Chapter 4: Laws of Motion

Ans: It is given that,

Initial velocity of the truck, $u=0$ (Initially at rest)

Acceleration, $a=2m{{s}^{-2}}$ 

Time, $t=10s$ 

From first equation of motion: $v=u+at$ 

$\Rightarrow v=0+2\times 10=20{m}/{s}\;$ 

Therefore, the final velocity of the truck and the stone is$20{m}/{s}\;$.

At $t=11s$:

The horizontal component $({{v}_{x}})$  of velocity, in the absence of air resistance, remains unchanged, i.e. \[{{v}_{x}}=20m/s\].

The vertical component of velocity $({{v}_{y}})$ of the stone is given by the first equation of motion as: 

${{v}_{y}}=u+{{a}_{y}}\delta t$ 

Where, 

$\delta t=11-10=1s$ 

$a=g=10m/{{s}^{2}}$ 

$\Rightarrow {{v}_{y}}=0+10\times 1=10{m}/{{{s}^{2}}}\;$ 

The resultant velocity $(v)$  of the stone is:

Two Velocities of Different Magnitude

$\Rightarrow v=\sqrt{v_{x}^{2}+v_{y}^{2}}$ 

$\Rightarrow v=\sqrt{{{20}^{2}}+{{10}^{2}}}=\sqrt{400+100}$

$\Rightarrow v=\sqrt{500}=22.36m/s$

Consider $\theta $  as the angle made by the resultant velocity with the horizontal component of velocity, ${{v}_{x}}$.

$\Rightarrow \tan \theta =\left( \frac{{{v}_{y}}}{{{v}_{x}}} \right)$ 

$\Rightarrow \theta ={{\tan }^{-1}}\left( \frac{{{v}_{y}}}{{{v}_{x}}} \right)$ 

$\Rightarrow \theta ={{\tan }^{-1}}\left( \frac{10}{20} \right)$

$\Rightarrow \theta ={{\tan }^{-1}}\left( 0.5 \right)$

$\Rightarrow \theta ={{26.57}^{\circ }}$

Therefore, the magnitude of resultant velocity is $22.36{m}/{s}\;$ making an angle of ${{26.57}^{\circ }}$ with the horizontal component of velocity.

1. acceleration of the stone at $t=11s$?

Ans: When the stone is dropped from the truck, the horizontal force acting on it becomes zero. However, the stone continues to move under the influence of gravity. 

Therefore, the acceleration of the stone is $10m{{s}^{-2}}$  and it acts vertically downward. 

Ans: If the string is cut when the bob is at one of its extremes then the bob will fall vertically on the ground.

Therefore, at the extreme position, the velocity of the bob becomes zero.

1. at its mean position. 

Ans: If the string is cut when the bob is at its mean position then the bob will trace a projectile path having the horizontal components of velocity only.

The direction of this velocity is tangential to the arc formed by the oscillating bob. At the mean position, the velocity of the bob is $1m/s$. 

Therefore, it will follow a parabolic path. 

Ans: It is given that,

Mass of the man, $m=70kg$ 

Acceleration, $a=0$(uniform speed)

From Newton’s second law: $R-mg=ma$ 

Where,

$ma$ is the net force acting on the man.

$\Rightarrow R-70\times 10=0$ 

$\Rightarrow R=700N$ 

Reading on the weighing scale$=\frac{700}{g}=\frac{700}{10}=70kg$ 

Therefore, the mass of the man, $m=70kg$ 

1. downwards with a uniform acceleration of $5m{{s}^{-2}}$,

Ans: Acceleration,$a=5{m}/{{{s}^{2}}}\;$ downward

From Newton’s second law: $R=m(g-a)$

$\Rightarrow R=70\left( 10-5 \right)=70\times 5$ 

$\Rightarrow R=350N$ 

Reading on the weighing scale$=\frac{350}{g}=\frac{350}{10}=35kg$ 

Therefore, the mass of the man, $m=35kg$ 

1. upwards with a uniform acceleration of $5m{{s}^{-2}}$.

Ans: Acceleration,$a=5{m}/{{{s}^{2}}}\;$ upward

From Newton’s second law: $R=m(g+a)$

$\Rightarrow R=70\left( 10+5 \right)=70\times 15$ 

$\Rightarrow R=1050N$ 

Reading on the weighing scale$=\frac{1050}{g}=\frac{1050}{10}=105kg$ 

Therefore, the mass of the man, $m=105kg$ 

1. What would be the reading if the lift mechanism failed and it hurtled down freely under gravity? 

Ans: When the lift moves freely under gravity,

Acceleration, $a=g=10m{{s}^{-2}}$ 

From Newton’s second law: $R=m(g-a)$

$\Rightarrow R=70\left( 10-10 \right)=0$ 

Reading on the weighing scale$=\frac{0}{g}=\frac{0}{10}=0kg$ 

Therefore, the man will be in a state of weightlessness.

Ans:  For $t<0$ :

From the given graph, it is observed that the position of the particle is coincident with the time axis. It indicates that the displacement of the particle in this time interval is zero. 

Therefore, the force acting on the particle is zero. 

For $t>4s$ :

From the given graph, it is observed that the position of the particle is parallel to the time axis. It indicates that the particle is at rest at a distance of $3m$ from the origin.

Therefore, no force acts on the particle. 

For $0<t<4$ :

From the given position-time graph, it is observed that it has a constant slope. Thus, the acceleration produced in the particle is zero. 

Therefore, the force acting on the particle is zero. 

1. impulse at $t=0$ and $t=4s$ ? 

Ans: At $t=0$:

\[Impulse=Change\text{ }in\text{ }momentum=mv-mu\] 

Mass of the particle, $m=4kg$

Initial velocity of the particle, $u=0$ 

Final velocity of the particle, $v=\frac{3}{4}m/s$ 

Impulse $=4\left( \frac{3}{4}-0 \right)=3kgm{{s}^{-1}}$ 

At $t=4s$:

Initial velocity of the particle, $u=\frac{3}{4}m/s$ 

Final velocity of the particle, $v=0$ 

Impulse\[=4\left( 0-\frac{3}{4} \right)=-3kgm{{s}^{-1}}\]

Therefore, the impulse at $t=0$ is $3kgm{{s}^{-1}}$ and at $t=4s$is $-3kgm{{s}^{-1}}$.

Ans: It is given that,

Horizontal force, $F=600N$ 

Mass of body A, ${{m}_{1}}=10kg$ 

Mass of body B, ${{m}_{2}}=20kg$

Total mass of the system, $m={{m}_{1}}+{{m}_{2}}=30kg$

From Newton’s second law of motion, 

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

$\Rightarrow a=\frac{F}{m}=\frac{600}{30}=20m{{s}^{-2}}$

When force $F$ is applied on body A: 

Two Bodies of Masses 

The equation of motion can be written as: $F-T=ma$

$\Rightarrow T=F-{{m}_{1}}a$ 

\[\Rightarrow T=600-10\times 20=400N\]

Therefore, the tension in the string is $400N$.

1. B along the direction of the string

Ans: When force $F$  is applied on body B: 

Two Bodies of Masses 

The equation of motion can be written as:  $F-T={{m}_{2}}a$ 

$\Rightarrow T=F-{{m}_{2}}a$ 

\[\Rightarrow T=600-20\times 20=200N\]

Therefore, the tension in the string is $200N$.

Ans: The given system of two masses and a pulley are represented in the following figure: 

Two Bodies of Masses 

Smaller mass, ${{m}_{1}}=8kg$ 

Larger mass, ${{m}_{2}}=12kg$

Tension in the string $=T$ 

Mass \[{{m}_{2}}\], owing to its weight, moves downward with acceleration $a$, and mass moves ${{m}_{1}}$upward.

Applying Newton’s second law of motion to the system of each mass: 

For mass ${{m}_{1}}$: 

The equation of motion can be written as: \[T-{{m}_{1}}g=ma\]    .......$(1)$ 

For mass ${{m}_{2}}$:

The equation of motion can be written as: \[{{m}_{2}}g-T={{m}_{2}}a\]   ........$(2)$ 

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

$({{m}_{2}}-{{m}_{1}})g=({{m}_{1}}+{{m}_{2}})a$ 

$\Rightarrow a=\left( \frac{{{m}_{2}}-{{m}_{1}}}{{{m}_{2}}+{{m}_{1}}} \right)g$        ........$(3)$ 

$\Rightarrow a=\left( \frac{12-8}{12+8} \right)\times 10=\frac{4}{20}\times 10$

$\Rightarrow a=2m{{s}^{-2}}$ 

Thus, the acceleration of the masses is $2m{{s}^{-2}}$ . Substituting the value of  $a$ in equation $(2)$:

\[\Rightarrow {{m}_{2}}g-T={{m}_{2}}\left( \frac{{{m}_{2}}-{{m}_{1}}}{{{m}_{2}}+{{m}_{1}}} \right)g\]

\[\Rightarrow T=\left( {{m}_{2}}-\frac{m_{2}^{2}-{{m}_{1}}{{m}_{2}}}{{{m}_{2}}+{{m}_{1}}} \right)g\]

\[\Rightarrow T=\left( \frac{2{{m}_{1}}{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}} \right)g\]

\[\Rightarrow T=\left( \frac{2\times 12\times 8}{12+8} \right)\times 10\]

\[\Rightarrow T=96N\]

Thus, the tension in the string is $96N$.

Ans: Consider $m$, ${{m}_{1}}$ and ${{m}_{2}}$ as the respective masses of the parent nucleus and the two daughter nuclei. The parent nucleus is at rest. 

Initial momentum of the system (parent nucleus) $=0$ 

Let ${{v}_{1}}$ and ${{v}_{2}}$  be the respective velocities of the daughter nuclei having masses ${{m}_{1}}$ and ${{m}_{2}}$.

Total linear momentum of the system after disintegration$={{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}$ 

From the law of conservation of momentum: 

\[\mathbf{Total}\text{ }\mathbf{initial}\text{ }\mathbf{momentum}=\mathbf{Total}\text{ }\mathbf{final}\text{ }\mathbf{momentum}\] 

 $\Rightarrow 0={{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}$ 

$\Rightarrow {{m}_{1}}{{v}_{1}}=-{{m}_{2}}{{v}_{2}}$

$\Rightarrow {{v}_{1}}=\frac{-{{m}_{2}}{{v}_{2}}}{{{m}_{1}}}$

The negative sign indicates that the fragments of the parent nucleus move in directions opposite to each other.

Ans: It is given that,

Mass of each ball$=0.05kg$ 

Initial velocity of each ball$=6m/s$ 

Magnitude of the initial momentum of each ball, ${{p}_{i}}=0.3kgm{{s}^{-1}}$ 

After collision, the balls change their directions of motion without changing the 

magnitudes of their velocity. 

Final momentum of each ball, ${{p}_{f}}=-0.3kgm{{s}^{-1}}$

Impulse imparted to each ball$=$ Change in the momentum of the system 

$\Rightarrow \operatorname{Im}pulse={{p}_{f}}-{{p}_{i}}$ 

$\Rightarrow \operatorname{Im}pulse=-0.3-0.3=-0.6kgm{{s}^{-1}}$

The negative sign indicates that the impulses imparted to the balls are opposite in direction. 

Ans: It is given that,

Mass of the gun, $M=100kg$ 

Mass of the shell, $m=0.020kg$ 

Muzzle speed of the shell, $v=80m/s$ 

Recoil speed of the gun  $=V$.

Both the gun and the shell are at rest initially. 

Initial momentum of the system$=0$

Final momentum of the system$=mv-MV$ 

Here, the negative sign appears because the directions of the shell and the gun are opposite to each other. 

From the law of conservation of momentum:

\[Final\text{ }momentum=Initial\text{ }momentum\] 

$mv-MV=0$

$\Rightarrow V=\frac{mv}{M}$ 

$\Rightarrow V=\frac{0.020\times 80}{100\times 1000}=0.016m/s$ 

Therefore, the recoil speed of the gun is $0.016m/s$.

Ans: The given situation can be represented as:

Deflection of a Ball Hit by a Batsman

Where, 

$AO=$ Incident path of the ball 

$OB=$ Path followed by the ball after a deflection 

$\angle AOB=$ Angle between the incident and deflected paths of the ball$={{45}^{\circ }}$ 

$\angle AOB=\angle BOP={{22.5}^{\circ }}=\theta $

Initial and final velocities of the ball$=v$ 

The horizontal component of the initial velocity$=v\cos \theta $ along $RO$ 

Vertical component of the initial velocity$=v\sin \theta $ along $PO$ 

The horizontal component of the final velocity$=v\cos \theta $ along $OS$ 

The vertical component of the final velocity $=v\sin \theta $ along $OP$ 

The horizontal components of velocities suffer no change. The vertical components of velocities are in opposite directions. 

It is known that Impulse imparted to the ball$=$ Change in the linear momentum of the ball.

$\operatorname{Im}pulse=mv\cos \theta -(-mv\cos \theta )=2mv\cos \theta $ 

It is given that,

Mass of the ball, $m=0.15kg$ 

Velocity of the ball, $v=54km/h=54\times \frac{5}{18}=15m/s$ 

Impulse$=2\times 0.15\times 15\cos {{22.5}^{\circ }}=4.16kgm{{s}^{-1}}$ 

Therefore, impulse imparted to the ball is $4.16kgm{{s}^{-1}}$.