Previous Year Questions with Solutions on Mechanics

1.0Introduction

Mechanics in JEE is a key part of mechanics, focusing on the relationship between forces and motion. It includes Newton's Laws of Motion, explaining inertia, force-mass-acceleration, and action-reaction principles. It covers Work, Energy, and Power, energy transfer, and conservation of mechanical energy. Friction addresses forces resisting motion, and Circular Motion deals with centripetal forces and acceleration. Conservation Laws focus on momentum and energy conservation. Mastering dynamics is essential for solving JEE Physics problems and is a crucial part of the syllabus.

2.0Key Concepts to Remember

Newton's First Law: The Law of Inertia:A body at rest will remain at rest, and a body in motion will continue in motion with a constant velocity unless acted upon by a net external force.

Newton's second Law:Net force acting on a system equals rate of change of linear momentum of system 

$F_{\text{net}}=\frac{dp}{dt}=\frac{d(mv)}{dt}=m\frac{dv}{dt}=ma$

Newton's third  Law:Forces always occur in pairs. If object A exerts a force on object B(called Action), object B simultaneously exerts a force (called Reaction) of equal magnitude but in the opposite direction on object A.

Work-Energy Theorem:The net force that acts on an object does work on the object equal to the change in its kinetic energy

$W=\Delta K=K_f-K_i$

Conservation of Linear Momentum:If net external force on the system is zero, then linear momentum of the system remains conserved.

${\vec{F}}_{\text{ext}}=\frac{d\vec{p}}{dt}$

$\text{If }{\vec{F}}_{\text{ext}}=0\Rightarrow \frac{d\vec{p}}{dt}=0\Rightarrow \vec{p}=\text{constant}\Rightarrow {\vec{p}}_{\text{initial}}={\vec{p}}_{\text{final}}$

Impulse Momentum Theorem: This theorem states that the impulse exerted on an object is equivalent to the change in its momentum.This theorem can be derived from Newton's Second Law of Motion. Since the rate of change of momentum is instantaneous, the impulse is effectively equal to the change in momentum.

$\vec{I}=\vec{F}t=\left({\vec{p}}_2-{\vec{p}}_1\right)$

Angular Impulse:The change in angular momentum of a system resulting from the action of a torque over an infinitesimal time interval is known as the angular impulse imparted to the system by the torque.

$\vec{J}=\int \vec{\tau }dt$

$\vec{J}={\tau }_{\text{avg}}\int dt$

Coefficient of Restitution:The ratio of magnitudes of impulse of restitution to that of deformation is called the coefficient of restitution and is denoted by e. 

$e=\frac{\text{impulse of recovery}}{\text{impulse of deformation}}=\frac{\int Rdt}{\int Ddt}$

$e=\frac{\text{velocity of separation along line of impact}}{\text{velocity of approach along line of impact}}=\frac{|v_B-v_A|}{|u_A-u_B|}$

Important Formulas

Newton's second Law	$F_{\text{net}}=\frac{dp}{dt}=\frac{d(mv)}{dt}=m\frac{dv}{dt}=ma$
Impulse	$d\vec{I}=\vec{F}dt=d\vec{P}$
Average Force	${\vec{F}}_{\text{avg}}=\frac{\Delta \vec{p}}{\Delta t}$
Impulse Momentum Theorem	$\vec{I}=\vec{F}t=\left({\vec{p}}_2-{\vec{p}}_1\right)$
Angular Impulse	$\vec{J}=\int \vec{\tau }dt$
Tension in Rod	$T=F\left(1-\frac{x}{L}\right)$
Variable Mass Problem	$F_{\text{gas}}=v_{\text{rel}}\left|\frac{dm}{dt}\right|$
Limiting Friction	$f_L={\mu }_{s}N$
Kinetic Friction	$f_k={\mu }_{k}N$
Contact Force
Angle of Friction	$\tan \lambda ={\mu }_s$
Angle of Repose or Angle of Sliding	${\mu }_s=\frac{f_L}{N}=\tan \theta \Rightarrow \theta ={\tan }^{-1}{\mu }_s$
Work	$W=Fscos\theta$
Work Done by Variable Force	$W={\int }_{r_1}^{r_2}\vec{F}\cdot d\vec{r}$
Work done by Spring Force	$W_s=\frac{1}{2}k(x_i^2-x_f^2)$
Gravitational Potential Energy	$U_f-U_i=mgh$
Relationship between conservative force field and potential energy	$\vec{F}=-\nabla U=-\text{grad}(U)=\frac{\partial U}{\partial x}\hat{i}-\frac{\partial U}{\partial x}\hat{j}-\frac{\partial U}{\partial x}\hat{k}$
Kinetic Energy	$K.E=\frac{1}{2}m{v}^2$
Power	$Power=\frac{Workdone}{Time}$
Angular Displacement	$\Delta \theta ={\theta }_f-{\theta }_i$
Angular Velocity	$\omega =\frac{\Delta \theta }{\Delta t}$
Moment of Inertia For Discrete System of Particles	$I=m{r}^2$
Moment Of Inertia(MOI) for Continuous Mass Distribution	$I_{AB}=\int r^{2}dm$
MOI of Rod about an axis passing through the centre	$I=\frac{M{L}^2}{12}$
MOI of Ring	$I=M{R}^2$
MOI of Disc	$I=\frac{1}{2}M{R}^2$
MOI of Hollow Cylinder	$I=M{R}^2$
MOI of Solid Cylinder	$I=\frac{M{R}^2}{2}$
MOI of  Solid Sphere about its diametric axis	$I=\frac{2}{5}M{R}^2$
MOI of Hollow Sphere about its diametric axis	$I=\frac{2}{3}M{R}^2$
Perpendicular Axis Theorem	$I_Z=I_X+I_Y$
Parallel Axis Theorem	$I=I_{CM}+M{d}^2$
Radius of Gyration	$K=\sqrt{\frac{I}{M}}$
Torque	$\tau =Fr\sin \theta$
Angular momentum	$$\begin{gathered} L=mv\times r\sin \theta \\ \vec{L}=\vec{r}\times \vec{p} \end{gathered}$$
Rotational Kinetic Energy	$K.E_r=\frac{1}{2}I{\omega }^2$

3.0Past Year Questions with Solutions on Mechanics:JEE (Mains)

Q-1.All surfaces shown in figure are assumed to be frictionless and the pulleys and the string are light. The acceleration of the block of mass 2 kg is :        

$(1)g(2)\frac{g}{3}(3)\frac{g}{2}(4)\frac{g}{4}$

Solution:

Ans (2)

$40-2T=4a$

$T-10=4a\Rightarrow 20=12a$

$\to a=\frac{5}{3}\to 2a=\frac{10}{3}=\frac{g}{3}$

Q-2.A spherical body of mass 100 g is dropped from a height of 10 m from the ground. After hitting the ground, the body rebounds to a height of 5m. The impulse of force imparted by the ground to the body is given by : (given g = 9.8 m/s²)

$(1)4.32kgm{s}^{-}1(2)43.2kgm{s}^{-}1(3)23.9kgm{s}^{-}1(4)2.39kgm{s}^{-}1$

Solution:

Ans (4)

$\vec{I}=\Delta \vec{P}={\vec{P}}_f-{\vec{P}}_i$

M=0.1 kg

$I=\Delta P=0.1\left(\sqrt{2}\times 9.8\times 5-(-\sqrt{2}\times 9.8\times 10)\right)=0.1(14+7\sqrt{2})\approx 2.39{\text{kgms}}^{-1}$

Q-3.A block of mass m is placed on a surface having a vertical cross section given by y = x²/4. If coefficient of friction is 0.5, the maximum height above the ground at which block can be placed without slipping is: 

$(1)\frac{1}{4}m(2)\frac{1}{2}m(3)\frac{1}{6}m(4)\frac{1}{3}m$

Solution:

Ans (1)

$\mu =\frac{1}{2}$

$y=\frac{x^2}{4}$

$\frac{dy}{dx}=\frac{x}{2}$

$\mu =\tan \theta =\frac{dy}{dx}$

$\frac{1}{2}=\frac{x}{2}\Rightarrow x=1$

$y=\frac{x^2}{4}=\frac{(1{)}^2}{4}=\frac{1}{4}$

Q-4.A light string passing over a smooth light fixed pulley connects two blocks of masses $m_{1}and{m}_2$. If the acceleration of the system is $\frac{g}{8}$, then the ratio of masses is 

$(1)\frac{9}{7}(2)\frac{8}{1}m(3)\frac{4}{3}(4)\frac{5}{3}$

Solution:

Ans (1)

$a=\frac{(m_1-m_2)g}{m_1+m_2}=\frac{g}{8}$

$8m_1-8m_2=m_1+m_2$

$7m_1=9m_2$

$\frac{m_1}{m_2}=\frac{9}{7}$

Q-5.In the given arrangement of a doubly inclined plane two blocks of masses M and m are placed. The blocks are connected by a light string passing over an ideal pulley as shown. The coefficient of friction between the surface of the plane and the blocks is 0.25. The value of m, for which M = 10 kg will move down with an acceleration of $2\frac{m}{s^2}$ is : (take $g=10\frac{m}{s^2}$ and tan 37° = 3/4)

(1) 9 kg      (2) 4.5 kg       (3)  6.5 kg          (4)  2.25 kg

Solution:

Ans (2)

For M block

$10g\sin 53^{\circ }-\mu (10g)\cos 53^{\circ }-T=10\times 2$

$T=80-15-20\Rightarrow \boxed{T=45N}$

For m block 

$T-mg\sin 37^{\circ }-\mu mg\cos 37^{\circ }=m\times 2$

45=10 m

m=4.5 kg

Q-6.Consider a block and trolley system as shown in figure. If the coefficient of kinetic friction between the trolley and the surface is 0.04, the acceleration of the system in ms^–2 is :(Consider that the string is massless and unstretchable and the pulley is also massless and frictionless) :

(1) 3          (2) 4        (3)  2          (4) 1.2

Solution:

Ans (3)

$f_k=\mu N=0.04\times 20g=\boxed{8\text{Newton}}$

$a=\frac{(60-8)}{26}=2m/s^2$

Q-7. Three blocks A, B and C are pulled on a horizontal smooth surface by a force of 80 N as shown in figure

The tensions T₁ and T₂ in the string are respectively:

(1) 40N, 64N      (2)60N, 80N    (3) 88N, 96N      (4) 80N, 100N  

Solution:

Ans(1) 40N,64N

$a_A=a_B=a_C=\frac{F}{5+3+2}=\frac{80}{10}=\boxed{8{\text{m/s}}^2}$

$T_1=5✕8=40N$

$T_2-T_1=3✕8$

$T_2-40=24\to T_2=64N$

Q-8.A 1 kg mass is suspended from the ceiling by a rope of length 4m. A horizontal force 'F' is applied at the midpoint of the rope so that the rope makes an angle of 45° with respect to the vertical axis as shown in figure. The magnitude of F is : 

$(1)\frac{10}{\sqrt{2}}N(2)1N(3)\frac{1}{10\times \sqrt{2}}N(4)10N$

Solution:

As system is in equilibrium

$T_1\cos 45^{\circ }=T_2$

$T_1\cos 45^{\circ }=10N$

$T_1\times \frac{1}{\sqrt{2}}=10N\Rightarrow T_1=10\sqrt{2}N$

$T_1\sin 45^{\circ }=F$

$10\sqrt{2}\times \frac{1}{\sqrt{2}}=F\Rightarrow \boxed{F=10N}$

Q-9.A light string passing over a smooth light pulley connects two blocks of masses m₁ and m₂ (where m₂ > m₁ ) . If the acceleration of the system is $\frac{g}{\sqrt{2}}$, then the ratio of the masses $\frac{m_1}{m_2}$ is :

$(1)\frac{\sqrt{2}-1}{\sqrt{2}+1}(2)\frac{1+\sqrt{5}}{\sqrt{5}-1}(3)\frac{1+\sqrt{5}}{\sqrt{2}-1}(4)\frac{\sqrt{3}+1}{\sqrt{2}-1}$

Solution:

Ans (1)

$a=\frac{(m_2-m_1)g}{(m_1+m_2)}$

$a=-\frac{g}{\sqrt{2}}=\left(\frac{m_2-m_1}{m_1+m_2}\right)g$

$(1+\sqrt{2})m_1=(\sqrt{2}-1)m_2$

$\frac{m_1}{m_2}=\frac{\sqrt{2}-1}{\sqrt{2}+1}$

Q-10.A body of weight 200 N is suspended from a tree branch through a chain of mass 10 kg. The branch pulls the chain by a force equal to (if g = 10 m/s²): 

(1) 150 N        (2) 300N        (3) 200 N        (4) 100 N

Solution:

Ans (300 N)

Chain block system is in equilibrium so,T=200+100=300 N

Q-11.If a rubber ball falls from a height h and rebounds upto the height of h/2 {\frac{h}{2}}. The percentage loss of total energy of the initial system as well as velocity ball before it strikes the ground, respectively, are :

$(1)50\%,\sqrt{gh}(2)50\%,\sqrt{gh}(3)40\%,\sqrt{2gh}(4)50\%,\sqrt{2gh}$

Solution:

Ans(4)

Velocity just before collision= $\sqrt{2gh}$

Velocity just after collision= $\sqrt{2g\left(\frac{h}{2}\right)}$

$\Delta KE=\frac{1}{2}m(2gh)-\frac{1}{2}mgh=\frac{1}{2}mgh$

% Loss in Energy

$\frac{\Delta KE}{K{E}_i}\times 100=\frac{\frac{1}{2}mgh}{\frac{1}{2}mg2h}=\boxed{50\%}$

Q-12. In a system two particles of masses m₁ = 3kg and m₂ = 2kg are placed at a certain distance from each other. The particle of mass m₁ is moved towards the center of mass of the system through a distance 2cm. In order to keep the center of mass of the system at the original position, the particle of mass m₂ should move towards the center of mass by the distance ____ cm.

Solution:

Ans (3)

$\Delta X_{C.O.M}=\frac{m_1\Delta x_1+m_2\Delta x_2}{m_1+m_2}$

$0=\frac{3\times 2+2\times (-x)}{3+2}$

x=3 cm

Q-13.A simple pendulum of length 1 m has a wooden bob of mass 1 kg. It is struck by a bullet of mass 10^–2 kg moving with a speed of 2 × 10² ms^–1.The bullet gets embedded into the bob. The height to which the bob rises before swinging back is. (use g = 10 m/s²)

(1)0.30 m          (2)0.20 m        (3)0.35 m            (4)0.40 m

Solution:

Ans(2)

Applying conservation of linear momentum

mv=M+mV

$10^{-2}\times \left(2\times 10^2\right)=(1+0.01)V\Rightarrow V=\frac{2}{1.01}$

KE of the block with the bullet in it,is converted into PE as it rises through a height h

$\frac{1}{2}(M+m)V^2=(M+m)gh\Rightarrow V^2=2gh$

$h=\frac{V^2}{2g}={\left(\frac{2}{1.01}\right)}^2\times \frac{1}{2\times 9.8}=0.2m$

Q-14.The identical spheres each of mass 2M are placed at the corners of a right angled triangle with mutually perpendicular sides equal to 4 m each. Taking point of intersection of these two sides as origin, the magnitude of position vector of the centre of mass of the system is $\frac{4\sqrt{2}}{x}$, where the value of x is ___________

Solution:

Ans (3)

Position Vector $r_{\text{com}}=\frac{m_1{\vec{r}}_1+m_2{\vec{r}}_2+m_3{\vec{r}}_3}{m_1+m_2+m_3}$

${\vec{r}}_{\text{com}}=\frac{2M\times 0+2M\times 4\hat{i}+2M\times 4\hat{j}}{6M}$

$\vec{r}=\frac{4}{3}\hat{i}+\frac{4}{3}\hat{j}$

$|\vec{r}|=\frac{4\sqrt{2}}{3}$

On comparing with $\frac{4\sqrt{2}}{x}$ we get,x=3

Q-15.An artillery piece of mass M₁ fires a shell of mass M₂ horizontally. Instantaneously after the firing, the ratio of kinetic energy of the artillery and that of the shell is :

$(1)\frac{M_1}{M_1+M_2}(2)\frac{M_2}{M_1}(3)\frac{M_2}{M_1+M_2}(4)\frac{M_1}{M_2}$

Solution:

Ans (2)

$|{\vec{p}}_1|=|{\vec{p}}_2|$

$KE=\frac{p^2}{2M}$

$KE\propto \frac{1}{M}$

$\frac{K{E}_1}{K{E}_2}=\frac{\frac{p^2}{2M_1}}{\frac{p^2}{2M_2}}=\frac{M_2}{M_1}$

Q-16.The acceleration due to gravity on the surface of earth is g. If the diameter of earth reduces to half of its original value and mass remains constant, then acceleration due to gravity on the surface of earth would be 

$(1)\frac{g}{4}(2)2g(3)\frac{g}{2}(4)4g$

Solution:

Ans(4)

$g=\frac{GM}{R^2}\Rightarrow g\propto \frac{1}{R^2}$

$\frac{g_2}{g_1}=\frac{R_1^2}{R_2^2}$

$g_2=4g_1\left(R_2=\frac{R_1}{2}\right)$

Q-17.At what distance above and below the surface of the earth a body will have the same weight, (take radius of earth as R.)

$(1)\sqrt{5}R-R(2)\frac{\sqrt{5}R-R}{2}(3)\frac{R}{2}(4)\frac{\sqrt{5}R-R}{2}$

Solution:

Ans(4)

$g_p=\frac{g}{{\left(1+\frac{h}{R}\right)}^2}$

$g_q=g\left(1-\frac{h}{R}\right)$

$g_p=g_q$

$\frac{g}{{\left(1+\frac{h}{R}\right)}^2}=g\left(1-\frac{h}{R}\right)$

$\left(1-\frac{h}{R}\right){\left(1+\frac{h}{R}\right)}^2=1$

Take

$\frac{h}{R}=x$

$(1-x)(1+x{)}_2=1$

$(1+x_2+2x)(1-x)=1$

$1+x_2+2x-x-x_3-2x_2=1$

$x-x_3-x_2=0$

$x(x_2+x-1=0)$

$x=0,\left(\frac{\sqrt{5}-1}{2}\right)$

$x=\frac{h}{R}\Rightarrow h=\frac{\sqrt{5}R-R}{2}=\left(\frac{\sqrt{5}-1}{2}\right)R$

Q-18.The gravitational potential at a point above the surface of earth is -5.12  ✕ 107 J/kg {-5.12  ✕ 10_7 J/kg} and the acceleration due to gravity at that point is 6.4 m/s². Assume that the mean radius of earth is  6400 km. The height of this point above the earth’s surface is : 

(1)1600 km         (2)540 km        (3)1200 km                (4)1000 km

Solution:Ans(1)

Let r be the distance of the given point from the centre of the earth. Then Gravitational potential

$-\frac{GM}{r}=-5.12\times 10^7\text{J/kg}\text{...........(1)}$

And acceleration due to gravity,

$g=\frac{GM}{r^2}=6.4{\text{m/s}}^2\text{...................(2)}$

Dividing (1) by (2) 

$r=\frac{5.12\times 10^7}{6.4}=8\times 10^{6}m=8000km$

Height of the point from earth's surface r-R=8000-6400=1600 km

Q-19.Escape velocity of a body from earth is 11.2 km/s. If the radius of a planet be one-third the radius of earth and mass be one-sixth that of earth, the escape velocity from the plate is:

(1)11.2 km/s               (2)8.4 km/s        (3)4.2 km/s                (4)7.9 km/s

Solution:

Ans (4)

$R_p=\frac{R_E}{3},M_e=\frac{M}{6}$

$V_e=\sqrt{\frac{2G{M}_e}{R_e}}$

$V_p=\sqrt{\frac{2G{M}_p}{R_p}}$

$\frac{V_e}{V_p}=\sqrt{2}$

$V_p=\frac{V_e}{2}=\frac{11.2}{\sqrt{2}}=7.9\text{km/sec}$

Q-20.A light planet is revolving around a massive star in a circular orbit of radius R with a period of revolution T. If the force of attraction between planet and star is proportional to $R^{\frac{-3}{2}}$ then choose the correct option :

$(1)T^2\propto R^{5/2}(2)T^2\propto R^{7/2}(3)T^2\propto R^{3/2}(4)T^2\propto R^3$

Solution:

Ans(1)

$F\propto \frac{1}{R^{3/2}}=F\propto \frac{K}{R^{3/2}}$

Gravitational force provides centripetal force

$\frac{K}{R^{3/2}}=m{\omega }^{2}R$

${\omega }^2=\frac{K}{m{R}^{5/2}}$

${\left(\frac{2\pi }{T}\right)}^2=\frac{K}{m{R}^{5/2}}\Rightarrow T^2=\frac{4{\pi }^{2}m{R}^{5/2}}{K}$

$T^2\propto R^{5/2}$

Q-21. A planet takes 200 days to complete one revolution around the Sun. If the distance of the planet from the Sun is reduced to one fourth of the original distance, how many days will it take to complete one revolution ?

(1)25               (2)50        (3)100               (4)20

Solution:

Ans (1)

$T^2\propto r^3$

$\frac{T_1^2}{r_1^3}=\frac{T_2^2}{r_2^3}$

${\left(\frac{200}{r^3}\right)}^2=\frac{T_2^2}{{\left(\frac{r}{4}\right)}^3}$

$\frac{200\times 200}{4\times 4\times 4}=T_2^2$

$T_2=\frac{200}{4\times 2}=25$

$T_2=25days$

Q-22.Four identical particles of mass m are kept at the four corners of a square. If the gravitational force exerted on one of the masses by the other masses is 

$\left(\frac{2\sqrt{2}+1}{32}\right)\frac{G{m}^2}{L^2}$ the length of the sides of the square is

$(1)\frac{L}{2}(2)4L(3)3L(4)2L$

Solution:

Ans (2)

$F_{\text{net}}=\sqrt{2}F+F^{\prime }$

$F=\frac{G{m}^2}{a^2},F^{\prime }=\frac{G{m}^2}{(\sqrt{2}a{)}^2}$

$F_{\text{net}}=\sqrt{2}\frac{G{m}^2}{a^2}+\frac{G{m}^2}{2a^2}$

$\left(\frac{2\sqrt{2}+1}{32}\right)\frac{G{m}^2}{L^2}=\frac{G{m}^2}{a^2}\left(\frac{2\sqrt{2}+1}{2}\right)$

a=4L

Q-23.Four particles each of mass 1 kg are placed at four corners of a square of side 2 m. Moment of inertia of a system about an axis perpendicular to its plane and passing through one of its vertex is _____ kgm².

Solution:

Ans (16)

$I=m{a}^2+m{a}^2+m(\sqrt{2}{)}^2$

$I=4m{a}_2=4✕1\; ✕2_2=16$

Q-24.A heavy iron bar of weight 12 kg is having its one end on the ground and the other on the shoulder of a man. The rod makes an angle 60° with the horizontal, the weight experienced by the man is 

(1) 6 kg               (2) 12 kg        (3) 3 kg            (4) 63 kg           

Solution:

Ans (3)

Torque about O=0        

$I=m{a}^2+m{a}^2+m(\sqrt{2}{)}^2$

Q-25.A ring and a solid sphere roll down the same inclined plane without slipping. They start from rest. The radii of both bodies are identical and the ratio of their kinetic energies is $\frac{7}{x}$ where x is 

Solution:

Ans. (7)

In pure rolling work done by friction is zero.  Hence potential energy is converted into kinetic energy.  Since initially the ring and the sphere have the same potential energy, finally they will have the same kinetic energy too.

Ratio of kinetic energies = 1 

$\to \frac{7}{x}=1\to x=7$

Q-26.A cylinder is rolling down on an inclined plane of inclination 60°. Its acceleration during rolling down will be $\frac{x}{\sqrt{3}}{\text{m/s}}^2$ where x = _________. (use g = 10 m/s²).

Solution:

Ans (10)

For Rolling motion

$a=\frac{g\sin \theta }{1+\frac{I_{\text{com}}}{M{R}^2}}\Rightarrow a=\frac{g\sin \theta }{1+\frac{1}{2}}=\frac{2\times 10\times \frac{\sqrt{2}}{2}}{3}$ =103=x=10

Q-27.A body of mass 5 kg moving with a uniform speed $3\sqrt{2}\text{m/s}$ in the X – Y plane along the line y = x + 4. The angular momentum of the particle about the origin will be ______ kg m²s^–1.

Solution:Ans(60  kg m²s^–1)

y-x-4=0

$d_1$ is the perpendicular distance of given line from origin

$d_1=\left|\frac{-4}{\sqrt{1^2+1^2}}\right|\Rightarrow 2\sqrt{2}\text{m}$

$|\vec{L}|=mv{d}_1=5\times 3\sqrt{2}\times 2\sqrt{2}\text{kg}{\text{m}}^2/\text{s}=60\text{kg}{\text{m}}^2/\text{s}$

Q-28.Consider a Disc of mass 5 kg, radius 2m, rotating with angular velocity of 10 rad/s about an axis perpendicular to the plane of rotation. An identical disc is kept gently over the rotating disc along the same axis. The energy dissipated so that both the discs continue to rotate together without slipping is _____ J.        

Solution:

Ans (250 J)

For the first disc

$I_1=\frac{1}{2}M{R}^2=\frac{1}{2}\times 5\times (2{)}^2=10\text{kg}{\text{m}}^2$

${\omega }_i=10\text{rad/s}$

$E_i=\frac{1}{2}{I}_1{\omega }_1^2=\frac{1}{2}\times 10\times (10{)}^2=500J$

When the second disc is placed on top combined moment of inertia becomes

$I_f=I_1+I_2=10+10=20\text{kg}{\text{m}}^2$

By using conservation of angular momentum

$I{\omega }_i=I{\omega }_f\Rightarrow 10\times 10=20\times {\omega }_f$

${\omega }_f=5\text{rad/s}$

Energy Dissipated

$E_i-E_f=500-250=250J$

Q-29.Two discs of moment of inertia I₁ = 4 kg m² and I₂ = 2 kg m² about their central axes & normal to their planes, rotating with angular speeds 10 rad/s & 4 rad/s respectively are brought into contact face to face with their axis of rotation coincident. The loss in kinetic energy of the system in the process is_____J.

Solution:Ans(24)

$$\begin{gathered} K.E_i=\frac{1}{2}{I}_1{\omega }_1^2+\frac{1}{2}{I}_2{\omega }_2^2 \\ =\left(\frac{1}{2}\times 4\times (10{)}^2\right)+\left(\frac{1}{2}\times 2\times (4{)}^2\right) \end{gathered}$$

$K.E_i=200+16=216J$

$I_1{\omega }_1+I_2{\omega }_2=(I_1+I_2)\omega$

$4\times 10+2\times 4=(4+2)\omega \Rightarrow \omega =\left(\frac{48}{6}\right)=8\text{rad/s}$

$$\begin{gathered} K.E=\frac{1}{2}(I_1+I_2){\omega }^2=\frac{1}{2}(4+2)(8{)}^2 \\ =\frac{1}{2}\times 6\times 64=192J \end{gathered}$$

Loss in Kinetic Energy

216-192= 24 J

Q-30.A solid circular disc of mass 50 kg rolls along a horizontal floor so that its center of mass has a speed of 0.4 m/s. The absolute value of work done on the disc to stop it is ______ J. 

Solution:Ans(6 )

By using work energy Theorem

$W=\Delta KE=0-\left(\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2\right)$

$W=0-\frac{1}{2}m{v}^2\left(1+\frac{k^2}{R^2}\right)$

$W=-\frac{1}{2}\times 50\times (0.4{)}^2\left(1+\frac{1}{2}\right)=-6J$

Absolute work=+6 J

Q-31. A body of mass 'm' is projected with a speed ‘u’ making an angle of 45° with the ground. The angular momentum of the body about the point of projection, at the highest point is expressed as $\frac{\sqrt{2}m{u}^3}{xg}$. The value of 'X' is_______. 

Solution:

Ans (8)

$H=\frac{u^2{\sin }^2\theta }{2g}=\frac{u^2}{4g}$

Angular momentum $=m.u2.u24g=mu342g✕22=2mu38g$

$=m\cdot \frac{u}{\sqrt{2}}\cdot \frac{u^2}{4g}=\frac{m{u}^3}{4\sqrt{2}g}=\frac{\sqrt{2}m{u}^3}{8g}$

x=8

Q-32.Two identical spheres each of mass 2 kg and radius 50 cm are fixed at the ends of a light rod so that the separation between the centers is 150 cm. Then, a moment of inertia of the system about an axis perpendicular to the rod and passing through its middle point is $\frac{x}{20}kg{m}^2$, where the value of x is ____. 

Solution:

Ans (53)

$I=\left(\frac{2}{5}m{R}^2+m{d}^2\right)\times 2$

$I=2\left(\frac{2}{5}\times 2\times {\left(\frac{1}{2}\right)}^2+2\times {\left(\frac{3}{4}\right)}^2\right)=\frac{53}{20}\text{kg}\cdot {\text{m}}^2$

x=53

Q-33.A disc of radius R and mass M is rolling horizontally without slipping with speed n. It then moves up an inclined smooth surface as shown in figure. The maximum height that the disc can go up the incline is :

$(1)\frac{v^2}{g}(2)\frac{3}{4}\cdot \frac{v^2}{g}(3)\frac{1}{2}\cdot \frac{v^2}{g}(4)\frac{2}{3}\cdot \frac{v^2}{g}$

Solution:

Ans(2)

$\frac{1}{2}M{v}^2+\frac{1}{2}I{\omega }^2=Mgh$

$\Rightarrow \frac{1}{2}M{v}^2+\frac{1}{2}M{R}^2{\left(\frac{v}{R}\right)}^2=Mgh\Rightarrow h=\frac{3v^2}{4g}$

Q-34.A uniform rod AB of mass 2 kg and Length 30 cm at rest on a smooth horizontal surface. An impulse of force 0.2 Ns is applied to end B. The time taken by the rod to turn through at right angles will be $\frac{\pi }{x}$ , where x = ____.

Solution:

Ans(4)

Impulse (J)= 0.2 N-s

$J=\int Fdt=0.2\text{N}\cdot \text{s}$

Angular Impulse $\vec{M}$

${\vec{M}}_e=\int \tau dt=\int F\cdot \frac{L}{2}dt$

$=\frac{L}{2}\int Fdt=\frac{L}{2}\times J=\frac{0.3}{2}\times 0.2=0.03$

$I_{CM}=\frac{M{L}^2}{12}=\frac{2\times (0.3{)}^2}{12}=\frac{0.09}{6}$

$M=I_{CM}({\omega }_f-{\omega }_i)$

$0.03=\left(\frac{0.09}{6}\right)({\omega }_f)\Rightarrow {\omega }_f=2\text{rad/s}$

${\omega }_f=2\frac{rad}{s}$

$\theta =\omega t\Rightarrow t=\frac{\theta }{\omega }=\frac{\pi }{2\times 2}=\frac{\pi }{4}\text{sec}$

x=4

Q-35.A particle moves in the x-y plane under the influence of a force  such that its linear momentum is $\vec{P}(t)=\hat{i}\cos (kt)-\hat{j}\sin (kt)$. If k is constant, the angle between F and P will be :

$(1)\frac{\pi }{2}(2)\frac{\pi }{6}(3)\frac{\pi }{4}(4)\frac{\pi }{3}$

Solution:

Ans(1)

$\vec{P}(t)=\cos (kt)\hat{i}-\sin (kt)\hat{j};|\vec{P}|=1$

$\vec{P}=m\vec{v}$

$\hat{P}=\hat{v}$

$\hat{v}=\cos (kt)\hat{i}-\sin (kt)\hat{j}$

$\hat{a}=\frac{-k\sin (kt)\hat{i}-k\cos (kt)\hat{j}}{k}$

$\Rightarrow \hat{a}=-\sin (kt)\hat{i}-k\cos (kt)\hat{j}$

$\hat{F}=\hat{a}=-\sin (kt)\hat{i}-k\cos (kt)\hat{j}$

$\cos \theta =\frac{\vec{F}\cdot \vec{P}}{|\vec{F}|}|\vec{P}|=\frac{-\sin (kt)\hat{i}-k\cos (kt)\hat{j}}{1\times 1}=0$

$\Rightarrow \theta =\frac{\pi }{2}$