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

Updated On May 15, 2024

By SaraNextGen

Numerical Problems
Question 1.
A force of $50 \mathrm{~N}$ act on the object of mass $20 \mathrm{~kg}$. shown in the figure. Calculate the acceleration of the object in $\mathrm{x}$ and $\mathrm{y}$ directions.
Answer:
Given $\mathrm{F}=50 \mathrm{~N}$ and $\mathrm{m}=20 \mathrm{~kg}$
(1) component of force along $x$ - direction
$
\begin{aligned}
& \mathrm{F}_{\mathrm{x}}=\mathrm{F} \cos \theta \\
& =50 \mathrm{x} \cos 30^{\circ}=43.30 \mathrm{~N} \\
& \mathrm{a}_{\mathrm{X}}=\frac{F_x}{m}=\frac{43.30}{20}=2.165 \mathrm{~ms}^{-2}
\end{aligned}
$
(2) Component of force along $y$ - direction
$
\begin{aligned}
& \mathrm{F}_{\mathrm{y}}=\mathrm{F} \sin \theta=50 \sin 30^{\circ}=25 \mathrm{~N} \\
& \mathrm{a}_{\mathrm{y}}=\frac{F_y}{m}=\frac{25}{20}=1.25 \mathrm{~ms}^{-2}
\end{aligned}
$

Question 2.
A spider of mass $50 \mathrm{~g}$ is hanging on a string of a cob web as shown in the figure. What is the tension in the string?
Answer:
Given $\mathrm{m}=50 \mathrm{~g}, \mathrm{~T}=$ ?
Tension in the string $\mathrm{T}=\mathrm{mg}$ $=50 \times 10^{-2} \times 9.8=0.49 \mathrm{~N}$
Question 3.
What is the reading shown in spring balance?

Answer:
When a spring balance hung on a rigid support and load is attached at its other end, the weight of the load exerts a force on the rigid support in turn support exerts equal and opposite force on that load, so that balance will be stretched. This is the principle of spring balance. Flence the answer is $4 \mathrm{~kg}$.
Given: $\mathrm{m}=2 \mathrm{~kg}, 0=30^{\circ}$.
Resolve the weight into its component as $\mathrm{mg} \sin \theta$ and $\mathrm{mg} \cos \theta$.
Here $\mathrm{mg} \sin \theta$ acts parallel to the surface
$
\begin{aligned}
& \therefore \mathrm{W}=\mathrm{mg} \sin \theta \\
& =2 \times 9.8 \times \sin 30^{\circ}=2 \times 9.8 \times \frac{1}{2}=9.8 \mathrm{~N}
\end{aligned}
$
Question 4.
The physics books are stacked on each other in the sequence: +1 volumes 1 and $2 ;+2$ volumes 1 and 2 on a table
(a) Identify the forces acting on each book and draw the free body diagram.
(b) Identify the forces exerted by each book on the other.
Answer:
Let $\mathrm{m}_1, \mathrm{~m}_2, \mathrm{~m}_3, \mathrm{~m}_4$, are the masses of +1 volume $\mathrm{I}$ and $I \mathrm{I}$ and +2 volumes I \& II

$\text { Let } \mathrm{m}_1, \mathrm{~m}_2, \mathrm{~m}_3, \mathrm{~m}_4 \text {, are the masses of }+1 \text { volume I and II and }+2 \text { volumes I \& II }$

(a) Force on book $m_4$
- Downward gravitational force acting downward $\left(\mathrm{m}_3 \mathrm{~g}\right)$
- Upward normal force $\left(\mathrm{N}_3\right)$ exerted by book of mass $\mathrm{m}_3$

(b) Force on book $\mathrm{m}_3$
- Downward gravitational force $\left(\mathrm{m}_3 \mathrm{~g}\right)$
- Downward force exerted by $\mathrm{m}_4\left(\mathrm{~N}_4\right)$
- Upward force exerted by $\mathrm{m}_2\left(\mathrm{~N}_2\right)$

(c) Force on book $\mathrm{m}_2$
- Downward gravitational force $\left(\mathrm{m}_2 \mathrm{~g}\right)$
- Downward force exerted by $\mathrm{m}_3\left(\mathrm{~N}_3\right)$
- Upward force exerted by $\mathrm{m}_1\left(\mathrm{CN}_1\right)$

(d) Force on book $\mathrm{m}_1$
- Downward gravitational force exerted by earth $\left(\mathrm{m}_1 \mathrm{~g}\right)$
- Downward force exerted by $\mathrm{m}_2\left(\mathrm{~N}_2\right)$
- Upward force exerted by the table $\left(\mathrm{N}_{\text {table }}\right)$

Question 5.
A bob attached to the string oscillates back and forth. Resolve the forces acting on the bob in to components. What is the acceleration experienced by the bob at an angle $\theta$.
Answer:
The gravitational force ( $\mathrm{mg}$ ) acting downward can be resolved into two components as $\mathrm{mg}$ $\cos \theta$ and $\mathrm{mg} \sin \theta$
$\mathrm{T}$ - tension exerted by the string.
Tangential force $\mathrm{F}_{\mathrm{T}}=\mathrm{ma}_{\mathrm{T}}=\mathrm{mg} \sin \theta$
$\therefore$ Tangential acceleration $\mathrm{a}_{\mathrm{T}}=\mathrm{g} \sin \theta$
Centripetal force $\mathrm{Fc}=\mathrm{ma}_{\mathrm{c}}=\mathrm{T}-\mathrm{mg} \cos \theta$
$\mathrm{a}_{\mathrm{c}}=\frac{T-m g \cos \theta}{m}$

Question 6.
Two masses $m_1$ and $m_2$ are connected with a string passing over a friction-less pulley fixed at the comer of the table as shown in the figure. The coefficient of static friction of mass $\mathrm{m}_1$ with the table is $\mu_{\mathrm{s}}$ Calculate the minimum mass $\mathrm{m}_3$ that may be placed on $\mathrm{m}_1$ to prevent it from sliding. Check if $\mathrm{m}_1=15 \mathrm{~kg}, \mathrm{~m}_2=10 \mathrm{~kg}, \mathrm{~m}_3=25$ and $\mu_{\mathrm{s}}=0.2$.

Solution:
Let $\mathrm{m}_3$ is the mass added on $\mathrm{m}_1$
Maximal static friction
$f_s^{\max }=\mu_{\mathrm{s}} \mathrm{N}=\mu_{\mathrm{s}}\left(\mathrm{m}_1+\mathrm{m}_3\right) \mathrm{g}$
Here
$\mathrm{N}=\left(\mathrm{m}_1+\mathrm{m}_3\right) \mathrm{g}$
Tension acting on string $=\mathrm{T}=\mathrm{m}_2 \mathrm{~g}$
Equate (1) and (2)
$\mu_{\mathrm{s}}\left(\mathrm{m}_1+\mathrm{m}_3\right)=\mathrm{m}_2 \mathrm{~g}$
$\mu_{\mathrm{s}} \mathrm{m}_1+\mu_{\mathrm{s}} \mathrm{m}_3=\mathrm{m}_2$
$\mathrm{m}_3=f_s^{\max }-\mathrm{m}_1$
(ii) Given,
$\mathrm{m}_1=15 \mathrm{~kg}, \mathrm{~m}_2=10 \mathrm{~kg}: \mathrm{m}_3=25 \mathrm{~kg}$ and $\mu_{\mathrm{s}}=0.2$
$\mathrm{m}_3=f_s^{\max }-\mathrm{m}_1$
$\mathrm{m}_3=\frac{10}{0.2}-15=50-15=35 \mathrm{~kg}$
The minimum mass $\mathrm{m}_3=35 \mathrm{~kg}$ has to be placed on $\mathrm{ml}$ to prevent it from sliding. But here $\mathrm{m}_3$ $=25 \mathrm{~kg}$ only
The combined masses $\left(\mathrm{m}_1+\mathrm{m}_3\right)$ will slide.
Question 7.
Calculate the acceleration of the bicycle of mass $25 \mathrm{~kg}$ as shown in Figures 1 and 2.

Answer:
Given:
Mass of bicycle $\mathrm{m}=25 \mathrm{~kg}$
Fig. I:
Net force acting in the forward direction, $\mathrm{F}=500-400=100 \mathrm{~N}$ acceleration a $=\frac{F}{m}=\frac{100}{25}=4 \mathrm{~ms}^{-2}$
Fig. II:
Net force acting on bicycle $F=400-400=0$
$\therefore$ acceleration $\mathrm{a}=\frac{F}{m}=\frac{0}{25}=0$
Question 8.
Apply Lami's theorem on sling shot and calculate the tension in each string?

Answer:
Given $\mathrm{F}=50 \mathrm{~N}, \theta=30^{\circ}$
Here $\mathrm{T}$ is resolved into its components as $\mathrm{T} \sin \theta$ and $\mathrm{T} \cos \theta$ as shown.
According to Lami's theorem,
$
\begin{aligned}
& \frac{F}{\sin \theta}=\frac{T}{\sin (180-\theta)}=\frac{T}{\sin (180-\theta)} \\
& \frac{F}{\sin \theta}=\frac{T}{\sin \theta} \\
& \frac{F}{2 \sin \theta \cos \theta}=\frac{T}{\sin \theta}\left[\mathrm{T}=\frac{T}{2 \cos \theta}\right] \\
& \mathrm{T}=\frac{T}{2 \cos \theta}=\frac{50}{2 \cos 30}=28.868 \mathrm{~N}
\end{aligned}
$
Question 9.
A football player kicks a $0.8 \mathrm{~kg}$ ball and imparts it a velocity $12 \mathrm{~ms}^{-1}$. The contact between the foot and ball is only for one - sixtieth of a second. Find the average kicking force.
Answer:
Given,
Mass of the ball $=0.8 \mathrm{~kg}$
Final velocity $(V)=12 \mathrm{~ms}^{-1}$ and time $\mathrm{t}=\frac{1}{60} \mathrm{~s}$
Initial velocity $=0$
We know the average kicking force
$
\begin{aligned}
& \mathrm{F}=\mathrm{ma}=\frac{m(v-u)}{t}=\frac{0.8(12-0)}{\left(\frac{1}{60}\right)} \\
& \mathrm{F}=576 \mathrm{~N}
\end{aligned}
$
Question 10.
A stone of mass $2 \mathrm{~kg}$ is attached to a string of length 1 meter. The string can withstand maximum tension $200 \mathrm{~N}$. What is the maximum speed that stone can have during the whirling motion?

Solution:
Given,
Mass of a stone $=2 \mathrm{~kg}$,
length of a string $=1 \mathrm{~m}$
Maximum tension $=200 \mathrm{~N}$
The force acting on a stone in the whirling motion is centripetal force. Which is provided by tension of the string.
$
\begin{aligned}
& \mathrm{T}_{\max }=\mathrm{F}_{\max }=\frac{m V_{\max }^2}{r} \\
& 200=v_{\max }^2=100 \\
& v_{\max }^2=10 \mathrm{~ms}^{-1}
\end{aligned}
$
Question 11.
Imagine that the gravitational force between Earth and Moon is provided by an invisible string that exists between the Moon and Earth. What is the tension that exists in this invisible string due' to Earth's centripetal force? (Mass of the Moon $=7.34 \times 1022 \mathrm{~kg}$, Distance between Moon and Earth $=3.84 \times 108 \mathrm{~m}$ ).

Solution:
Given,
Mass of the moon $=7.34 \times 1022 \mathrm{~kg}$
Distance between moon and earth $=3.84 \times 108 \mathrm{~m}$
Centripetal force $=\mathrm{F}=\frac{m \mathrm{~V}^2}{r}=\frac{7.34 \times 10^{22} \times\left(1.023 \times 10^3\right)^2}{3.84 \times 10^8}=2 \times 10^{20}$
Question 12.
Two bodies of masses $15 \mathrm{~kg}$ and $10 \mathrm{~kg}$ are connected with light string kept on a smooth surface. A horizontal force $\mathrm{F}=500 \mathrm{~N}$ is applied to a $15 \mathrm{~kg}$ as shown in the figure. Calculate the tension acting in the string.

Answer:
Given,
$
\mathrm{m}_1=15 \mathrm{~kg}, \mathrm{~m}_2=10 \mathrm{~kg}, \mathrm{~F}=500 \mathrm{~N}
$
Tension acting in the string $\mathrm{T}=\frac{m_2}{m_1+m_2} \mathrm{~F}$
$
\mathrm{T}=\frac{10}{25} \times 500=200 \mathrm{~N}
$
Question 13.
People often say "For every action there is an equivalent opposite reaction". Here they meant 'action of a human'. Is it correct to apply Newton's third law to human actions? What is meant by 'action' in Newton third law? Give your arguments based on Newton's laws.
Answer:
Newton's third law is applicable to only human's physical actions which involves physical force. Third law is not applicable to human's psychological actions or thoughts.
Question 14.
A car takes a turn with velocity $50 \mathrm{~ms}^{-1}$ on the circular road of radius of curvature To $\mathrm{m}$. Calculate the centrifugal force experienced by a person of mass $60 \mathrm{~kg}$ inside the car?
Answer:
Given,
Mass of a person $=60 \mathrm{~kg}$
Velocity of the car $=50 \mathrm{~ms}^{-1}$
Radius of curvature $=10 \mathrm{~m}$
Centrifugal force $\mathrm{F}=\frac{m \mathrm{~V}^2}{r}=\frac{60 \times(50)^2}{10}=15,000 \mathrm{~N}$
Question 15.
A long stick rests on the surface. A person standing $10 \mathrm{~m}$ away from the stick. With what minimum speed an object of mass $0.5 \mathrm{~kg}$ should he thrown so that it hits the stick. (Assume the coefficient of kinetic friction is 0.7 ).
Answer:

Given,
Distance $(\mathrm{s})=10 \mathrm{~m}$
Mass of the object $(\mathrm{m})=0.5 \mathrm{~kg}$
Coefficient of kinetic friction $(\mu)=0.7$
Work done in moving a body in horizontal surface $\omega=\mu_{\mathrm{R}} \times \mathrm{s}=\mu \mathrm{mg} \mathrm{x} \mathrm{s}$
This work done is equal to initial kinetic energy of the object
$
\begin{aligned}
& \frac{1}{2} m v^2=\mu \mathrm{mg} \mathrm{s} \\
& \left|v^2\right|=2 \mu \mathrm{gs}=2 \times 0.7 \times 9.8 \times 10 \\
& \mathrm{v}^2=14 \times 9.8=137.2 \\
& \mathrm{v}=11.71 \mathrm{~ms}^{-1}
\end{aligned}
$