Kamal A.A. 1000 Solved Problems in Classical Physics: An Exercise Book

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46 1 Kinematics and Statics

= (mg)(BD) = (mg)





= mg

cos α (1)

The friction with the ground, which acts toward right produces a counterclock-

wise torque τ

= (μ mg)AC = μmg

AB = μmg L sin α (2)

For limiting equilibrium τ

= τ

∴ mg

cos α = μmg L sin α

∴ μ =

cot α

Chapter 2

Particle Dynamics

Abstract Chapter 2 is concerned with motion of blocks on horizontal and inclined

planes with and without friction, work, power and energy. Elastic, inelastic and

partially elastic collisions in both one dimension and two dimensions are treated.

Problems on variable mass cover rocket motion, falling of rain drops, etc.

2.1 Basic Concepts and Formulae

Internal and External Forces

Forces acting upon a system due to external agencies are called external forces. As

an example a body placed on a surface is acted by earth’s gravitation which is an

external force.

Forces that act between pairs of particles which constitute the body or a system

are all internal to the system and are called internal forces. The size of a system

is entirely arbitrary and is deﬁned by the convenience of the situation. If a system

is made sufﬁciently extensive then all forces become internal forces. By Newton’s

third law of motion internal forces between pairs of particles get cancelled. Hence

net internal force is zero. Internal forces cannot cause motion.

Inertial and Gravitational Mass

If mass is determined by Newton’s second law, that is, m = F/a, then it is called

inertial mass.

If the mass is determined by the gravitational force exerted on it by another body,

say the earth of mass M, that is, m



= Fr

/GM, then it is called the gravitational

mass. It turns out that m = m



Frames of Reference

A frame of reference (coordinate system) is necessary in order to measure the

motion of particles. A reference frame is called an inertial frame if Newton’s laws

48 2 Particle Dynamics

are found to be valid in that frame. It is found that inertial frames move with constant

velocity with respect to one another.

Conservation Laws

(i) If the total force F is zero, then linear momentum p is conserved.

(ii) If the total external torque is zero, then the angular momentum J is conserved.

(iii) If the forces acting on a particle are conservative, then the total mechanical

energy (kinetic + potential) of the particle is conserved.

Conservative Force

If the force ﬁeld is such that the work done around a closed orbit is zero, i.e.



F · ds = 0 (2.1)

then the force and the system are said to be conservative. A system cannot be con-

servative if a dissipative force like friction is present. Since the quantity Fds due to

friction will always be negative and the integrand cannot vanish, by Stokes theorem

the condition for conservative forces given by (2.1) becomes

∇×F = 0 (2.2)

Since the curl of a gradient always vanishes, it follows that F must be the gradient

of the scalar quantity V , i.e.

F =−∇V (2.3)

V is called the potential energy.

Centre of Mass



i=1

(2.4)

where r

is the position vector of the centre of mass from the origin and M = m

is the total mass. The centre of mass moves as if it were a single particle of mass

equal to the total mass of the system, acted upon by the total external force and

independent of the nature of the internal forces.

The reduced mass (μ) of two bodies of mass m

and m

is given by

μ =

+ m

(2.5)

2.1 Basic Concepts and Formulae 49

A two-body problem is reduced to a one-body problem through the introduction

of the reduced mass μ.

Motion of a Body of a Variable Mass

It is well known that in relativistic mechanics the mass of a particle increases with

increasing velocity. However, in Newtonian mechanics too one can give meaning

to variable mass as in the following example. Consider an open wagon moving on

rails on a horizontal plane under steady heavy shower. As rain is collected the mass

of the wagon increases at constant rate. Other examples are rocket, motion of jet

propelled vehicles, an engine taking water on the run.

F =

(mv) = m

+ v

(2.6)

Motion of a Rocket

If m is the mass of the rocket plus fuel at any time t and v

the velocity of the ejected

gases relative to the rocket then

Resultant force on rocket = (upward thrust on rocket) – (weight of the rocket)

= v

− mg (2.7)

Therefore, acceleration of the rocket

a =

− g (2.8)

Assuming that v

and g remain constant and at t = 0, v = 0 and m = m

= v





− gt (2.9)

where m

is the initial mass of the system and m

the mass at burn-out velocity v

(the velocity at which all the fuel is burnt out is called the burn-out velocity).

Now

m = m

−v/v

(2.10)

Time taken for the rocket to reach the burn-out velocity is given by

t = t

− m

(2.11)

where α =−dm/dt is a positive constant.

50 2 Particle Dynamics

Elastic Collisions (One-Dimensional Head-On)

By deﬁnition total kinetic energy is conserved. If u

and u

be the respective initial

velocities of m

and m

, v

and v

being the corresponding ﬁnal velocities, then

− u

= v

− v

(2.12)

Thus, the relative velocity of approach before the collision is equal to the relative

velocity of separation after the collision:



− m

+ m



+ m

(2.13)

+ m



− m

+ m



(2.14)

Inelastic Collisions, Direct Impact

The bodies stick together in the course of collision and are unable to separate

out. After the collision they travel as one body with common velocity v given by

v =

+ m

(2.15)

Energy wasted =

μ(u

− u

)

(2.16)

where μ is the reduced mass.

Ballistic pendulum is a device for measuring t he velocity of a bullet. The pen-

dulum consists of a large wooden block of mass M which is supported vertically

by two cords. A bullet of mass m hits the block horizontally with velocity v and

is lodged within it. As a result of collision the block is raised through maximum

height h (see prob. 2.44). Applying momentum conservation for the initial collision

process, and energy conservation for the subsequent motion, it can be shown that

v =



1 +





2gh (2.17)

Partially elastic collisions are collisions which fall in between perfectly elas-

tic collisions and totally inelastic collisions. The coefﬁcient of restitution e which

deﬁnes the degree of inelasticity is given by

e =−

relative velocity of separation

relative velocity of approach

− v

− u

(2.18)

For perfectly elastic collisions e = 1, for totally inelastic collisions e = 0 and for

partially elastic collisions 0 < e < 1.

2.1 Basic Concepts and Formulae 51

Relations of Quantities in the Lab System (LS) and Centre of Mass System

(CMS)

Quantities that are unprimed refer to LS and primed refer to CMS.

Lab system CM system

: u

; u

∗

+ m

(2.19)

: u

= 0; u

∗

+ m

(2.20)

Scattering Angle

Because of elastic scattering

∗

= u

∗

; v

∗

= u

∗

The centre of mass velocity

=−

+ m

=−u

∗

(2.21)

tan θ =

sin θ

∗

cos θ

∗

+ m

(2.22)

tan θ

∗

sin θ

cos θ − m

(2.23)

If m

< m

, all scattering angles for m

in LS are possible.

If m

= m

, scattering only in the forward hemisphere (θ ≤ 90

◦

) is possible.

If m

> m

, the maximum scattering angle is possible, θ

max

, being given by

max

= sin

−1

) (2.24)

Recoil Angle

tan ϕ =

sin ϕ

∗

cos ϕ

∗

+ 1

= tan

∗

(2.25)

Or φ =

∗

(2.26)

Recoiling angle is limited to ϕ ≤ 90

◦

52 2 Particle Dynamics

2.2 Problems

2.2.1 Motion of Blocks on a Plane

2.1 Three blocks of mass m

, m

and m

interconnected by cords are pulled by a

constant force F on a frictionless horizontal table, Fig. 2.1.Find

(a) Common acceleration ‘a’

(b) Tensions T

and T

Fig. 2.1

2.2 A block of mass M on a rough horizontal table is driven by another block of

mass m connected by a thread passing over a frictionless pulley. Assuming

that the coefﬁcient of friction between the mass M and the table is μ, ﬁnd

(a) acceleration of the masses (b) tension in the thread (Fig. 2.2).

Fig. 2.2

2.3 A block of mass m

sits on a block of mass m

, which rests on a smooth table,

Fig. 2.3. If the coefﬁcient of friction between the blocks is μ, ﬁnd the maximum

force that can be applied to m

so that m

may not slide.

Fig. 2.3

2.4 Two blocks m

and m

are in contact on a frictionless table. A horizontal force

F is applied to the block m

,Fig.2.4. (a) Find the force of contact between

the blocks. (b) Find the force of contact between the blocks if the same force is

applied to m

rather than to m

,Fig.2.5.

Fig. 2.4

2.2 Problems 53

Fig. 2.5

2.5 A box of weight mg is dragged with force F at an angle θ above the horizontal.

(a) Find the force exerted by the ﬂoor on the box. (b) Find the acceleration

of the box if the coefﬁcient of friction with the ﬂoor is μ. (c) How would the

results be altered if the box is pushed with the same force?

2.6 A uniform chain of length L lies on a table. If the coefﬁcient of friction is μ,

what is the maximum length of the part of the chain hanging over the table such

that the chain does not slide?

2.7 A uniform chain of length L and mass M is lying on a smooth table and one-

third of its length is hanging vertically down over the edge of the table. Find

the work required to pull the hanging part on the table.

2.8 A block of metal of mass 2 kg on a horizontal table is attached to a mass of

0.45 kg by a light string passing over a frictionless pulley at the edge of the

table. The block is subjected to a horizontal force by allowing the 0.45 kg mass

to fall. The coefﬁcient of sliding friction between the block and table is 0.2.

Calculate (a) the initial acceleration, (b) the tension in the string, (c) the distance

the block would continue to move if, after 2 s of motion, the string should break

(Fig. 2.6).

[University of New Castle]

Fig. 2.6

2.2.2 Motion on Incline

2.9 A block of mass of 2 kg slides on an inclined plane that makes an angle of

◦

with the horizontal. The coefﬁcient of friction between the block and the

surface is

√

3/2.

(a) What force should be applied to the block so that it moves down without

any acceleration?

(b) What force should be applied to the block so that it moves up without any

acceleration?

[Indian Institute of Technology 1976]

54 2 Particle Dynamics

2.10 A block is placed on a ramp of parabolic shape given by the equation y =

/20, Fig. 2.7.Ifμ

= 0.5, what is the maximum height above the ground at

which the block can be placed without slipping?

Fig. 2.7

2.11 A block slides with constant velocity down an inclined plane that has slope

angle θ = 30

◦

(a) Find the coefﬁcient of kinetic friction between the block and the plane.

(b) If the block is projected up the same plane with initial speed v

2.5m/s, how far up the plane will it move before coming to rest? What

fraction of the initial kinetic energy is transformed into potential energy?

What happens to the remaining energy?

your answer.

2.12 Consider a ﬁxed inclined plane at angle θ . Two blocks of mass M

and M

are

attached by a string passing over a pulley of radius r and moment of inertia I

as in Fig. 2.8:

(a) Find the net torque acting on the system comprising the two masses, pul-

ley and the string.

(b) Find the total angular momentum of the system about the centre of the

pulley when the blocks are moving with speed v.

Fig. 2.8

2.13 A box of mass 1 kg rests on a frictionless inclined plane which is at an angle

of 30

◦

to the horizontal plane. Find the constant force that needs to be applied

parallel to the incline to move the box

(a) up the incline with an acceleration of 1 m/s

(b) down the incline with an acceleration of 1 m/s

[University of Aberystwyth, Wales 2008]

2.2 Problems 55

2.14 A wedge of mass M is placed on a horizontal ﬂoor. Another mass m is placed

on the incline of the wedge. Assume that all surfaces are frictionless, and the

incline makes an angle θ with the horizontal. The mass m is released from rest

on mass M, which is also initially at rest. Find the accelerations of M and m

(Fig. 2.9).

Fig. 2.9

2.15 Two smooth inclined planes of angles 45

◦

and hinged together back to back.

Two masses m and 3m connected by a ﬁne string passing over a light pul-

ley move on the planes. Show that the acceleration of their centre of mass is

√

5/8 g at an angle tan

−1

½ to the horizon (Fig. 2.10).

Fig. 2.10

2.16 Two blocks of masses m

and m

are connected by a string of negligible mass

which passes over a pulley of mass M and radius r mounted on a frictionless

axle. The blocks move with an acceleration of magnitude a and direction as

shown in the diagram. The string does not slip on the pulley, so the tensions

and T

are different. You can assume that the surfaces of the inclines are

frictionless. The moment of inertia of the pulley is given by I = ½Mr

(a) Draw free body diagrams for the two blocks and the pulley.

(b) Write down the equations for the translational motion of the two blocks

and the rotational motion of the pulley.

a =

√

− m

)

M + 2(m

+ m

)

2.17 Two masses in an Atwood machine are 1.9 and 2.1 kg, the vertical distance of

the heavier body being 20 cm above the lighter one. After what time would the