Chapter 4: Laws of Motion

Ans: The net force is zero.

As the speed of the rain drop falling down is constant, its acceleration is zero.

Therefore, from Newton’s second law of motion, the net force acting on the rain drop is zero.

1. a cork of mass $10g$ floating on water, 

Ans: The net force is zero.

It is known that the weight of a cork floating on water acts downward. 

The weight of the cork is balanced by buoyant force exerted by the water in the upward direction.

Therefore, no net force acts on the floating cork.

1. a kite skilfully held stationary in the sky, 

Ans: The net force is zero.

The kite is stationary in the sky indicates that it is not moving.

Therefore, from Newton’s first law of motion, the net force acting on the kite is zero.

1. a car moving with a constant velocity of $30{km}/{h}\;$ on a rough road, 

Ans: The net force is zero.

As the car is moving with constant velocity, its acceleration is zero.

Therefore, from Newton’s second law of motion, net force acting on the car is equal to zero.

1. a high-speed electron in space far from all material objects, and free of electric and magnetic fields. 

Ans: The net force is zero.

As the high speed electron is free from the influence of all the fields, no net force acts on the electron.

Ans: It is known that,

Acceleration due to gravity always acts downward irrespective of the direction of motion of an object. The only force that acts on the pebble thrown vertically upward during its upward motion is the gravitational force.

From Newton’s second law of motion: $F=m\times a$ 

Where,

$F$ is the net force 

$m$ is the mass of the pebble, $m=0.05kg$ 

a is the acceleration due to gravity, $a=g=10{m}/{{{s}^{-2}}}\;$ 

$\Rightarrow F=0.05\times 10=0.5N$ 

Therefore, the net force on the pebble is $0.5N$ and this force acts in the downward direction. 

1. during its downward motion, 

Ans: The only force that acts on the pebble during its downward motion is the gravitational force.

Therefore, the net force on the pebble in its downward direction is same as in upward direction i.e., $0.5N$ and this force acts in the downward direction. 

1. at the highest point where it is momentarily at rest. Do your answers change if the pebble was thrown at an angle of ${{45}^{\circ }}$ with the horizontal direction? 

Ans: When the pebble is thrown at an angle of ${{45}^{\circ }}$with the horizontal, it will have both the horizontal and vertical components of velocity. 

At the highest point, only the vertical component of velocity becomes zero. However, the pebble will have the horizontal component of velocity throughout its motion. This component of velocity produces no effect on the net force acting on the pebble. 

Therefore, the net force on the pebble is $0.5N$.

Ans: It is given that,

Mass of the stone, \[\mathbf{m}=\mathbf{0}.\mathbf{1kg}\] 

Acceleration of the stone, \[\mathbf{a}=g=10{m}/{{{s}^{2}}}\;\] 

From Newton’s second law of motion, 

The net force acting on the stone is $F=ma=mg$ 

$\Rightarrow F=0.1\times 10=1N$ 

It is known that acceleration due to gravity always acts in the downward direction. 

Therefore, the magnitude of force is $1N$ and its direction is vertically downward.

1. just after it is dropped from the window of a train running at a constant velocity of $36{km}/{h}\;$, 

Ans: It is given that,

The train is moving with a constant velocity. 

Therefore, its acceleration is zero in the direction of its motion, i.e. in the horizontal direction. 

Thus, no force is acting on the stone in the horizontal direction. 

The net force acting on the stone is because of acceleration due to gravity and it always acts vertically downward. 

Therefore, the magnitude of force is $1N$ and its direction is vertically downward.

1. just after it is dropped from the window of a train accelerating with $1m{{s}^{-2}}$ 

Ans: It is given that, 

The train is accelerating at the rate of $1m{{s}^{-2}}$.

Therefore, the net force acting on the stone is $F'=ma=0.1\times 1=1N$ 

This force is acting in the horizontal direction. Now, when the stone is dropped, the horizontal force stops acting on the stone. This is because of the fact that the force acting on a body at an instant depends on the situation at that instant and not on earlier situations. 

Therefore, the net force acting on the stone is given only by acceleration due to gravity i.e., $F=mg=1N$.

Therefore, the magnitude of force is $1N$ and its direction is vertically downward.

1. lying on the floor of a train which is accelerating with $1m{{s}^{-2}}$, the stone being at rest relative to the train. 

Ans: It is known that,

The weight of the stone is balanced by the normal reaction of the floor. The only acceleration is provided by the horizontal motion of the train. 

Acceleration of the train, $a=1m{{s}^{-2}}$

The net force acting on the stone will be in the direction of motion of the train. 

Magnitude: $F=ma=0.1\times 1=0.1N$ 

Therefore, the magnitude of force is $0.1N$ and its direction is in the direction of motion of the train.

Ans: (i) $T$ 

The centripetal force of a particle connected to a string revolving in a circular path around a centre is provided by the tension produced in the string.

Therefore, the net force on the particle is the tension $T$ , i.e., 

$F=T=\frac{m{{v}^{2}}}{l}$ 

Where $F$ is the net force acting on the particle. 

Ans: It is given that,

Retarding force, \[F=50N\] 

Mass of the body, $m=20kg$ 

Initial velocity of the body, \[u=15m/s\] 

Final velocity of the body, $v=0$ 

From Newton’s second law of motion,

The acceleration $(a)$ produced in the body: $F=ma$ 

$\Rightarrow -50=20\times a$ 

$\Rightarrow a=\frac{-50}{20}=-2.5m{{s}^{-2}}$ 

From the first equation of motion, 

The time $(t)$ taken by the body to come to rest: $v=u+at$ 

$\Rightarrow 0=15+(-2.5)t$ 

$\Rightarrow t=\frac{-15}{-2.5}=6s$ 

Therefore, the time taken by the body to stop is $6s$.

Ans: It is given that,

Mass of the body, $m=3kg$ 

Initial speed of the body, \[u=2m/s\] 

Final speed of the body, \[v=3.5m/s\]

Time, $t=25s$ 

From the first equation of motion, 

The acceleration $(a)$ produced in the body: $v=u+at$

$\Rightarrow a=\frac{v-u}{t}$ 

$\Rightarrow a=\frac{3.5-2}{25}=\frac{1.5}{25}=0.6m{{s}^{-2}}$ 

From Newton’s second law of motion, 

Force, $F=ma$ 

\[\Rightarrow F=3\times 0.06=0.18N\] 

As the application of force does not change the direction of the body, the net force acting on the body is in the direction of its motion. 

Therefore, the magnitude of force is $0.18N$ and direction is along the direction of motion.

Ans: It is given that,

Mass of the body, $m=5kg$ 

Representation of given data:

Law of Vectors

Resultant of two forces $8N$ and $6N$, $R=\sqrt{{{\left( 8 \right)}^{2}}+{{\left( -6 \right)}^{2}}}$ 

$\Rightarrow R=\sqrt{64+36}$ 

$\Rightarrow R=10N$ 

Angle made by $R$ with the force of $8N$ 

$\theta ={{\tan }^{-1}}\left( \frac{-6}{8} \right)=-{{36.87}^{\circ }}$ 

The negative sign indicates that $\theta $ is in the clockwise direction with respect to the force of magnitude $8N$. 

From Newton’s second law of motion,

The acceleration $(a)$ produced in the body: $F=ma$ 

$a=\frac{F}{m}=\frac{10}{5}=2m{{s}^{-2}}$ 

Therefore, the magnitude of acceleration is $2m{{s}^{-2}}$ and direction is ${{37}^{\circ }}$ with a force of $8N$.

Ans: It is given that,

Initial speed of the three-wheeler, \[u=36\text{ }km/h\] 

Final speed of the three-wheeler, \[v=0m/s\] 

Time, $t=4s$ 

Mass of the three-wheeler, $m=400kg$ 

Mass of the driver, $m'=65kg$ 

Total mass of the system, \[M=400+65=465kg\] 

From the first law of motion, 

The acceleration $(a)$ of the three-wheeler can be calculated from: $v=u+at$ 

$\Rightarrow a=\frac{v-u}{t}=\frac{0-10}{4}$ 

$\Rightarrow a=-2.5{m}/{{{s}^{2}}}\;$ 

The negative sign indicates that the velocity of the three-wheeler is decreasing with time. 

From Newton’s second law of motion, 

The net force acting on the three-wheeler can be calculated as: $F=ma$ 

$\Rightarrow F=465\times (-2.5)$ 

$\Rightarrow F=-1162.5N$ 

The negative sign indicates that the force is acting against the direction of motion of the three-wheeler. 

Therefore, the average retarding force on the vehicle is $-1162.5N$.

Ans: It is known that,

Mass of the rocket, $m=20,000kg$ 

Initial acceleration, $a=5m{{s}^{-2}}$ 

Acceleration due to gravity, $g=10m{{s}^{-2}}$

By using Newton’s second law of motion, 

The net force (thrust) acting on the rocket can be written as:

$F-mg=ma$ 

$\Rightarrow F=m(g+a)$ 

$\Rightarrow F=20000(10+5)=20000\times 15$ 

$\Rightarrow F=3\times {{10}^{5}}N$  

Therefore, the initial thrust(force) of the blast is $3\times {{10}^{5}}N$.

Ans: It is given that, 

Mass of the body, $m=0.40kg$ 

Initial speed of the body, \[u=10m/s\] due north 

Force acting on the body, $F=-8.0N$ 

Acceleration produced in the body, $a=\frac{F}{m}$

At $t=0$ 

$\Rightarrow a=\frac{-8}{0.4}=-20m{{s}^{-2}}$ 

At $t=-5s$ 

Acceleration, $a'=0$ and $u=10m/s$ 

$s=ut+\frac{1}{2}a'{{t}^{2}}$ 

$\Rightarrow s=10\times (-5)=-50m$ 

At $t=25s$

Acceleration, $a=-20m{{s}^{-2}}$ and $u=10m/s$ 

$s'=ut'+\frac{1}{2}a{{(t')}^{2}}$ 

$\Rightarrow s'=10\times (25)+\frac{1}{2}\times (-20){{(25)}^{2}}$ 

$\Rightarrow s'=250+(-6250)$ 

$\Rightarrow s'=-6000m$ 

At $t=100s$

For $0\le t\le 30s$, $a=-20m{{s}^{-2}}$ and $u=10m/s$

${{s}_{1}}=ut+\frac{1}{2}a{{t}^{2}}$

$\Rightarrow {{s}_{1}}=10\times 30+\frac{1}{2}\times (-20)\times {{(30)}^{2}}$

$\Rightarrow {{s}_{1}}=300-9000$

$\Rightarrow {{s}_{1}}=8700m$

For $30s\le t\le 100s$

First equation of motion: $v=u+at$ 

$\Rightarrow v=10+(-20)\times 30$ 

$\Rightarrow v=-590m{{s}^{-1}}$ 

Velocity of body after $30s=-590m/s$ 

For motion between $30s$ to $100s$ i.e., in $70s$ 

${{s}_{2}}=vt+\frac{1}{2}a'{{t}^{2}}$

${{s}_{2}}=-590\times 70=-41300m$

Total distance, $s''={{s}_{1}}+{{s}_{2}}$ 

$\Rightarrow s''=-8700+(-41300)=-50000m$ 

Therefore, the position of the body at $t=-5s$ is $-50m$ at $t=25s$ is  $-6000m$ and at $t=100s$ is$-50,000m$.