1.163 Acceleration of rocket

Problem 1.163:A small test rocket launched from rest travels $14.4 \:m$ upward in $2.0 \:s$. Determine acceleration of the rocket.

Solution:

The rocket was initially at rest. Then it speeds upward and covers a distance of $14.4 \:m$ in $2.0 \:s$.

We will assume that it was uniformly accelerated.

We can use kinematic equation $S\:=\:ut\:+\:\frac{1}{2}at^2$ to solve the problem.

For the rocket, initial velocity, $u\:=\:0$.

Time, $t\:=\:2.0\:s$.

Displacement, $S\:=\:14.4\:m$.

Then, $14.4\:=\:0\:+\:\frac{1}{2}\times a \times (2)^2$.

Or, $14.4\:=\:\frac{1}{2}\times a \times 2 \times 2$

Or, $14.4\:=\: a \times 2$.

From this acceleration, $a\:=\:\frac{14.4}{2}\:=\:7.2\:ms^{-2}$.

So the rocket has upward acceleration of $7.2\:ms^{-2}$.

1.162 Acceleration and velocity vector in projectile motion

1.162 Acceleration and velocity vector in projectile motion

Problem 1.162: A projectile is fired into the air at an angle with the ground. It follows the parabolic path as shown in the diagram, landing on the right. Neglect air resistance. At any instant, the projectile has a velocity $\vec{v}$ and an acceleration $\vec{a}$  . Which one or more of the drawings could not represent the directions for $\vec{v}$and $\vec{a}$  at any point on the trajectory?

 

Solution:

This is projectile motion. When air resistance is neglected, only one force acts on the body. It is the gravitational pull of earth which is always towards the ground in a vertical direction. So the acceleration of the projectile must be always in the downward vertical direction.

Velocity at any instant will be along the direction of motion of the projectile. Therefore it must be tangential to the path.

Keeping these facts in mind, let us examine each of the four cases given to us.

In diagram $1$.

Velocity is inclined and upward. This must represent the upward rise of the projectile just as it starts its motion. See the acceleration vector is vertically downward. So this is a possible situation.

 

In diagram $2$.

Velocity is perfectly horizontal. This is is possible at the highest point on the flight path. Acceleration vector is vertically downward. So this is a also possible .

In diagram $3$.

Velocity is inclined and downward. This  represents  the downward motion  of the projectile just as it starts falling. Ahe acceleration vector is vertically downward.. This is shown in the following diagram.

In diagram $4$.

In this the velocity vector is horizontal which is possible at the highest point on the path. But acceleration is shown pointed upward. This is not a possible cases.

Answer:

Out of the four situations, option $4$ is not practically correct.

4.101 Speed of car – Doppler effect

4.101 Speed of car – Doppler effect

Problem 4.101: A policeman on duty detects a drop of $30\: \%$ in the pitch of the horn of a motor-car as it crosses him. Calculate the speed of car, if the velocity of sound is $330m/s$.

Solution:

How to solve this problem? First try to visualize….There is a car and a policeman. The car is honking its horn. The policeman naturally hears the sound. But the car is moving towards him. He will hear an increase in pitch or frequency. This is due to Doppler effect. Now the car will go past him and will move away from him. He hears the sound with a lower frequency again due to Doppler effect.

The problem says  there is a drop of $30\%$ in the pitch. This is the percentage difference between the high frequency heard while the car approached him and the lower frequency heard while the car move past him.

Let $f$ be the actual frequency of the sound. Policeman hear this as frequency $f_1$ as th car moves towards him and as $f_2$ as the car moves away from him. Then second frequency is $30\:\%$ less than the first frequency. It is only $70\:\%$ of $f_1$.

Mathematically, $f_2$ =  $70\:\%$ of $f_1$.

Or, $f_2$ = $\frac{70}{100}\:f_1$.

Then, $\frac{f_2}{f_1}\:=\:\frac{70}{100}\:=\:\frac{7}{10}$------$(1)$.

Now we need to calculate $f_1$ and $f_2$ separately.

Calculation of $f_1$.

In Doppler effect the apparent frequency $f_1$ of a source moving towards a stationary observer is given by , $f_1\:=\:f\frac{V}{V\:-\:V_s}$------$(2)$.

Calculation of $f_2$.

The apparent frequency $f_2$ of a source moving away a stationary observer is given by , $f_2\:=\:f\frac{V}{V\:+\:V_s}$------$(3)$.

From $(1)$, $(2)$ and $(3)$,

$\frac{f_2}{f_1}\:=\:\frac{f\frac{V}{V\:+\:V_s}}{f\frac{V}{V\:-\:V_s}}\:=\:\frac{7}{10}$

Or, $\frac{V\:-\:V_s}{V\:+\:V_s}\:=\:\frac{7}{10}$.

Then, $10V\:-\:10V_s\:=\:7V\:+\:7V_s$.

Or, $7V_s\:+\:10V_s\:=\:10V\:-\:7V$.

That is, $17V_s\:=\:3V$

From this, $V_s\:=\:\frac{3V}{17}$.

Given that speed of sound in air is $330\:ms^{-1}$.

Therefore, speed of car,

$V_s\:=\:\frac{3\times 330}{17}\:=\:\frac{990}{17}\:=\:58.24\:ms^{-1}$.

So the car was traveling with a speed of $58.24\:ms^{-1}$.

1.161:Tension in strings connected to moving blocks

1.161:Tension in strings connected to moving blocks

Problem 1.161:A system consists of three blocks of masses $m_1$, $m_2$ and $m_3$ connected by a string passing over a pulley $P$. The mass $m_1$ hangs freely and $m_2$ and $m_3$ are on a rough horizontal table $($the coefficient of friction = $\mu$$)$. The pulley is frictionless and of negligible mass. Find the downward acceleration of mass $m_1$. Assume $m_1\:=\:m_2\:=\:m_3\:=m$.

Solution:

When released from rest, the hanging block will speed up downward, pulling the other to with it. This is possible due to the force exerted by the string which is called tension.

There will be frictional force from the rough surface, opposing the sliding motion of the other two blocks.Different forces acting on the system are as shown in the diagram below.

Since the blocks are connected by strings, they will have acceleration of  same magnitude.

Unbalanced force in any block will accelerate it.

For first block

Normal force from the floor, $N\:=\:mg$.

Force of friction, $f\:=\:\mu\:mg$.

The block is moving to the right side. Therefore force $T_2$ must be greater than force of friction.

So $T_2\:-\:f\:=\:ma$.

Or, $T_2\:-\:\mu mg\:=\:ma$.....$(1)$

For second  block

Normal force from the floor, $N\:=\:mg$.

Force of friction, $f\:=\:\mu\:mg$.

The block is moving to the right side. Therefore force $T_1$ must be greater than the sum of  force of friction and tension $T_2$.

So $T_1\:-\:(T_2\:+\:f)\:=\:ma$.

Or, $T_1\:-\:T_2\:-\:\mu mg\:=\:ma$......$(2)$

For third block

The block is moving downward. Therefore weight $mg$ must be greater than tension $T_1$.

So $mg\:-\:T_1\:=\:ma$.......$(3)$

Adding equations $(1)$, $(2)$ and $(3)$ we get,

$T_2\:-\:\mu mg\:+\:T_1\:-\:T_2\:-\:\mu mg\:+\:\:mg-\:T_1\:=\:ma\:+\:ma\:+\:ma$.

Or, $mg\:-\:2\mu mg =\:3ma $

From this acceleration, $a\:=\:\frac{mg\:-\:2\mu mg}{3m}$.

Or, $a\:=\:\frac{(1\:-\:2\mu)mg}{3m}$.

Then, $a\:=\:\frac{(1\:-\:2\mu)g}{3}$.

Tension $T_2$.

We have, from equation $(1)$

$T_2\:-\:\mu mg\:=\:ma$

Or, $T_2\:=\:ma\:+\:\mu mg$

Using this value of acceleration, $a\:=\:\frac{(1\:-\:2\mu)g}{3}$,

We get, $T_2\:=\:m(\frac{(1\:-\:2\mu)g}{3})\:+\:\mu mg$

Or, $T_2\:=\:mg(\frac{(1\:-\:2\mu)}{3}\:+\:\mu)$

Or, $T_2\:=\:mg\frac{(1\:-\:2\mu+\:3\mu)}{3}$

Then Tension, $T_2\:=\:mg\frac{(1\:+\mu)}{3}\:$.

Tension $T_1$

We have, from equation $(3)$

$mg\:-\:T_1\:=\:ma$

From this, $mg\:-\:ma\:=\:T_1$

Or $T_1\:=\:mg\:-\:ma$.

Using this value of acceleration, $a\:=\:\frac{(1\:-\:2\mu)g}{3}$,

We get, $T_1\:=\:mg\:-\:m(\frac{(1\:-\:2\mu)g}{3})$.

Or, $T_1\:=\:mg(1\:-\:\frac{(1\:-\:2\mu)}{3})$

Then, $T_1\:=\:mg\frac{(3\:-\:1\:+\:2\mu)}{3}$

or, $T_1\:=\:mg\frac{(2\:+\:2\mu)}{3}$

Then Tension $T_1\:=\:2mg\frac{(1\:+\:\mu)}{3}$

Tension in the vertical string is twice that in the horizontal  string.

 

1.160: Acceleration of connected blocks

1.160: Acceleration of connected blocks

Problem 1.160:A system consists of three blocks of masses $m_1$, $m_2$ and $m_3$ connected by a string passing over a pulley $P$. The mass $m_1$ hangs freely and $m_2$ and $m_3$ are on a rough horizontal table $($the coefficient of friction = $\mu$$)$. The pulley is frictionless and of negligible mass. Find the downward acceleration of mass $m_1$. Assume $m_1\:=\:m_2\:=\:m_3\:=m$. Take acceleration due to gravity as $10\:ms^{-2}$.

Solution:

When released from rest, the hanging block will speed up downward, pulling the other to with it. This is possible due to the force exerted by the string which is called tension.

There will be frictional force from the rough surface, opposing the sliding motion of the other two blocks.

Different forces acting on the system are as shown in the diagram below.

Since the blocks are connected by strings, they will have acceleration of  same magnitude.

Unbalanced force in any block will accelerate it.

For first block

Normal force from the floor, $N\:=\:mg$.

Force of friction, $f\:=\:\mu\:mg$.

The block is moving to the right side. Therefore force $T_2$ must be greater than force of friction.

So $T_2\:-\:f\:=\:ma$.

Or, $T_2\:-\:\mu mg\:=\:ma$.....$(1)$

For second  block

Normal force from the floor, $N\:=\:mg$.

Force of friction, $f\:=\:\mu\:mg$.

The block is moving to the right side. Therefore force $T_1$ must be greater than the sum of  force of friction and tension $T_2$.

So $T_1\:-\:(T_2\:+\:f)\:=\:ma$.

Or, $T_1\:-\:T_2\:-\:\mu mg\:=\:ma$......$(2)$

For third block

The block is moving downward. Therefore weight $mg$ must be greater than tension $T_1$.

So $mg\:-\:T_1\:=\:ma$.......$(3)$

Adding equations $(1)$, $(2)$ and $(3)$ we get,

$T_2\:-\:\mu mg\:+\:T_1\:-\:T_2\:-\:\mu mg\:+\:\:mg-\:T_1\:=\:ma\:+\:ma\:+\:ma$.

Or, $mg\:-\:2\mu mg =\:3ma $

From this acceleration, $a\:=\:\frac{mg\:-\:2\mu mg}{3m}$.

Or, $a\:=\:\frac{(1\:-\:2\mu)mg}{3m}$.

Then, $a\:=\:\frac{(1\:-\:2\mu)g}{3}$.

All blocks will accelerate at the same rate.

Problem 3.102 Heating a gas at constant volume

Problem 3.102:A cylinder of fixed capacity $44.8$ liters contains helium gas at standard temperature and pressure. What is the amount of heat needed to raise the temperature of the gas in the cylinder by $15\:^oC$. Take $R$ as $8.314\:Jmol^{-1}$

Solution:

The gas is in a cylinder of fixed capacity. So the process is isochoric in nature.

The cylinder has a capacity of $44.8$ liters.

The gas is at standard temperature and pressure.

From Avogadro's law, $22.4$ liters of gas at standard temperature and pressure will be one mole.

So the given gas has $2$ mole.

Helium is monoatomic with degree of freedom $f\:=\:3$.

Specific heat capacity at constant volume, $C_v\:=\:\frac{f}{2}R$.

With $f\:=\:3$, $C_v\:=\:\frac{3}{2}R$.

The required change in temperature of the gas in the cylinder is $15\:^oC$

This is equal to change of $15\:K$.

Heat required for this constant volume process can be estimated using the relation $Q\:=\:nC_v\delta T$.

Then, $Q\:=\:2\times\frac{3}{2}R\times 15$.

Or, $Q\:=\:3\times R\times 15\:=\:3\times 8.314\times 15\:=\:374.13\:J$.

Therefore, the amount of heat needed to raise the temperature of  $44.8$ liters helium gas without changing its volume is $374.13\:J$.