Additional Questions - Chapter 3 Laws of Motion 11th Science Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Additional Questions Solved
Multiple Choice Questions
Question 1.
The concept "force causes motion" was given by -
(a) Galileo
(b) Aristotle
(c) Newton
(d) Joule
Answer:
(b) Aristotle
Question 2.
Who decoupled the motion and force?
(a) Galileo
(b) Aristotle
(c) Newton
(d) Joule
Answer:
(a) Galileo
Question 3.
The inability of objects to move on its own or change its state of motion is called as -
(a) force
(b) momentum
(c) inertia
(d) impulse
Answer:
(c) inertia

Question 4.
Inertia means -
(a) inability
(b) resistance to change its state
(c) movement
(d) inertial frame
Answer:
(b) resistance to change its state
Question 5.
When a bus starts to move from rest, the passengers experience a sudden backward push is an example for -
(a) Inertia of motion
(b) Inertia of direction
(c) Inertia of rest
(d) back pull
Answer:
(c) Inertia of rest
Question 6.
If the brake is applied in the moving bus suddenly, passengers move forward is an example
for -
(a) Inertia of motion
(b) Inertia of direction
(c) Inertia of rest
(d) back pull
Answer:
(a) Inertia of motion

Question 7.
In whirling motion, if the string is cut suddenly, the stone moves tangential to circle is an -
(a) Inertia of motion
(b) Inertia of direction
(c) Inertia of rest
(d) back pull
Answer:
(b) Inertia of direction
Question 8.
Newtons laws are applicable in -
(a) Inertial frame
(b) non inertial frame
(c) in any frame
(d) none
Answer:
(a) Inertial frame
Question 9.
The accelerated train is an example for -
(a) inertial frame
(b) non-inertial frame
(c) both (a) and (b)
(d) none of the above
Answer:
(b) non-inertial frame
Question 10.
Rate of change of momentum of an object is equal to -
(a) acceleration
(b) work done
(c) force
(d) impulse
Answer:
(c) force
Question 11.
The product of mass and velocity is -

(a) force
(b) impulse
(c) momentum
(d) acceleration
Answer:
(c) momentum
Question 12 .
Unit of momentum -
(b) $\mathrm{kg} \mathrm{ms}^{-1}$
(c) $\mathrm{MLT}^{-2}$
(d) $\mathrm{MLT}^{-1}$
Answer:
(b) $\mathrm{kg} \mathrm{ms}^{-1}$
Question 13 .
According to Newton's third law -
(a) $\mathrm{F}_{12}=\mathrm{F}_{21}$
(*) $\mathrm{F}_{12}=-\mathrm{F}_{21}$
(c) $\mathrm{F}_{12}+\mathrm{F}_{21}=0$
(d) $\mathrm{F}_{12} \times \mathrm{F}_{21}=0$
Answer:
(a) $\mathrm{F}_{12}=\mathrm{F}_{21}$
Question 14 .
According to Newton's third law -
(a) $\overrightarrow{\mathrm{F}_{12}}=\stackrel{\mathrm{F}_{21}}{\longrightarrow}$
(b) $\overrightarrow{\mathrm{F}_{12}}=-\overrightarrow{\mathrm{F}_{21}}$
(c) $\mathrm{F}_{12}+\mathrm{F}_{21}=0$
(d) $\mathrm{F}_{12} x \mathrm{~F}_{21}=0$
Answer:
(b) $\overrightarrow{\mathrm{F}_{12}}=-\overrightarrow{\mathrm{F}_{21}}$
Question 15 .
The law which is valid in both inertial and non-inertial frame is -
(a) Newton's first law
(b) Newton's second law
(c) Newton's third law
(d) none

Answer:
(c) Newton's third law
Question 16.
When a force is applied on a body, it can change -
(a) velocity
(b) momentum
(c) direction of motion
(d) all the above
Answer:
(d) all the above
Question 17.
The rate of change of velocity is $1 \mathrm{~ms}^{-2}$ when a force is applied on the body of mass $75 \mathrm{gm}$ the force is -
(a) $75 \mathrm{~N}$
(b) $0.75 \mathrm{~N}$
(c) $0.075 \mathrm{~N}$
(d) $0.75 \times 10^{-3} \mathrm{~N}$
Answer:
(c) Force is given by
$\mathrm{F}=\mathrm{ma}$
$=75 \mathrm{gm} 1 \mathrm{~cm} \mathrm{~s}^{-2}=75 \times 10^{-3} \times 1=75 \times 10^{-3}=0.075 \mathrm{~N}$
Question 18 .
The action and reaction forces acting on -
(a) same body
(b) different bodies
(c) either same or different bodies
(d) none of the above
Answer:
(b) different bodies

Question 19.
Newton's first law of motion gives the concept of -
(a) velocity
(b) energy
(c) momentum
(d) Inertia
Answer:
(d) Inertia
Question 20.
Inertia of a body has direct dependence on -
(a) velocity
(b) area
(c) mass
(d) volume
Answer:
(c) mass
Question 21.
If a car and a scooter have the same momentum, then which one is having greater speed?
(a) scooter
(b) car
(c) both have same velocity
(d) data insufficient
Answer:
(a) scooter
Question 22.
Newton's second law gives -

(a) $\overrightarrow{\mathrm{F}} \propto \frac{d \overrightarrow{\mathrm{P}}}{\mathrm{dt}}$
(b) $\overrightarrow{\mathrm{F}}=\frac{\overrightarrow{d \mathrm{P}}}{\mathrm{dt}}$
(c) $\overrightarrow{\mathrm{F}}=m \vec{a}$
(d) all the above
Answer:
(d) all the above
Question 23.
1 dyne is -
(a) $10^5 \mathrm{~N}$
(b) $10^{-5} \mathrm{~N}$
(c) $1 \mathrm{~N}$
(d) $10^{-3} \mathrm{~N}$
Answer:
(b) $10^{-5} \mathrm{~N}$
Question 24.
If same force is acting on two masses $\mathrm{m}_1$ and $\mathrm{m}_2$, and the accelerations of two bodies are $\mathrm{a}_1$ and $a_2$ respectively, then -
(a) $\frac{a_2}{a_1}=\frac{m_2}{m_1}$
(b) $\frac{a_1}{a_2}=\frac{m_1}{m_2}$
(c) $\frac{a_1}{a_2}=\frac{m_2}{m_1}$
(d) $\mathrm{m}_1 \mathrm{a}_1+\mathrm{m}_2 \mathrm{a}_2=0$
Answer:
(c) $\frac{a_1}{a_2}=\frac{m_2}{m_1}$

Question 25.
If a force $\overline{\mathrm{F}}=3 \hat{i}-4 \hat{j} \mathrm{~N}$ produces an acceleration of $10 \mathrm{~ms}^{-2}$ on a body, then the mass of a body is -
(a) $10 \mathrm{~kg}$
(b) $9 \mathrm{~kg}$
(c) $0.9 \mathrm{~kg}$
(d) $0.5 \mathrm{~kg}$
Answer:
$
\begin{aligned}
& \overline{\mathrm{F}}=3 \hat{i}-4 \hat{j} \\
& \text { Magnitude: } \\
& |\overline{\mathrm{F}}|=\sqrt{9+16}=\sqrt{25}=5 \mathrm{~N} \\
& \mathrm{~F}=\mathrm{ma} \\
& \Rightarrow \mathrm{m}=\frac{|\mathrm{F}|}{a}=\frac{5}{10}=\frac{1}{2}=0.5 \mathrm{~kg}
\end{aligned}
$
Question 26.
A constant retarding force of $50 \mathrm{~N}$ is applied to a body of mass $20 \mathrm{~kg}$ moving initially with a speed of $15 \mathrm{~ms}^{-1}$. How long does the body take to stop?
(a) $0.75 \mathrm{~s}$
(b) $1.33 \mathrm{~s}$
(c) $6 \mathrm{~s}$
(d) $35 \mathrm{~s}$
Answer:
Acceleration $\mathrm{a}=\frac{-F}{m}=\frac{50}{20}=-2.5 \mathrm{~ms}^{-2}$
$
\begin{aligned}
& \mathrm{u}=15 \mathrm{~ms}^{-1} \\
& \mathrm{v}=0 \\
& \mathrm{t}=? \\
& \mathrm{v}=\mathrm{u}+\mathrm{at} \\
& 0=15-2.5 \mathrm{t} \\
& \mathrm{t}=\frac{15}{2.5}=6 \mathrm{~s}
\end{aligned}
$
Question 27.
Rain drops come down with -
(a) zero acceleration and non zero velocity
(b) zero velocity with non zero acceleration
(c) zero acceleration and non zero net force
(d) none
Answer:
(a) zero acceleration and non zero velocity
Question 28 .
If force is the cause then the effect is -

(a) mass
(b) potential energy
(c) acceleration
(d) Inertia
Answer:
(c) acceleration
Question 29.
In free body diagram, the object is represented by a -
(a) line
(b) arrow
(c) circle
(d) point
Answer:
(d) point
Question 30.
When an object of mass $m$ slides on a friction less surface inclined at an angle 0 , then normal force exerted by the surface is -
(a) $g \cos \theta$
(b) $\mathrm{mg} \cos \theta$
(c) $g \sin \theta$
(d) $\operatorname{mg} \tan \theta$
Answer:
(b) $m g \cos \theta$
Question 31.
The acceleration of the sliding object in an inclined plane -
(a) $\mathrm{g} \cos \theta$
(b) $\mathrm{mg} \cos \theta$
(c) $g \sin \theta$
(d) $\mathrm{mg} \sin \theta$

Answer:
(c) $\mathrm{g} \sin \theta$
Question 32 .
The speed of an object sliding in an inclined plane at the bottom is -
(a) $\mathrm{mg} \cos \theta$
(b) $\sqrt{2 \operatorname{sgsin} \theta}$
(c) $\sqrt{2 \operatorname{sgcos} \theta}$
(d) $\sqrt{2 \operatorname{sgtan} \theta}$
Answer:
(b) $\sqrt{2 \operatorname{sgsin} \theta}$
Question 33.
The acceleration of two bodies of mass $\mathrm{m}_1$ and $\mathrm{m}_2$ in contact on a horizontal surface is -
(a) $a=\frac{\mathbf{F}}{m_1}$
(b) $a=\frac{F}{m_2}$
(c) $a=\frac{\mathrm{F}}{m_1+m_2}$
(d) $a=\frac{\mathrm{F}}{m_1 m_2}$
Answer:
(c) $a=\frac{\mathrm{F}}{m_1+m_2}$
Question 34.
Two blocks of masses $m_1$ and $m_2\left(m_1>m_2\right)$ in contact with each other on frictionless, horizontal surface. If a horizontal force $\mathrm{F}$ is given on $\mathrm{m}_1$, set into motion with acceleration a, then reaction force on mass $\mathrm{m}_1$ by $\mathrm{m}_2$, is -
(a) $\frac{\mathrm{F} m_1}{m_1+m_2}$
(b) $\frac{m_1 m_2}{\mathrm{~F} m_1}$
(c) $\frac{m_1 m_2}{F m_2}$

(d) $\frac{\mathrm{F} m_2}{m_1+m_2}$
Answer:
(d) $\frac{\mathrm{F} m_2}{m_1+m_2}$
Question 35.

If two masses $\mathrm{m}_1$ and $\mathrm{m}_2\left(\mathrm{~m}_1>\mathrm{m}_2\right)$ tied to string moving over a frictionless pulley, then acceleration of masses -
(a) $\frac{\left(m_1-m_2\right)}{m_1+m_2} \mathrm{~g}$
(b) $\frac{m_1+m_2}{\left(m_1-m_2\right)} \mathrm{g}$
(c) $\frac{2 m_1 m_2}{m_1+m_2} \mathrm{~g}$
(d) $\frac{m_1 m_2}{2 m_1 m_2} \mathrm{~g}$
Answer:
(a) $\frac{\left(m_1-m_2\right)}{m_1+m_2} \mathrm{~g}$
Question 36.
if two masses $m_1$ and $m_2\left(m_1>m_2\right)$ tied to string moving over a frictionless pulley, then acceleration of masses -
(a) $\frac{\left(m_1-m_2\right)}{m_1+m_2} \mathrm{~g}$
(b) $\frac{m_1+m_2}{\left(m_1-m_2\right)}$ g
(c) $\frac{2 m_1 m_2}{m_1+m_2} \mathrm{~g}$
(d) $\frac{m_1 m_2}{2 m_1 m_2} \mathrm{~g}$
Answer:
(a) $\frac{\left(m_1-m_2\right)}{m_1+m_2} \mathrm{~g}$

Question 37.
Three massses is in contact as shown. If force $\mathrm{F}$ is applied to mass $\mathrm{m}_1$, the acceleration of three masses is -
(a) $\frac{\mathrm{F}}{m_1+m_2+m_3}$
(b) $\frac{m_1 F}{\left(m_1+m_2+m_3\right)}$
(c) $\frac{\left(m_2+m_3\right) F}{\left(m_1+m_2+m_3\right)}$
(d) $\frac{m_3 \mathrm{~F}}{m_1+m_2+m_3}$

Answer:
(a) $\frac{\mathrm{F}}{m_1+m_2+m_3}$
Question 38.
Three masses in contact is as shown above. If force $\mathrm{F}$ is applied to mass $\mathrm{m}_1$ then the contact force acting on mass $\mathrm{m}_2$ is -
(a) $\frac{\mathrm{F}}{m_1+m_2+m_3}$
(b) $\frac{m_1 F}{\left(m_1+m_2+m_3\right)}$
(c) $\frac{\left(m_2+m_3\right) F}{\left(m_1+m_2+m_3\right)}$
(d) $\frac{m_3 \mathrm{~F}}{m_1+m_2+m_3}$
Answer:
(c) $\frac{\left(m_2+m_3\right) F}{\left(m_1+m_2+m_3\right)}$
Question 39.
Three masses is contact as shown. It force $\mathrm{F}$ is applied to mass $\mathrm{m}_1$, then the contact force acting on mass $\mathrm{m}_3$ is -
(a) $\frac{\mathrm{F}}{m_1+m_2+m_3}$
(b) $\frac{m_1 F}{\left(m_1+m_2+m_3\right)}$
(c) $\frac{\left(m_2+m_3\right) F}{\left(m_1+m_2+m_3\right)}$
(d) $\frac{m_3 \mathrm{~F}}{m_1+m_2+m_3}$
Answer:
(d) $\frac{m_3 \mathrm{~F}}{m_1+m_2+m_3}$

Question 40.
Two masses connected with a string. When a force $\mathrm{F}$ is applied on mass $\mathrm{m}_2$. The acceleration produced is -

(a) $\frac{\mathrm{F}}{m_1+m_2}$
(c) $\frac{m_1+m_2}{\mathrm{~F}}$
(d) $\frac{m_3 \mathrm{~F}}{m_1+m_2+m_3}$
Answer:
(a) $\frac{\mathrm{F}}{m_1+m_2}$
Question 41.
Two masses connected with a string. When a force $\mathrm{F}$ is applied on mass $\mathrm{m}_2$. The force acting on $\mathrm{m}_1$ is -
(a) $\frac{m_1 \mathrm{~F}}{m_1+m_2}$
(b) $\frac{m_2 \mathrm{~F}}{m_1+m_2}$
(c) $\frac{m_1+m_2}{m_1} \mathbf{F}$
(d) $\frac{m_1+m_2}{m_2} \mathbf{F}$
Answer:
(b) $\frac{m_2 \mathrm{~F}}{m_1+m_2}$
Question 42 .
If a block of mass $\mathrm{m}$ lying on a frictionless inclined plane of length $\mathrm{L}$ height $\mathrm{h}$ and angle of inclination $\theta$, then the velocity at its bottom is -
(a) $g \sin \theta$
(b) $g \cos \theta$
(c) $\sqrt{2 g h}$
(d) $\sqrt{2 \operatorname{asin} \theta}$
Answer:
(c) $\sqrt{2 g h}$
Question 43.
If a block of mass $\mathrm{m}$ lying on a frictionless inclined plane of length $\mathrm{L}$, height $\mathrm{h}$ and angle of inclination $\theta$, then the time take taken to reach the bottom is -
(a) g $\operatorname{sing} \theta$

(b) $\sin \theta \sqrt{\frac{2 h}{g}}$
(c) $\sin \theta \sqrt{\frac{g}{h}}$
(d) $\frac{1}{\sin \theta} \sqrt{\frac{2 h}{g}}$
Answer:
(d) $\frac{1}{\sin \theta} \sqrt{\frac{2 h}{g}}$
Question 44.
A rocket works on the principle of conservation of -
(a) energy
(b) mass
(c) angular momentum
(d) linear momentum
Answer:
(b) mass
Question 45.
A bomb at rest explodes. The total momentum of all its fragments is -
(a) zero
(b) infinity
(c) always 1
(d) always greater then 1
Answer:
(a) zero
Question 46.
A block of mass $m_1$ is pulled along a horizontal friction-less surface by a rope of mass $\mathrm{m}_2$ If a force $F$ is given at its free end. The net force acting on the block is -
(a) $\frac{m_1 \mathrm{~F}}{m_1-m_2}$

(b) F
(c) $\frac{m_2 \mathrm{~F}}{\left(m_1+m_2\right)}$
(d) $\frac{m_1 \mathrm{~F}}{\left(m_1+m_2\right)}$
Answer:
(b) $\mathrm{F}$
Question 47.
A block of mass $\mathrm{m}$ is pulled along a horizontal surface by a rope. The tension in the rope will be same at all the points -
(a) if the rope is accelerated
(b) if the rope is mass less
(c) always
(d) none of the above
Answer:
(b) if the rope is mass less
Question 48.
The lines of forces act at a common point is called as -
(a) concurrent forces
(b) co-planar forces
(c) equilibrium
(d) resultant
Answer:
(a) concurrent forces
Question 49.
If the lines of forces act in the same plane, they can be -
(a) concurrent forces
(b) coplanar forces
(c) either concurrent force or coplanar forces

(d) Lami's force
Answer:
(d) concurrent forces
Question 50.
Lami's theorem is applicable only when the system of forces are is -
(a) same plane
(b) different plane
(c) equilibrium
(d) none of the above
Answer:
(c) equilibrium
Question 51.
Due to the action of internal forces of the system, the total linear momentum of the system is
(a) a variable
(b) a constant
(c) always zero
(d) always infinity
Answer:
(c) always zero
Question 52.
The velocity with which a gun suddenly moves backward after firing is -
(a) linear velocity
(b) positive velocity
(c) recoil velocity
(d) $\mathrm{v}_1+\mathrm{v}_2$
Answer:
(c) recoil velocity

Question 53 .
If a very large force acts on an object for a very short duration, then the force is called as -
(a) Newtonian force
(b) impulsive force
(c) concurrent force
(d) coplanar force
Answer:
(A) impulsive force
Question 54
The unit of impulse is -
(a) $\mathrm{Nm}$
(b) $\mathrm{Ns}$
(c) $\mathrm{Nm}^2$
(d) $\mathrm{Ns}^{-2}$
Answer:
(b) Ns
Question 55.
The force which always opposes the relative motion between an object and the surface where it is placed is -
(a) concurrent force
(b) frictional force
(c) impulsive force
(d) coplanar force
Answer:
(b) frictional force
Question 56.
The force which opposes the initiation of motion of an object on the surface is -
(a) static friction
(b) kinetic friction
(c) friction
(d) zero
Answer:
(d) static friction
Question 57.
When the object is at rest, the resultant of gravitational force and upward normal force is -
(a) Static force
(b) zero
(c) one
(d) infinity

Answer:
(b) zero
Question 58.
The magnitude of static frictional force $\mathrm{d}$ lies between -
(a) $0 \leq \mathrm{f} \leq \mu_{\mathrm{s}} \mathrm{N}$
(b) $0 \geq f \geq \mu_{\mathrm{s}} \mathrm{N}$
(c) 0 and 1
(d) 0 and minimal static frictional force.
Answer:
(a) $0 \leq \mathrm{f} \leq \mu_{\mathrm{s}} \mathrm{N}$
Question 59.
The unit of co-efficient of static friction is -
(a) $\mathrm{N}$
(b) $\mathrm{Nm}$
(c) $\mathrm{N} \mathrm{s}$
(d) no unit
Answer:
(d) no unit
Question 60.
If the object is at rest and no external force is applied on the object, the static friction acting on the object is -
(a) $\mu_{\mathrm{S}} \mathrm{N}$
(b) zero
(c) one
(d) infinity
Answer:
(d) no unit
Question 61.
When object begins to slide, the static friction acting on the object attains -
(a) zero
(b) minimum
(c) maximum
(d) infinity
Answer:
(c) maximum
Question 62.
The static friction does not depend upon-
(a) the area of contact

(b) normal force
(c) the magnitude of applied force
(d) none of the above
Answer:
(a) the area of contact
Question 63.
Which of the following pairs of materials has minimum amount of coefficient of static friction is -
(a) Glass and glass
(b) wood and wood
(c) ice and ice
(d) steel and steel
Answer:
(c) ice and ice
Question 64.
Kinetic friction is also called as -
(a) sliding friction
(b) dynamic friction
(c) both (a) and (b)
(d) static friction
Answer:
(c) both (a) and (b)
Question 65.
The unit of coefficient of kinetic friction is/has -
(a) $\mathrm{Nm}$
(b) $\mathrm{Ns}$
(c) $\mathrm{Nm}^2$
(d) no unit
Answer:

(d) no unit
Question 66.
The nature of materials in mutual contact decides -
(a) $\mu \mathrm{s}$
(b) $\mu \mathrm{k}$
(c) $\mu \mathrm{s}$ or $\mu \mathrm{k}$
(d) none
Answer:
(c) $\mu \mathrm{s}$ or $\mu \mathrm{k}$
Question 67.
Coefficient of kinetic friction is less than -
(a) $\mathrm{O}$
(b) one
(c) $\mu \mathrm{s}$
(d) $\mu \mathrm{sN}$
Answer:
(c) $\mu \mathrm{s}$
Question 68.
The static friction -
(a) increases linearly
(b) is constant
(c) zero
(d) varies parabolically
Answer:
(a) increases linearly
Question 69.
The kinetic friction -
(a) increases linearly
(b) is constant

(c) zero
(d) varies parabolically
Answer:
(b) is constant
Question 70.
Kinetic friction is independent of -
(a) nature of materials
(b) temperature of the surface
(c) applied force
(d) none of the above
Answer:
(c) applied force
Question 71.
The angle between the normal force and the resultant force of normal force and maximum frictional force is -
(a) angle of friction
(b) angle of repose
(c) angle of inclination
(d) none of the above
Answer:
(a) angle of friction
Question 72 .
The angle friction $\theta$ is given by -
(a) $\tan \mu_{\mathrm{s}}$
(b) $\tan ^{-1} \mu_{\mathrm{s}}$

(c) $\frac{f S^{\max }}{N}$
(d) $\sin ^{-1} \mu_{\mathrm{s}}$
Answer:
(b) $\tan ^{-1} \mu_{\mathrm{s}}$
Question 73.
The angle of inclined plane with the horizontal such that an object placed on it begins to slide is -
(a) angle of friction
(b) angle of repose
(c) angle of response
(d) angle of retardation
Answer:
(b) angle of repose
Question 74.
Comparatively, which of the following has lesser value than others?
(a) static friction
(b) kinetic friction
(c) Rolling friction
(d) skiping friction
Answer:
(c) Rolling friction
Question 75.
The origin of friction is -
(a) electrostatic interaction
(b) electromagnetic interaction magnetic
(c) photon interaction
(d) interaction
Answer:
(b) electromagnetic interaction
Question 76.
Friction can be reduced by -
(a) polishing
(b) lubricating
(c) using ball bearings
(d) all the above
Answer:
(c) using ball bearings

Question 77.
For a particle revolving in a circular path, the acceleration of the particle is -
(a) along the tangent
(b) along the radius
(c) along the circumference of the circle
(d) zero
Answer:
(b) along the radius
Question 78.
A particle moves along a circular path under the action of a force. The work done by the force is -
(a) Positive and non zero
(b) zero
(c) Negative and non zero
(d) none of the above
Answer:
(b) zero
Question 79.
A bullet hits and gets embedded in a solid block resting on a horizontal frictionless table. Which of the following is conserved?
(a) Momentum and kinetic energy
(b) kinetic energy alone
(c) Momentum alone
(d) potential energy alone
Answer:
(c) Momentum alone
Question 80 .
The origin of the centripetal force can be -
(a) gravitational force
(b) frictional force

(c) coulomb force
(d) all the above
Answer:
(d) all the above
Question 81 .
Centripetal acceleration is -
(a) $\frac{m v^2}{r}$
(b) $\frac{v^2}{r}$
(c) $r v^2$
(d) $r \omega$
Answer:
(b) $\frac{v^2}{r}$
Question 82 .
Centripetal acceleration is -
(a) $\frac{m v^2}{r}$
(b) $\mathrm{r} \omega^2$
(c) $1 \mathrm{v}^2$
(d) $r \omega$
Answer:
(c) $\mathrm{r} \omega^2$
Question 83.
The centripetal force is -
(a) $\frac{m v^2}{r}$
(b) $r \omega^2$
(c) both (a) and (b)
(d) none

Answer:
(c) both (a) and (b)
Question 84.
When a car is moving on a circular track the centripetal force is due to -
(a) gravitational force
(b) frictional force
(c) magnetic force
(d) elastic force
Answer:
(b) frictional force
Question 85 .
If the road is horizontal then the normal force and gravitational force are -
(a) equal and along the same direction
(b) equal and opposite
(c) unequal and along the same direction
(d) unequal and opposite
Answer:
(b) equal and opposite
Question 86.
The velocity of a car for safe turn on leveled circular road -
(a) $v \leq \sqrt{\mu_s r g}$
(b) $v \geq \sqrt{\mu_s r g}$
(c) $v=\sqrt{\mu_s r g}$
(d) $v \leq \mu_s r g$
Answer:
(a) $v \leq \sqrt{\mu_s r g}$

Question 87.
In a leveled circular road, skidding mainly depends on -
(a) $\mu_{\mathrm{s}}$
(b) $\mu_{\mathrm{k}}$
(c) acceleration
(d) none
Answer:
(a) $\mu_{\mathrm{s}}$
Question 88 .
The speed of a car to move on the banked road so that it will have safe turn is -
(a) $\mu_{\mathrm{s}} \mathrm{rg}$
(b) $\sqrt{r g \tan \theta}$
(c) $r g \tan \theta$
(d) $r^2 g \tan \theta$
Answer:
(b) $\sqrt{r g \tan \theta}$
Question 89.
Centrifugal force is a-
(a) pseudo force
(b) real force
(c) forced acting towards center
(d) none of the above
Answer:
(a) pseudo force
Question 90 .
Origin of centrifugal force is due to -
(a) interaction between two
(b) inertia
(c) electromagnetic interaction
(d) inertial frame
Answer:
(b) inertia
Question 91.
Centripetal force acts in -
(a) inertial frame
(b) non inertial frame
(c) both (a) and (b)

(d) linear motion
Answer:
(c) both (a) and (b)
Question 92.
Centrifugal force acts in -
(a) inertial frame
(b) non inertial frame
(c) both (a) and (b)
(d) linear motion
Answer:
(b) non inertial frame
Question 93.
A cricket ball of mass loo $\mathrm{g}$ moving with a velocity of $20 \mathrm{~ms}^1$ is brought to rest by a player in $0.05 \mathrm{~s}$ the impulse of the ball is -
(a) $5 \mathrm{Ns}$
(b) $-2 \mathrm{Ns}$
(c) $-2.5 \mathrm{Ns}$
(d) zero
Answer:
(b) $-2 \mathrm{Ns}$
mass $=0.1 \mathrm{~kg}$
Initial velocity $\mathrm{t}=20 \mathrm{~ms}^{-1}$
Final velocity $\mathrm{y}=0$
Change in momentum in impulse $=\mathrm{m}(\mathrm{v}-\mathrm{u})=0.1(0-20)=-2 \mathrm{Ns}$
Question 94.
If a stone tied at the one end of a string of length $0.5 \mathrm{~m}$ is whirled in a horizontal circle with a constant speed $6 \mathrm{~ms}^{-1}$ then the acceleration of the shone is -
(a) $12 \mathrm{~ms}^{-2}$
(b) $36 \mathrm{~ms}^{-2}$
(c) $2 \pi^2 \mathrm{~ms}^{-2}$
(d) $72 \mathrm{~ms}^{-2}$
Answer:
(d) Centripetal acceleration $=\frac{v^2}{r}=\frac{6^2}{0.5}=\frac{36}{0.5}=72 \mathrm{~ms}^{-2}$
Question 95.
A block of mass $3 \mathrm{~kg}$ is at rest on a rough inclined plane with angle of inclination $30^{\circ}$ with horizontal. If .is 0.7 , then the frictional force is -
(a) $17.82 \mathrm{~N}$

(b) $1.81 \mathrm{~N}$
(c) $3.63 \mathrm{~N}$
(d) $2.1 \mathrm{~N}$
Answer:
(a) Frictional force $=\mu \mathrm{mg} \cos \theta=0.7 \times 3 \times 9.8 \cos 30^{\circ}=17.82 \mathrm{~N}$
Question 96.
Two masses $2 \mathrm{~kg}$ and $4 \mathrm{~kg}$ are tied at the ends of a mass less string and which is passing over a friction-less pulley. The tension in the string is -
(a) $3.68 \mathrm{~N}$
(b) $78.4 \mathrm{~N}$
(c) $26 \mathrm{~N}$
(d) $13.26 \mathrm{~N}$
Answer:
(c) Tension in the string $\mathrm{T}=\frac{2 m_1 m_2}{\left(m_1+m_2\right)}$ g
$
\mathrm{T}=\frac{2 x 2 x 4}{2+4} \times 9.8=\frac{16}{6} \times 9.8=26.13 \mathrm{~N}
$
Question 97.
A bomb of $10 \mathrm{~kg}$ at rest explodes into two pieces of mass $4 \mathrm{~kg}$ and $6 \mathrm{~kg}$. if the velocity of $4 \mathrm{~kg}$ mass is $6 \mathrm{~ms}^{-1}$ then the velocity of $6 \mathrm{~kg}$ is -
(a) $-4 \mathrm{~ms}^{-1}$
(b) $-6 \mathrm{~ms}^{-1}$
(c) $-24 \mathrm{~ms}^{-1}$
(d) $-2.2 \mathrm{~ms}^{-1}$
Answer:
(a) According to law of conservation of momentum
$
\begin{aligned}
& \mathrm{m}_1 \mathrm{v}_1+\mathrm{m}_2 \mathrm{v}_2=0 \\
& \mathrm{v}_2=-\frac{m_1 v_1}{m_2}=\frac{4 x 6}{6}=-4 \mathrm{~ms}^{-1}
\end{aligned}
$

Question 98.
A body is subjected under three concurrent forces and it is in equilibrium. The resultant of any two forces is -
(a) coplanar with the third force
(b) is equal and opposite to third force
(c) both (a) and (b)
(d) none of the above
Answer:
(c) both (a) and (b)
Question 99.
An impulse is applied to a moving object with the force at an angle of $20^{\circ}$ with respect to velocity vector. The angle between the impulse vector and the change in momentum vector is
(a) $0^{\circ}$
(b) $30^{\circ}$
(c) $60^{\circ}$
(d) $120^{\circ}$
Answer:
(a) Impulse and change in momentum are in same direction. So the angle is zero.
Question 100.
A bullet of mass $m$ and velocity $v_1$ is fired into a large block of wood of mass $M$. The final velocity of the system is-
(a) $\frac{v_1}{m+\mathrm{M}}$
(b) $\frac{m v_1}{m+\mathrm{M}}$
(c) $\frac{m+m}{m} v_1$
(d) $\frac{m+m}{m-M} v_1$
Answer:
(b) $\frac{m v_1}{m+\mathrm{M}}$

Question 101.
A block of mass $2 \mathrm{~kg}$ is placed on the floor. The co - efficient of static friction is 0.4 . The force of friction between the block and floor is -
(a) $2.8 \mathrm{~N}$
(b) $7.8 \mathrm{~N}$
(c) $2 \mathrm{~N}$
(d) zero
Answer:
(b) The force required to move $==\mu \mathrm{R}=\mu \mathrm{mg}=0.4 \times 2 \times 9.8=7.84 \mathrm{~N}$
Question 102 .
A truck weighing $1000 \mathrm{~kg}$ is moving with velocity of $50 \mathrm{~km} / \mathrm{h}$ on smooth horizontal roads. A mass of $250 \mathrm{~kg}$ is dropped into it. The velocity with which it moves now is -
(a) $12.5 \mathrm{~km} / \mathrm{h}$
(b) $20 \mathrm{~km} / \mathrm{h}$
(c) $40 \mathrm{~km} / \mathrm{h}$
(d) $50 \mathrm{~km} / \mathrm{h}$
Answer:
(c) According to law of conservation of linear momentum
$\mathrm{m}_2 \mathrm{v}_2=\left(\mathrm{m}_1+\mathrm{m}_2\right) \mathrm{v}_2$
$
\mathrm{v}_2=\frac{m_1 v_1}{m_1+m_2}=\frac{1000 \times 50}{1250}=40 \mathrm{~km} / \mathrm{h}
$
Question 103.
A body of mass loo $\mathrm{g}$ is sliding from an inclined plane of inclination $30^{\circ}$. if $\mathrm{u}=1.7$, then the frictional force experienced is -
(a) $\frac{3.4}{\sqrt{3}} \mathrm{~N}$
(b) $1.47 \mathrm{~N}$
(c) $\frac{\sqrt{3}}{3.4} \mathrm{~N}$
(d) $1.38 \mathrm{~N}$

Answer:
(b) Frictional force $\mathrm{F}=\mu \mathrm{mg} \cos \theta=1.7 \times 0.1 \times 10 \cos 30^{\circ}=\frac{1.7}{2} \times \sqrt{3}=1.47 \mathrm{~N}$
Short Answer Questions (1 Mark)
Question 1.
A passenger sitting in a car at rest, pushes the car from within. The car doesn't move, why?
Answer:
For motion, there should be external force.
Question 2.
Give the magnitude and directions of the net force acting on a rain drop falling with a constant speed.
Answer:
as $\overline{\mathrm{a}}=0$ so $\overline{\mathrm{F}}=0$.
Question 3 .
Why the passengers in a moving car are thrown outwards when it suddenly takes a turn?
Answer:
Due to inertia of direction.
Question 4.
You accelerate your car forward. What is the direction of the frictional force on a package resting on the floor of the car?
Answer:
The package in the accelerated car (a non inertial frame) experiences a Pseudo force in a direction opposite to that of the motion of the car. The frictional force on the package which acts opposite to this pseudo force is thus in the same direction (forward) as that of the car.
Question 5.
What is the purpose of using shockers in a car?
Answer:
To decrease the impact of force by increasing the time for which force acts.
Question 6.
Why are types made of rubber not of steel?
Answer:
Since coefficient of friction between rubber and road is less than the coefficient of friction between steel and road.
Question 7.
Wheels are made circular. Why?
Answer:

Rolling friction is less than sliding friction.
Question 8.
If a ball is thrown up in a moving train, it comes back to the thrower's hands. Why?
Answer:
Both during its upward and downward motion, the ball continues to move inertia of motion with the same horizontal velocity as the train. In this period, the ball covers the same horizontal distance as the train and so it comes back to the thrower's hand.
Question 9.
Calculate the force acting on a body which changes the momentum of the body at the rate of 1 $\mathrm{kg}-\mathrm{m} / \mathrm{s}^2$.
Answer:
As $\mathrm{F}=$ rate change of momentum
$\mathrm{F}=1 \mathrm{~kg}-\mathrm{m} / \mathrm{s}^2=1 \mathrm{~N}$
Question 10.
On a rainy day skidding takes place along a curved path. Why?
Answer:
As the friction between the types and road reduces on a rainy day.
Question 11.
Why does a gun recoils when a bullet is being fired?
Answer:
To conserve momentum.
Question 12 .
Why is it difficult to catch a cricket ball than a tennis ball even when both are moving with the same velocity?
Answer:
Being heavier, cricket ball has higher rate of change of momentum during motion so more force sumed.

Question 13.
The distance travelled by a moving body is directly proportional to time. Is any external force acting on it?
Answer:
As $s \propto t$, so acceleration $a=0$, therefore, no external force is acting on the body.
Question 14.
Calculate the impulse necessary to stop a $1500 \mathrm{~kg}$ car moving at a speed of $25 \mathrm{~ms}^{-1}$.
Answer:
Use formula $I=$ change in momentum $=\mathrm{m}(\mathrm{v}-\mathrm{u})($ Impulse $-37500 \mathrm{Ns})$
Question 15 .
Lubricants are used between the two parts of a machine. Why?
Answer:
To reduce friction and so to reduce wear and tear.
Question 16.
What provides the centripetal force to a car taking a turn on a level road?
Answer:
Force of friction between the type and road provides centripetal force.
Question 17.
A body is acted upon by a number of external forces. Can it remain at rest?
Answer:
Yes, if the external forces acting on the body can be represented in magnitude and direction by the sides of a closed polygon taken in the same order.
Question 18.
Bodies of larger mass need greater initial effort to put them in motion. Why?
Answer:
As $\mathrm{F}=$ ma so for given a, more force will be required to put a large mass in motion.
Question 19.
An athlete runs a certain distance before taking a long jump Why?

Answer:
So that inertia of motion may help him in his muscular efforts to take a longer jump.
Question 20.
Action and reaction forces do not balance each other. Why?
Answer:
As they acts on different bodies.
Question 21.
The wheels of vehicles are provided with mudguards. Why?
Answer:
When the wheel rotates at a high speed, the mud sticking to the wheel flies off tangentially, this is due to inertia of direction. If order that the flying mud does not spoil the clothes of passer by the wheels are provided with mudguards.
Question 22.
China wares are wrapped in straw paper before packing. Why?
Answer:
The straw paper between the China ware increases the Time of experiencing the jerk during transportation. Hence impact of force reduces on China wares.
Question 23.
Why is it difficult to walk on a sand?
Answer:
Less reaction force.
Question 24.
The outer edge of a curved road is generally raised over the inner edge Why?
Answer:
In addition to the frictional force, a component of reaction force also provides centripetal force.

Question 25 .
Explain why the water doesn't fall even at the top of the circle when the bucket full of water is upside down rotating in a vertical circle?
Answer:
Weight of the water and bucket is used up in providing the necessary centripetal force at the top of the circle.
Question 26.
Why does a speedy motor cyclist bends towards the center of a circular path while taking a turn on it?
Answer:
So that in addition of the frictional force, the horizontal component of the normal reaction also provides the necessary centripetal forces.
Question 27.
An impulse is applied to a moving object with a force at an angle of $20^{\circ}$ wr.t. velocity vector, what is the angle between the impulse vector and change in momentum vector ?
Answer:
Impulse and change in momentum are along the same direction. Therefore angle between these two vectors is zero.
Short Answer Questions (2 Marks)
Question 28.
A man getting out of a moving bus runs in the same direction for a certain distance.
Comment.
Answer:
Due to inertia of motion.
Question 29.
If the net force acting upon the particle is zero, show that its linear momentum remains constant.
Answer:
As $\mathrm{F} \times \frac{d p}{d t}$
when $\mathrm{F}=0, \frac{d p}{d t}=0$ so $\mathrm{P}=$ constant
Question 30.
A force of 36 dynes is inclined to the horizontal at an angle of $60^{\circ}$. Find the acceleration in a mass of $18 \mathrm{~g}$ that moves in a horizontal direction.
Answer:
$F=36$ dyne at an angle of $60^{\circ}$

$
\begin{aligned}
& \mathrm{F}_{\mathrm{x}}=\mathrm{F} \cos 60^{\circ}=18 \text { dyne } \\
& \mathrm{F}_{\mathrm{x}}=\mathrm{ma}_{\mathrm{x}}
\end{aligned}
$
So $\mathrm{a}_{\mathrm{x}}=\frac{F_x}{m}=1 \mathrm{~cm} / \mathrm{s}^2$
Question 31.
The motion of a particle of mass $m$ is described by $h=u t+\frac{1}{2} \mathrm{gt}^2$. Find the force acting on particle.
Answer:
$
\mathrm{a}=u \mathrm{ut}+\frac{1}{2} \mathrm{gt}^2
$
find a by differentiating $h$ twice w.r.t.
$
\mathrm{a}=\mathrm{g}
$
As $\mathrm{F}=\mathrm{ma}$ so $\mathrm{F}=\mathrm{mg}($ answer $)$
Question 32.
A particle of mass $0.3 \mathrm{~kg}$ is subjected to a force of $\mathrm{F}=-\mathrm{kx}$ with $\mathrm{k}=15 \mathrm{Nmr}^{-1}$. What will be its initial acceleration if it is released from a point $20 \mathrm{~cm}$ away from the origin?
Answer:
As $\mathrm{F}=\mathrm{ma}$ so $\mathrm{F}=-\mathrm{kx}=\mathrm{ma}$
$
\mathrm{a}=\frac{-k x}{m}
$
for $\mathrm{x}=20 \mathrm{~cm}, \Rightarrow \mathrm{a}=-10 \mathrm{~m} / \mathrm{s}^2$.
Question 33.
A $50 \mathrm{~g}$ bullet is fired from a $10 \mathrm{~kg}$ gun with a speed of $500 \mathrm{~ms}^{-1}$. What is the speed of the recoil of the gun?
Answer:
Initial momentum $=0$
Using conservation of linear momentum $\mathrm{mv}+\mathrm{MV}=0$
$
\mathrm{V}=\frac{-m v}{M} \Rightarrow \mathrm{V}=2.5 \mathrm{~m} / \mathrm{s}
$
Question 34.
Smooth block is released at rest on a $45^{\circ}$ incline and then slides a distance $\mathrm{d}$. If the time taken of slide on rough incline is $n$ times as large as that to slide than on a smooth incline. Show that coefficient of friction, $\mu=\left(1-\frac{1}{n^2}\right)$
Answer:
When there is no friction, the block slides down the inclined plane with acceleration. $\mathrm{a}=\mathrm{g} \sin$ $\theta$
when there is friction, the downward acceleration of the block is a' $=\mathrm{g}(\sin \theta-\mu \cos \theta)$
As the block Slides a distance $d$ in each case so
$
d=\frac{1}{2} a t^2=\frac{1}{2} a t^{\prime} t^{\prime 2}
$

$
\frac{a}{a^{\prime}}=\frac{t^{\prime 2}}{t^2}=\frac{(n t)^2}{t^2}=\mathrm{n}^2
$
or $\frac{g \sin \theta}{g(\sin \theta-\mu \cos \theta)}=\mathrm{n}^2$
Solving, we get (Using $\theta=45^{\circ}$ )
$
\mu=1-\frac{1}{n^2}
$
Question 35 .
A spring balance is attached to the ceiling of a lift. When the lift is at rest spring balance reads $49 \mathrm{~N}$ of a body hang on it. If the lift moves:
1. Downward
2. upward, with an acceleration of $5 \mathrm{~ms}^2$
3. with a constant velocity.
What will be the reading of the balance in each case?
Answer:
$
\begin{aligned}
& \text { 1. } \mathrm{R}=\mathrm{m}(\mathrm{g}-\mathrm{a})=49 \mathrm{~N} \\
& \text { so }=\mathrm{m}=\frac{49}{9.8}=5 \mathrm{~kg} \\
& \mathrm{R}=5(9.8-5) \\
& \mathrm{R}=24 \mathrm{~N} \\
& \text { 2. } \mathrm{R}=\mathrm{m}(\mathrm{g}+\mathrm{a}) \\
& \mathrm{R}=5(9.8+5) \\
& \mathrm{R}=74 \mathrm{~N}
\end{aligned}
$
3. as a $=0$ so $\mathrm{R}=\mathrm{mg}=49 \mathrm{~N}$
Question 36.
A bob of mass $0.1 \mathrm{~kg}$ hung from the ceiling of room by a string $2 \mathrm{~m}$ long is oscillating. At its mean position the speed of a bob is $1 \mathrm{~ms}^{-1}$. What is the trajectory of the 'oscillating bob if the string is cut when the bob is -
1. At the mean position
2. At its extreme position.
Answer:
1. Parabolic
2. vertically downwards
Question 37.
A block placed on a rough horizontal surface is pulled by a horizontal force $F$. Let $f$ be the force applied by the rough surface on the block. Plot a graph of $f$ versus $F$.
Answer:

Unto point $\mathrm{A}, \mathrm{f}=\mathrm{F}$ ( 50 Long as block is stationary) beyond $\mathrm{A}$, when $\mathrm{F}$ increases, block starts moving $f$ remains constant.
Question 38.
A mass of $2 \mathrm{~kg}$ is suspended with thread $\mathrm{AB}$. Thread $\mathrm{CD}$ of the same type is attached to the other end of $2 \mathrm{~kg}$ mass.
- Lower end of the lower thread is pulled gradually, hander and hander is the downward direction so as to apply force on AB Which of the thread will break \& why?
- If the lower thread is pulled with a jerk, what happens?
Answer:
- Thread AB breaks down
- CD will break.
Question 39.
A block of mass $M$ is held against a rough vertical wall by pressing it with a finger. If the coefficient of friction between the block and the wall is $\mathrm{p}$ and the acceleration due to gravity is $g$, calculate the minimum force required to be applied by the finger to held the block against the wall?
Answer:

For the block not to fall $\mathrm{f}=\mathrm{Mg}$
But $\mathrm{f}=\mu \mathrm{R}=\mu \mathrm{F}$ so
$
\begin{aligned}
& \mu \mathrm{F}=\mathrm{Mg} \\
& \mathrm{F}=\frac{M g}{\mu}
\end{aligned}
$
Short Answer Questions (3 Marks) \& Numericals
Question 40.
A block of mass $500 \mathrm{~g}$ is at rest on a horizontal table. What steady force is required to give the block a velocity of $200 \mathrm{~cm} \mathrm{~s}^{-2}$ in $4 \mathrm{~s}$ ?
Answer:
Use $\mathrm{F}$ - ma
$\mathrm{a}=\frac{v-u}{t}=\frac{200-0}{4}=50 \mathrm{~cm} / \mathrm{s}^2$ $\mathrm{~F}=500 \times 50=25,000$ dyne.
Question 41.
A force of $98 \mathrm{~N}$ is just required to move a mass of $45 \mathrm{~kg}$ on a rough horizontal surface. Find the coefficient of friction and angle of friction?
Answer:
$
\begin{aligned}
& \mathrm{F}=48 \mathrm{~N}, \mathrm{R}=45 \times 9.8=441 \mathrm{~N} \\
& \mu=\frac{F^{\prime}}{R}=0.22
\end{aligned}
$
Angle of friction $\theta=\tan ^{-1} 0.22=12^{\circ} 24^{\prime}$
Question 42.
Calculate the force required to move a train of 2000 quintal up on an incline plane of 1 in 50 with an acceleration of $2 \mathrm{~ms}^{-2}$. The force of friction per quintal is $0.5 \mathrm{~N}$.
Answer:
Force of friction $=0.5 \mathrm{~N}$ per quintal
$
f=0.5 \times 2000=1000 \mathrm{~N}
$

$
\begin{aligned}
& \mathrm{m}=2000 \text { quintals }=2000 \times 100 \mathrm{~kg} \\
& \sin \theta=\frac{1}{50}, \mathrm{a}-2 \mathrm{~m} / \mathrm{s}^2
\end{aligned}
$
In moving up an inclined plane, force required against gravity $\mathrm{mg} \sin \theta=39200 \mathrm{~N}$
And force required to produce acceleration $=m a$
$
=2000 \times 100 \times 2=40,0000 \mathrm{~N}
$
Total force required $=1000+39,200+40,0000=440200 \mathrm{~N}$.
Question 43.
A force of $100 \mathrm{~N}$ gives a mass $\mathrm{m}_1$, an acceleration of $10 \mathrm{~ms}^{-2}$ and of $20 \mathrm{~ms}^{-2}$ to a mass $\mathrm{m}_2$. What acceleration must be given to it if both the masses are tied together?
Answer:
Suppose, $\mathrm{a}=$ acceleration produced if $\mathrm{m}_1$ and $\mathrm{m}_2$ are tied together,
$\mathrm{F}=100 \mathrm{~N}$
Let $\mathrm{a}_1$ and $\mathrm{a}_2$ be the acceleration produced in $\mathrm{m}_1$ and $\mathrm{m}_2$ respectively.
$\therefore \mathrm{a}_1$ and $\mathrm{a}_2=20 \mathrm{~ms}^{-2}$ (given)
Again $\mathrm{m}_1=\frac{\mathrm{F}}{a_1}$ and $\mathrm{m}_2=\frac{\mathrm{F}}{a_2}$
$\Rightarrow \mathrm{m}_1=\frac{100}{10}=10 \mathrm{~kg}$
and $\mathrm{m}_2=\frac{100}{20}=5 \mathrm{~kg}$
$
\begin{aligned}
& \therefore \mathrm{m}_1+\mathrm{m}_2=10+5=15 \\
& \text { so, } \mathrm{a}=\frac{\mathrm{F}}{m_1+m_2}=\frac{100}{15}=\frac{20}{3}=6.67 \mathrm{~ms}^2
\end{aligned}
$
Question 44.
The pulley arrangement of figure are identical. The mass of the rope is negligible. In (a) mass $\mathrm{m}$ is lifted up by attaching a mass ( $2 \mathrm{~m}$ ) to the other end of the rope. In (b), $\mathrm{m}$ is lifted up by pulling the other end of the rope with a constant downward force $F=2 \mathrm{mg}$. In which case, the acceleration of $\mathrm{m}$ is more?

Answer:
Case (a):
$
\mathrm{a}=\frac{2 m-m}{2 m+m} \mathrm{~g}=\mathrm{a}=\frac{g}{3}
$
Case (b):
FBD of mass $\mathrm{m}$
$
\begin{aligned}
& \mathrm{ma}=\mathrm{T}-\mathrm{mg} \\
& \mathrm{ma}=2 \mathrm{mg}-\mathrm{mg} \\
& \Rightarrow \mathrm{ma}=\mathrm{mg} \\
& \mathrm{a}^{\prime}=\mathrm{g}
\end{aligned}
$
So in case (b) acceleration of $\mathrm{m}$ is more.
Question 45.
Figure shows the position-time graph of a particle of mass $4 \mathrm{~kg}$. What is the 

(a) Force on the particle for $t<0, t>4$ s, $0<\mathrm{t}<4$ s?
(b) Impulse at $t=0$ and $t=4 s$ ?
(Consider one dimensional motion only)
Answer:
(a) For $\mathrm{t}<0$. No force as Particles is at rest. For $\mathrm{t}>4 \mathrm{~s}$, No force again particle comes at rest. For $0<\mathrm{t}<4 \mathrm{~s}$, as slope of $\mathrm{OA}$ is constant so velocity constant i.e., $\mathrm{a}=0$, so force must be zero.
(b) Impulse at $\mathrm{t}=0$
Impulse $=$ change in momentum
$\mathrm{I}=\mathrm{m}(\mathrm{v}-\mathrm{w})=4(0-0.75)=3 \mathrm{~kg} \mathrm{~ms}^{-1}$
Impulse at $t=4 \mathrm{~s}$
$1=\mathrm{m}(\mathrm{v}-\mathrm{u})=4(0-0.75)=-3 \mathrm{~kg} \mathrm{~ms}^{-1}$
Question 46.
What is the acceleration of the block and trolley system as the figure, if the coefficient of kinetic friction between the trolley and the surface is 0.04 ? Also Calculate friction in the string: Take $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$, mass of the string is negligible.

Answer:
Free body diagram of the block $30-\mathrm{T}=3 \mathrm{a}$
Free body diagram of the trolley

$\mathrm{T}-\mathrm{f}_{\mathrm{k}}=20 \mathrm{a}$
where $\mathrm{f}_{\mathrm{k}}=\mu_{\mathrm{k}}=0.04 \times 20 \times 10=8 \mathrm{~N}$
Solving (i) \& (ii), $\mathrm{a}=0.96 \mathrm{~m} / \mathrm{s}^2$ and $\mathrm{T}=27.2 \mathrm{~N}$
Question 47.
Three blocks of masses $\mathrm{ml}=10 \mathrm{~kg}, \mathrm{~m}^2=20 \mathrm{~kg}$ are connected by strings on smooth horizontal surface and pulled by a force of $60 \mathrm{~N}$. Find the acceleration of the system and frictions in the string.

Solution:
All the blocks more with common acceleration a under the force $\mathrm{F}=60 \mathrm{~N}$.
$
\begin{aligned}
& \mathrm{F}=\left(\mathrm{m}_1+\mathrm{m}_2+\mathrm{m}_3\right) \mathrm{a} \\
& \mathrm{a}=\frac{\mathrm{F}}{\left(m_1+m_2+m_3\right)}=1 \mathrm{~m} / \mathrm{s}^2
\end{aligned}
$
to determine, $\mathrm{T}_1 \rightarrow$ Free body diagram of $\mathrm{m}_1$.

$
\mathrm{T}_1=\mathrm{m}_1 \mathrm{a}=10 \times 1=10 \mathrm{~N}
$
to determine, $\mathrm{T}_2 \rightarrow$ Free body diagram of $\mathrm{m}_3$

$\mathrm{F}-\mathrm{T}_2=\mathrm{m}_3 \mathrm{a}$
Solving, we get $\mathrm{T}_2=30 \mathrm{~N}$
Question 48.
The rear side of a truck is open and a box of $40 \mathrm{~kg}$ mass is placed $5 \mathrm{~m}$ away from the open end. The coefficient of friction between the box and the surface below it is 0.15 on a straight road, the truck starts from rest and accelerates with $2 \mathrm{~m} / \mathrm{s}^2$. At what distance from the starting point does the box fall off the truck? (ignore the size of the box)
Answer:
Force on the box due to accelerated motion of the truck
$\mathrm{F}=\mathrm{ma}=40 \times 2=80 \mathrm{~N}$ (in forward direction)
Reaction on the box, $F^{\prime}=\mathrm{F}=80 \mathrm{~N}$ (in backward direction)
Force of limiting friction, $\mathrm{f}=\mu \mathrm{R}=0.15 \times 40 \times 10=60 \mathrm{~N}$
Net force on the box in backward direction is $P=F ' f=80-60=20 \mathrm{~N}$
Backward acceleration in the box $=\mathrm{a}=\frac{p}{m}=\frac{20}{40}=0.5 \mathrm{~ms}^{-2}$
$\mathrm{t}=$ time taken by the box to travel $\mathrm{s}=5 \mathrm{~m}$ and falls off the truck, then from
$\mathrm{s}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^2$
$5=0 \times \mathrm{t}+\frac{1}{2} \times 0.5 \mathrm{xt}^2$
$\mathrm{t}=4.47$
If the truck travels a distance $\mathrm{x}$ during this time
then $\mathrm{x}=0 \times 4.34+\frac{1}{2} \times 2 \times(4.471)^2$
$\mathrm{x}=19.98 \mathrm{~m}$

Question 49.
A block slides down as incline of $30^{\circ}$ with the horizontal. Starting from rest, it covers $8 \mathrm{~m}$ in the first 2 seconds. Find the coefficient of static friction.
Use $s=u t+\frac{1}{2} \mathrm{at}^2$
$\mathrm{a}=\frac{2 s}{t^2}$ at $^2$ as $\mathrm{u}=0$
$\mu=\frac{g \sin \theta-a}{g \operatorname{Cos} \theta}$
Putting the value and solving, $\mu=0.11$
Question 50.
A helicopter of mass $2000 \mathrm{~kg}$ rises with a vertical acceleration of $15 \mathrm{~m} / \mathrm{s}^2$. The total mass of the crew and passengers is $500 \mathrm{~kg}$. Give the magnitude and direction of the:
(a) Force on the floor of the helicopter by the crew and passenger.
(b) Action of the rotor of the helicopter on the surrounding air
(c) Force on the helicopter due to the surrounding air $\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2\right)$
Answer:
(a) Force on the floor of the helicopter by the crew and passengers
$=$ apparent weight of crew and passengers
$=500(10+15)$
$=12500 \mathrm{~N}$
(b) Action of rotor of helicopter on surrounding air is Obviously vertically downwards, because helicopter rises on account of reaction of this force. Thus force of action
$
\begin{aligned}
& =(2000+500)(10+15) \\
& =2500 \times 25 \\
& =62,500 \mathrm{~N}
\end{aligned}
$
(c) Force on the helicopter due to surrounding air is obviously a reaction. As action and reaction are equal and opposite, therefore Force of reaction $F^{\prime}=62,500$ vertically upwards.
Question 51.
A rectangular box lies on a rough inclined surface. The coefficient of friction between the surface and the box is $(\mu)$. Let the mass of the box be $\mathrm{m}$.
1. At what angle of inclination $\theta$ of the plane to the horizontal will the box just start to slide down the plane?
2. What is the force acting on the box down the plane, if the angle of inclination of the plane is increased to a $>\theta$.
3. What is the force needed to be applied upwards along the plane to make the box either remain stationary or just move up with uniform speed ?
4. What is the force needed to be applied upwards along the plane to $\mathrm{pk} \mathrm{kg} \mathrm{f}$ make the box move up the plane with acceleration a?
Answer:
1. When the box just starts sliding
$
\mu=\tan \theta
$
or $0=\tan ^{-1} \mu$
2. Force acting on the box down the plane
$=m g(\sin a-\mu \cos a)$
3. Force needed $\mathrm{mg}(\sin a+\mu \cos a)$
4. Force needed $=m g(\sin a+\mu \cos a)+m a$.

Question 52.
Two masses of $5 \mathrm{~kg}$ and $3 \mathrm{~kg}$ are suspended with help of mass less in extensible string as shown. Calculate $T_1$ and $T_2$ when system is going upwards with acceleration $\mathrm{m} / \mathrm{s}^2$. (Use $g 9.8$ $\left.\mathrm{m} / \mathrm{s}^2\right)$
Answer:
According Newton's second law of motion
$
\begin{aligned}
& \text { (1) } \mathrm{T}_1-\left(\mathrm{m}_1+\mathrm{m}_2\right) \mathrm{g}=\left(\mathrm{m}_1+\mathrm{m}_2\right) \mathrm{a} \\
& \mathrm{T}_1=\left(\mathrm{m}_1+\mathrm{m}_2\right)(\mathrm{a}+\mathrm{g})=(5+3)(2+9.8) \\
& \mathrm{T}_1=94.4 \mathrm{~N}
\end{aligned}
$
(2) $\mathrm{T}_2-\mathrm{m}_2 \mathrm{~g}=\mathrm{m}_2 \mathrm{a}$
$\mathrm{T}_2=\mathrm{m}_2(\mathrm{a}+\mathrm{g})$
$\mathrm{T}_2=3(2+9.8)$
$\mathrm{T}_2=35.4 \mathrm{~N}$

Question 53.
There are few forces acting at a Point $\mathrm{P}$ produced by strings as shown, which is at rest. Find the forces $\mathrm{F}_1 \& \mathrm{~F}_1$

Answer:

Using Resolution of forces $\mathrm{IN}$ and $2 \mathrm{~N}$ and then applying laws of vector addition. Calculate for $\mathrm{F}_1 \& \mathrm{~F}_1$.
$
\mathrm{F}_1=\frac{1}{\sqrt{2}} \mathrm{~N}, \mathrm{~F}_2=\frac{3}{\sqrt{2}} \mathrm{~N}
$
Question 54.
A hunter has a machine gun that can fire $50 \mathrm{~g}$ bullets with a velocity of $150 \mathrm{~ms} \mathrm{~A} 60 \mathrm{~kg}$ tiger springs at him with a velocity of $10 \mathrm{~ms}^{-1}$. How many bullets must the hunter fire into the target so as to stop him in his track?
Answer:
Given $\mathrm{m}=$ mass of bullet $=50 \mathrm{gm}=0.50 \mathrm{~kg}$
$\mathrm{M}=$ mass of tiger $=60 \mathrm{~kg}$
$\mathrm{v}=$ Velocity of bullet $-150 \mathrm{~m} / \mathrm{s}$
$\mathrm{V}=$ Velocity of tiger $=-10 \mathrm{~m} / \mathrm{s}$
(v It is coming from opposite direction $\mathrm{n}=$ no. of bullets fired per second at the tiger so as to stop it.)
$\mathrm{P}_{\mathrm{i}}=0$, before firing .......(i)
$\mathrm{P}_{\mathrm{f}}=\mathrm{n}(\mathrm{mv})+\mathrm{MV}$.
(ii)
$\therefore$ From the law of conservation of momentum, $\mathrm{P}_{\mathrm{i}}=\mathrm{P}_{\mathrm{f}}$
$
\begin{aligned}
& \Rightarrow 0=\mathrm{n}(\mathrm{mv})+\mathrm{MV} \\
& \mathrm{n}=\frac{M V}{m v}=\frac{-60 \times(-10)}{0.05 \times 150}=80
\end{aligned}
$
Question 55.
Two blocks of mass $2 \mathrm{~kg}$ and $5 \mathrm{~kg}$ are connected by an ideal string passing over a pulley. The block of mass $2 \mathrm{~kg}$ is free to slide on a surface inclined at an angle of $30^{\circ}$ with the horizontal whereas $5 \mathrm{~kg}$ block hangs freely. Find the acceleration of the system and the tension in the string.

Let a be the acceleration of the system and $\mathrm{T}$ be the Tension in the string. Equations of motions for $5 \mathrm{~kg}$ and $2 \mathrm{~kg}$ blocks are
$5 \mathrm{~g}-\mathrm{T}=5 \mathrm{a}$
$\mathrm{T}-2 \mathrm{~g} \sin \theta-\mathrm{f}=2 \mathrm{a}$
where $\mathrm{f}=$ force of limiting friction
$=\mu \mathrm{R}=\mu \mathrm{mg} \cos \theta=0.3 \times 2 \mathrm{~g} \mathrm{x} \cos 30^{\circ}$
Solving (1) \& (2)
$
\mathrm{a}=4.87 \mathrm{~m} / \mathrm{s}^2
$