\hat j=\left(R\cos\omega t-vt\right)\hat i+R\sin\omega t\,\hat j,\\ \vec v\,'&=(\dot x-v)\hat i+\dot y\hat j=\left(-R\omega\sin\omega t-v\right)\hat i+R\omega\cos\omega t\,\hat j . \end{align*} \begin{align*} \vec L'=m\,\vec r\,'\times\vec v\,'&=m\left[\left(R\cos\omega t-vt\right)R\omega\cos\omega t+R\sin\omega t\left(R\omega\sin\omega t+v\right)\right]\hat k\\ &=m\left(R^2\omega-vtR\omega\cos\omega t+vR\sin\omega t\right)\hat k . \end{align*} At \(t=2\pi/\omega\): \(\cos\omega t=1\), \(\sin\omega t=0\) and \(vt=2\pi R\), so \[ \vec L'=mR^2\omega\left(1-2\pi\right)\hat k=-5.28\,mR^2\omega\,\hat k\;\Rightarrow\;\left|\vec L'\right|=5.28\,mR^2\omega,\qquad\boxed{x=5.28} \]

Answer: 5.28 (range 5.25 to 5.30)

CM01.13IIT JAM 2017, Q53NAT | 2 marks

In planar polar co-ordinates, an object's position at time \(t\) is given as \((r,\theta)=\left(e^t\,m,\ \sqrt8\,t\ \text{rad}\right)\). The magnitude of its acceleration in \(m/s^2\) at \(t=0\) (to the nearest integer) is

Show solution

\(r=e^t\), \(\dot r=\ddot r=e^t\); \(\dot\theta=\sqrt8\), \(\ddot\theta=0\). At \(t=0\), \(r=\dot r=\ddot r=1\). \[ a_r=\ddot r-r\dot\theta^2=1-8=-7,\qquad a_\theta=r\ddot\theta+2\dot r\dot\theta=0+2\sqrt8=\sqrt{32}. \] \[ |\vec a|=\sqrt{49+32}=\sqrt{81}=\boxed{9\ \text{m/s}^2} \]

Answer: 9 (range 9 to 9)

CM01.14IIT JAM 2017, Q60NAT | 2 marks

At \(t=0\), a particle of mass \(m\) having velocity \(v_0\) starts moving through a liquid kept in a horizontal tube and experiences a drag force \(\left(F_d=-k\dfrac{dx}{dt}\right)\). It covers a distance \(L\) before coming to rest. If the times taken to cover the distances \(L/2\) and \(L/4\) are \(t_2\) and \(t_4\) respectively, then the ratio \(t_2/t_4\) (ignoring gravity) is

(Specify your answer to two digits after the decimal point)

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