Exemplar MCQs

Hint

(c) The ceiling pulls the chain with a force equal to the sum of the weights of the chain and the body. So the pulling force \(=(w_{1}+w_{2})\)

Hint

(b) The horse pushes the ground in the backward direction and, in turn, the ground pushes the horse in the forward direction, according to Newton’s third law of motion.

Hint

A car accelerates on a horizontal road due to the force exerted by the road on the car.

Hint

From the free-body diagram
\(
\begin{aligned}
&\mathrm{K}_{1} \mathrm{x}_{1}=\mathrm{mg}=10 \times 9.8=98 \mathrm{~N} \\
&\mathrm{~K}_{2} \mathrm{x}_{2}=\mathrm{K}_{1} \mathrm{x}_{1}
\end{aligned}
\)
So, \(\mathrm{K}_{1} \mathrm{x}_{1}=\mathrm{K}_{2} \mathrm{x}_{2}=98 \mathrm{~N}\)
Therefore, both the spring balances will read the same mass, i.e. \(10 \mathrm{~kg}\).

Hint

(c) Here the force excreted by the plane on the block is the normal force \(\mathrm{N}\) of the block. so from the free body diagram, \(N=m g \cos \theta\)

Hint

(b) As there is no relative motion between block and wedge:
\(m a \cos \theta=m g \sin \theta\)
\(\Rightarrow \mathrm{a}=\mathrm{g} \tan \theta\)
From the free body diagram shown below,  normal force by wedge on the block is
\(
\begin{aligned}
&\mathrm{N}=\mathrm{mg} \cos \theta+m a \sin \theta \\
&=\mathrm{mg} \cos \theta+\mathrm{m}(\mathrm{g} \tan \theta) \sin \theta \\
&=\mathrm{mg} \cos \theta+\mathrm{mg} \frac{\sin ^{2} \theta}{\cos \theta} \\
&=\mathrm{mg}\left(\frac{\cos ^{2} \theta+\sin ^{2} \theta}{\cos \theta}\right) \\
&=\frac{\mathrm{mg}}{\cos \theta}
\end{aligned}
\)

Hint

(d) remain standing
If the earth suddenly stops attracting objects placed near its surface, the net force on the person will become zero and according to the first law of motion, the person will remain standing.

Hint

(a) Using, \(F_{\text {net }}=m a\),
\(
a=0 \Rightarrow F_{\text {net }}=0
\)
As the whole system is at rest, the resultant force on the charged particle at \(A\) is zero.

Hint

(b) To decelerate a body, a force must be applied opposite to the direction of velocity, irrespective of its magnitude. According to the condition of motion under two forces, \(F_{1}\) and \(F_{2}\) must be acting in opposite directions and may be equal, not necessarily must be equal.

Hint

(b) Let \(\mathrm{m}_{\mathrm{A}}\) a be the mass of object \(\mathrm{A}\) and \(\mathrm{m}_{\mathrm{B}}\) be the mass of the object \(B\). Let \(\mathrm{a}_{\mathrm{A}}\) and \(\mathrm{a}_{\mathrm{B}}\) be the acceleration of the two bodies downwards due to the air resistance. Since the bodies are acted upon by a constant air resistance, from the newton’s law we can write that: \(\mathrm{a}_{\mathrm{A}}=\frac{\mathrm{F}}{\mathrm{m}_{\mathrm{A}}}\) and \(\mathrm{a}_{\mathrm{B}}=\frac{\mathrm{F}}{\mathrm{m}_{\mathrm{B}}}\) Given that \(\mathrm{m}_{\mathrm{A}}>\mathrm{m}_{\mathrm{B}}\), Hence we can say that \(\mathrm{a}_{\mathrm{A}} \)< \(\mathrm{a}_{\mathrm{B}}\). If \(S_{A}\) and \(S_{B}\) are the heights attained by the objects \(A\) and \(B\) respectively when thrown upwards According to the equation of motion:
\(
S_{A}=u t-\frac{1}{2} a_{A} t^{2}; S_{B}=u t-\frac{1}{2} a_{B} t^{2}\)

Since \(\mathrm{a}_{\mathrm{A}} \)< \(\mathrm{a}_{\mathrm{B}}\), \(S_{A} > S_{B}\).

Hence we can say that object \(A\) rises to a greater height than object B.

Hint

(c) Downward gravitational force will be balanced by the upward pseudo force (because of the motion of the wedge in a downward direction). The block will remain at its position, as both the box and the inclined plane are falling with the same acceleration \(g\).

Hint

(b) Let acceleration due to air resistance force be \(a\).
Let \(\mathrm{h}\) be the maximum height attained by the particle.
The direction of air resistance force is in the direction of motion.
In the upward direction of motion,
\(
\begin{aligned}
&a_{e f f}=|g-a| \\
&t_{1}=\sqrt{\frac{2 h}{|g-a|} \ldots(1)}
\end{aligned}\)
In the downward direction of motion,
\(
\begin{aligned}
&a_{e f f}=g+a \\
&t_{2}=\sqrt{\frac{2 h}{g+a} \ldots(2)}
\end{aligned}
\)
So, \(t_{1}>t_{2}\).

Hint

(a) \(
\mathrm{t}_{1}=\mathrm{t}_{2}
\)
After the coin is dropped, the only force acting on it is gravity, which is the same for both cases. So \(t_{1}=t_{2}\).

Hint

As both the nucleus and the passengers are in the train so they will move with uniform velocity and there is no relative motion in between them. Thus, the distance between the alpha particle and the recoiling nucleus will be the same as before he has measured i.e \(x\)

Hint

(b) & (d) A reference frame attached to the earth cannot be an inertial frame because the earth is revolving around the sun and also rotating about its axis.

Hint

(c) & (d)

According to Newton’s second law which says that net force acting on the particle is equal to rate of change of momentum ( or mathematically \(\mathrm{F}=\mathrm{ma})\), so if a particle is at rest then \(\mathrm{F}_{\text {net }}=\mathrm{ma}=\mathrm{m} \frac{d v}{d t}=m \frac{d(0)}{d t}=m \times 0=0\)
Now, if the frame is inertial, then the resultant force on the particle is zero.
If the frame is non-inertial,
the vector sum of all the forces plus a pseudo force is zero.
i.e. \(F_{\text {net }} \neq 0\).

Hint

(a) & (b)

\(S_{1}\) is moving with constant velocity with respect to the frame \(S_{2}\). So, if \(S_{1}\) is inertial, then \(S_{2}\) will be inertial and if \(S_{1}\) is non-inertial, then \(S_{2}\) will be non-inertial.

Hint

(a) & (c)

The slope of the \(x\)-t graph gives velocity. In regions A B and C D, the slope or velocity is constant, i.e. acceleration is zero. Hence, from the second law, force is zero in these regions.

Hint

(c) & (d) It means the normal force exerted by the floor of the elevator on the person is greater than the weight of the person. i.e. \(\mathrm{N}>\mathrm{mg}\)
(i) Going up and speeding up:
\(
\begin{aligned}
&a_{e f f}=g+a \\
&N=m a_{e f f}=m g+m a(N>m g)
\end{aligned}
\)
(ii) Going down and speeding up:
\(
\begin{aligned}
&a_{e f f}=g-a \\
&N=m g-m a(N<m g)
\end{aligned}
\)
(iii) Going down and slowing down:
\(
\begin{aligned}
&a_{e f f}=g-(-a)=g+a \\
&N=m g+m a(N>m g)
\end{aligned}
\)
(iv) Going up and slowing down:
\(
\begin{aligned}
&a_{e f f}=g-a \\
&N=m g-\operatorname{ma}(N<m g)
\end{aligned}
\)

Hint

(c) & (d)

Since the tension in the cable = Weight of the elevator
Therefore, total upward force = total downward force
That is, there’s no acceleration or uniform velocity.
So, the elevator is going up/down with uniform speed.

Hint

Let us assume there a two frames \(S_{1}\) and \(S_{2}\) moving with acceleration \(a_{1}\) and \(a_{2}\). Now we know \((a 1-a 2)=a\) as relative acceleration is \(a\).

Now, the magnitude of pseudo forces \(F_{1}\) and \(F_{2}\) will be \(m a_{1}\) and \(m_{2}\) irrespective of where the object is.
\(
F_{1}=m a_{1}, F_{2}=m a_{2}
\)
when \(a_{1}=0\), then \(F_{1}=0, F_{2} \neq 0\)
\(a_{2}=0\), then \(F_{1} \neq 0, F_{2}=0\)
when, \(a_{1} \neq 0, a_{2} \neq\) O then, \(F_{1} \neq 0, F_{2} \neq 0\)
but it is impossible for both \(F_{1}=0\) and \(F_{2}=0\) as for this to happen,
\(\mathrm{a}_{1}=0\) and \(\mathrm{a}_{2}=0\) which is not the case as there is relative acceleration.

Hint

(c) When a body moves in such a way that the linear distance covered by each particle of the body is the same during the motion, then the motion is said to be translatory or translation motion.

Translatory motion can be, again of two types viz., curvilinear (shown in fig. (a)) or rectilinear (shown in fig. (b)), accordingly as the paths of every constituent particle are similarly curved or straight line paths. Here it is important that the body does not change its orientation. Here we can also define it further in uniform and non-uniform translatory motion. Here figure (b) is uniformly translatory motion.

Hint

Key concept: To solve these types of problems we have to apply Newton’s second law of motion! Newton’s Second Law of Motion According to this law: The rate of change of linear momentum of a body is directly proportional to the external force applied on the body and this change takes place always in the direction of the force applied.
We know that \(F=d f / d t\)
According to the question that the meter scale is moving with uniform velocity, hence, the change in momentum will be zero, i.e. \(\mathrm{dp}=0\)
This implies momentum will remain the same. So, Force \(=F=0\).
So, we can say that all parts of the meter scale is moving with uniform velocity because the total force is zero and if there is any torque acting on the body this means that the body will be in rotational motion which means that the direction of velocity will be changing continuously. So, the torque acting about centre of mass of the scale is also zero.

Hint

(c) According to the problem, \(u=(3 \hat{i}+4 \hat{j}) \mathrm{m} / \mathrm{s}\) and \(v=-(3 \hat{i}+4 \hat{j}) \mathrm{m} / \mathrm{s}\)
Mass of the ball \(=150 \mathrm{~g}=0.15 \mathrm{~kg}\)
Change in momentum will be
\(
\begin{aligned}
\overrightarrow{\Delta p} & =\overrightarrow{p_f}-\overrightarrow{p_l} \\
& =m \vec{v}-m \vec{u} \\
& =m(\vec{v}-\vec{u}) \\
& =(0.15)[-(3 i+4 j)-(3 \hat{i}+4 \hat{j})] \\
& =(0.15)[-6 \hat{i}-8 \hat{j}] \\
& =-[0.15 \times 6 \hat{i}+0.15 \times 8 j] \\
& =-[0.9 \hat{i}+1.20 \hat{j}]
\end{aligned}
\)
Hence, \(\overrightarrow{\Delta p}=-[0.9 \hat{i}+1.2 \hat{j}]\)
Important point: Change in velocity = final velocity – initial velocity

Hint

(c) By previous solution, \(\overrightarrow{\Delta p}=-(0.9 \hat{i}+1.2 \hat{j})\)
\(
\begin{aligned}
\text { Magnitude } & =|\Delta p|=\sqrt{(0.9)^2+(1.2)^2} \\
& =\sqrt{0.81+1.44}=1.5 \mathrm{~kg}-\mathrm{ms}^{-1}
\end{aligned}
\)

Hint

(d) Key concept: If no external force acts on a system (called isolated) of constant mass, the total momentum of the system remains constant with time.
According to this law for a system of particles \(\vec{F}=\frac{d \vec{p}}{d t}\)
In the absence of external force \(\vec{F}=0\), then \(\vec{p}=\) constant i.e., \(\vec{p}=\vec{p}_1+\vec{p}_2+\vec{p}_3+\cdots=\) constant
or, \(\quad m_1 \vec{v}_1+m_2 \vec{v}_2+m_3 \vec{v}_3+\cdots=\) constant
This equation shows that in absence of external force for a closed system the linear momentum of individual particles may change but their sum remains unchanged with time.
Conservation of linear momentum is equivalent to Newton’s third law of motion.
For a system of two particles in absence of external force by law of conservation of linear momentum.
\(
\vec{p}_1+\vec{p}_2=\text { constant. } \quad \therefore m_1 \vec{v}_1+m_2 \vec{v}_2=\text { constant }
\)
Differentiating above with respect to time,
\(
\begin{aligned}
& m_1 \frac{d \vec{v}_1}{d t}+m_2 \frac{d \vec{v}_2}{d t}=0 \Rightarrow m_1 \vec{a}_1+m_2 \vec{a}_2=0 \\
\Rightarrow & \vec{F}_1+\vec{F}_2=0 \\
\therefore & \vec{F}_2=-\vec{F}_1
\end{aligned}
\)
i.e., for every action there is equal and opposite reaction which is Newton’s third law of motion.
In case of collision between particles equal and opposite forces will act on individual particles by Newton’s third law.
Hence total force on the system will be zero.
Important point: We should not confuse with system and individual particles. As total force on the system of both particles is zero but force acts on individual particles.

The law of conservation of linear momentum is independent of the frame of reference though linear momentum depends on the frame of reference.

Hint

(a) First we have to calculate acceleration and then Newton’s second law will be applied.

Explanation:
\(
\begin{aligned}
& x(t)=p t+q t^2+r t^3 \\
& p=3 m s^{-1}, q=4 m s^{-2} r=5 m s^{-3} \\
& x(t)=3 t+4 t^2+5 t^3 \\
& \text { thus, } \frac{d x(t)}{d t}=3+8 t+15 t^2 \\
& \frac{d^2 x(t)}{d t^2}=0+8+30 t \\
& {\left[\frac{d^2 x(t)}{d t^2}\right]_{t=2}=8+30 \times 2=68 \mathrm{~ms}^{-2}}
\end{aligned}
\)
Now, we know that,
\(
\vec{F}=m \vec{a}=m \cdot \frac{d^2 x}{d t^2}
\)
Thus,
\(
\begin{aligned}
& \vec{F}=2 \times 68 \\
& =136 \mathrm{~N}
\end{aligned}
\)

Hint

Key concept: According to Newton’s second law of motion only external forces can change linear momentum of the system. The internal forces cannot change linear momentum of system under consideration. If we take hockey player as a system, the external force which can change the direction of motion of the player is the force must be friction between the ground and shoes of player. The muscle force is the internal force, this cannot change the linear momentum of the player. According to Newton’s Second Law, The rate of change of linear momentum of a body is equal to the external force applied on the body or \(\mathrm{F}=\mathrm{dp} / \mathrm{dt}\). So, the external force must be in the direction of change in momentum.

\(
\begin{aligned}
\text { Change in momentum } & =p_2-p_1 \\
& =A B-O A=A B+(-O A)
\end{aligned}
\)
\(=\) Clearly the change in momentum is OR will be along south-west, so direction of force will also be along south-west

Hint

The answer is the option (b) 10 s
Explanation:
Initial velocity \((u)=(6 \widehat{i}+12 \widehat{j}) \mathrm{ms}^{-1}\)
Force, \(F=(-3 \widehat{i}+4 \widehat{j}) N\), Mass \((\mathrm{m})=5 \mathrm{~kg}\)
\(
\vec{a}=\frac{\vec{F}}{m}=\left(\frac{-3}{5} \widehat{i}+\frac{4}{5} \widehat{j}\right) \mathrm{ms}^{-2}
\)
Since there is only \(Y\) component in final velocity and the \(X\) component is zero,
\(
\begin{aligned}
& v_x=u_x+a_x t \\
& 0=6+\frac{-3}{5} t \rightarrow \frac{3}{5} t=6 \\
& \mathrm{t}=10 \mathrm{~s} .
\end{aligned}
\)

Hint

Let us assume the eastward direction as \(x\)-axis.
A car is able to move towards due to friction acting between its tires and the road.
The force of friction of the road on the tire acts in the forward direction and is equal but in the opposite direction to the force of friction of the tire on the road.
Mass of the car \(=m\)
As the car starts from rest, its initial velocity \(u=0\) Velocity acquired along east \(=v i\) Time interval (in which car acquired that velocity) \(t=2 \mathrm{~s}\).
As acceleration is uniform, so by applying the kinematic equation \((v=u+a t)\), we get
\(
\begin{aligned}
& v=u+a t \\
\Rightarrow & v \hat{i}=o+a \times 2 \\
\Rightarrow \quad & a=\frac{v}{2} \hat{i}
\end{aligned}
\)
Force, \(F=m a=\frac{m v}{2} \hat{i}\)
Hence, the force acting on the car is \(\frac{m v}{2}\) towards the east. As the external force on the system is only friction, hence the force \(\frac{m v}{2}\) is by friction. Hence, force by the engine is the internal force.

Hint

(a) The force at \(t=\frac{1}{8} \mathrm{~s}\) on the particle is \(-16 \pi^2 \mathrm{Am}\).
(b) The particle is acted upon by on impulse of magnitude \(4 \pi^2 \mathrm{Am}\). at \(\mathrm{t}=0 \mathrm{~s}\) and \(\left(t=\frac{1}{4}\right) \mathrm{s}\)
(d) The particle is not acted upon by a constant force.
Explanation: Here, mass \(=m\)
Thus, \(x(t)=0\), for \(t<0\)
\(
x(t)=A \sin 4 \pi t, \text { for } 0<t<\left(\frac{1}{4}\right) s
\)
and
\(
x(t)=0, \text { fort }>\left(\frac{1}{4}\right) s
\)
Now,
For
\(
\begin{aligned}
& 0<t<\left(\frac{1}{4}\right) s, x(t)=A \sin 4 \pi t \\
& v=\frac{d x}{d t}=4 A \pi \cos 4 \pi t \\
& a=\frac{d v}{d t}=-16 \pi^2 A \sin 4 \pi t \\
& F(t)=m a(t)=-16 \pi^2 A m \sin 4 \pi t
\end{aligned}
\)
Now, it’s clear that force is a function of time, hence opt (d) is verified.
\(
\begin{aligned}
& \text { (a) At } t=\left(\frac{1}{8}\right) s \\
& a=-16 \pi A \sin \frac{\pi}{2} \\
& a(t)=-16 \pi^2 A
\end{aligned}
\)
thus, force at \(t=\frac{1}{8}\)
\(
F=m a=m\left(-16 \pi^2 A\right)
\)
Hence, \(F=-16 \pi^2 A m N\).
Hence, opt (a)
(b) Impulse = Change in momentum between the values \(t=0\) s and \(\left(t=\frac{1}{4}\right) \mathrm{s}\)
Now we already know from above that, \(F(t)\) varies from 0 at \(t=0\) s to the max. value at
\(
\begin{aligned}
& F(t)=-16 \pi^2 A m N \cdot \text { at }\left(t=\frac{1}{4}\right) s \text { \& by the eq. } \vec{I}=\vec{F} t \text { at }\left(t=\frac{1}{4}\right) s . \\
& |I|=16 \pi^2 A m \times \frac{1}{4}=4 \pi^2 A m
\end{aligned}
\)
Hence opt (b).

Hint

Explanation: By opt (e) the max. force by which bodies move together is \(0.45 \mathrm{mg}\) Newton.
\(
\mathrm{mA}=\frac{m}{2}=\mathrm{mB}=\mathrm{m}
\)
Consider the acceleration of the bodies A and B to be ‘ \(a\) ‘.
Both the bodies A & B will keep moving by force F until the force of friction between their surface is larger or equal to 0 than the force acting on \(A\).
\(
a=\frac{F-f_1}{m_A-m_B}=\frac{F-f_1}{\frac{m}{2}+m}=\frac{2\left(F-f_1\right)}{3 m}
\)
Thus, the force on
\(
A=m_A a=\frac{m}{2} \frac{\left(F-f_1\right)}{3 m}
\)
Hence, the force on \(A\) is
\(
F_{A B}=\frac{\left(F-f_1\right)}{3}
\)
The body \(A\) will move along with body \(B\) only if \(F_{A B}\) is equal or smaller than \(f_2\).
Hence, \(F_{A B}=f_2\)
\(
\mu N=\frac{\left(F-f_1\right)}{3}
\)
\(0.2 \times m_A g=\frac{\left(F-f_1\right)}{3} \dots(i)\)
\(\mathrm{N}\) is the reaction force by \(\mathrm{B}\) on \(\mathrm{A}\)
\(f_1=\mu N B=\mu\left(m_A+m_B\right) g\) (here, \(\mathrm{N}_{\mathrm{B}}\) is the normal reactn on \(\mathrm{B}\) along with \(\mathrm{A}\) by the surface)
\(
0.1 \times\left(m_A+m_B\right) g
\)
\(
f_1=0.1 \times \frac{3}{2} m g=0.15 m g \dots(ii)
\)
Now, \(F-f_1=3 \times 0.2 \mathrm{mAg}\) [from (i)]
\(
F-0.15 \mathrm{mg}=0.6 \times \frac{\mathrm{m}}{2} g
\)
\(
F_{\max }=0.3 \mathrm{mg}+0.15 \mathrm{mg}=0.45 \mathrm{mg} \dots(iii)
\)
Thus, the max force on \(B\) is \(0.45 \mathrm{mg}\), therefore, \(A\) & B can move together.
Hence, opt (e).
Both the bodies A & B will keep moving by force F until the force of friction between their surface is larger or equal to 0 than the force acting on A, i.e., \(0.45 \mathrm{mg}\) Newton, hereby opt (c) is rejected.
Now, for opt (d), the minimum force which can move A & B together is,
\(
\begin{aligned}
& F_{\text {in }} \geq f_1+f_2 \\
& \geq 0.15 \mathrm{mg}+0.2 \times \frac{\mathrm{m}}{2} g \quad \ldots . .[\text { [from (i) & (ii)] } \\
& F_{\text {in }} \geq 0.25 \mathrm{mg} \text { Newton }
\end{aligned}
\)
Since \(0.1 \mathrm{mg}<0.25 \mathrm{mg}\), opt (d) is verified which states that the body will be at rest if \(\vec{F}=0.1 \mathrm{mg}\).

Hint

(b, d)
Key concept: When a mass \(m x\) is placed on a rough inclined plane: Another mass \(m 2\) hung from the string connected by ftictionless pulley, the tension IT) produced in string will try to start the motion of mass W.. At limiting condition. From Fig(a)
For \(m_2, T=m_2 g \dots(i)\)
For \(m_1, T=m_1 g \sin \theta+F\)
\(
\begin{array}{ll}
\Rightarrow & T=m_1 g \sin \theta+\mu R \\
\Rightarrow & T=m_1 g \sin \theta+\mu m_1 g \cos \theta \dots(ii)
\end{array}
\)
From equation (i) and (ii), \(m_2=m_1[\sin \theta+\mu \cos \theta]\) This is the minimum value of \(m_2\) to start the motion.
\(
\text { Note: In the above condition, Coefficient of friction } \mu=\left[\frac{m_2}{m_1 \cos \theta}-\tan \theta\right]
\)
Simplified situation is shown in the diagram(b).
Let \(\mathrm{m} 1\) moves up the plane. Different forces involved are shown in the diagram.
\(N=\) Normal reaction between wedge and block \(m_1\)
\(f=\) Frictional force between wedge and block \(m_1\)
\(T=\) Tension in the string
\(f=\mu N=\mu m_1 g \cos \theta\)
For the system \(\left(m_1+m_2\right)\), condition for \(m_1\) to move up the inclined plane.
\(
\begin{aligned}
& m_2 g-\left(m_1 g \sin \theta+f\right)>0 \\
\Rightarrow & m_2 g-\left(m_1 g \sin \theta+\mu m_1 g \cos \theta\right)>0 \\
\Rightarrow & m_2>m_1(\sin \theta+\mu \cos \theta)
\end{aligned}
\)
Hence, option (b) is correct.
Condition for \(m_1\) to move down the inclined plane. In this case, friction \(f\) acts up the plane.
Hence, \(m_1 g \sin \theta-f>m_2 g\)
\(
\begin{aligned}
& \Rightarrow \quad m_1 g \sin \theta-\mu m_1 g \cos \theta>m_2 g \\
& \Rightarrow \quad m_1(\sin \theta-\mu \cos \theta)>m_2 \\
& \Rightarrow \quad m_2<m_1(\sin \theta-\mu \cos \theta)
\end{aligned}
\)
Hence, option (d) is correct.

Hint

(b,c)

In the fig. below \(\mu=\) coefficient of friction of \(\mathrm{A}\) whereas \(\mathrm{B}\) is on a frictionless surface.
(i) when \(\mathrm{A}\) is about to start
\(
\begin{aligned}
& f=\mu N_1=\mu m g \cos \theta_1 \\
& m g \sin \theta_1+f=m g \sin \theta_2
\end{aligned}
\)
When A just start moving upward
\(
\begin{aligned}
& m g \sin \theta_1+\mu m g \cos \theta_1=m g \sin \theta_2 \\
& \mu=\frac{\sin \theta_2-\sin \theta_1}{\cos \theta_1}
\end{aligned}
\)
If on the plane, the body A moves upward, and B moves downward
\(
\begin{aligned}
& m g \sin \theta_2-m g \sin \theta_1>0 \\
& \sin \theta_2-\sin \theta_1>0 \\
& \theta_2-\theta_1>0
\end{aligned}
\)
Hence, opt (c) is verified and opt (a) is rejected.
(ii) If the body B moves upward and A moves downward then,
\(
\begin{aligned}
& m g \sin \theta_2-f-m g \sin \theta_2>0 \\
& m g \sin \theta_1-\mu m g \cos \theta_1-m g \sin \theta_2>0 \\
& \sin \theta_1-\sin \theta_1>\mu \cos \theta_1
\end{aligned}
\)
Since \(\Theta_1\) and \(\Theta_2\) are acute angles, \(\Theta_1>\Theta_2\), thus, \(\sin \theta>\mu \cos \theta_1\) may be true
Since it is now clear that body B can also move upward, opt (d) is also rejected.

Hint

Let \(\mathrm{m}\) be the mass of each ball, \(\mathrm{m}=0.05 \mathrm{~kg}\)
Let \(v\) be the speed of each ball, \(v=5 \mathrm{~m} / \mathrm{s}\)
We know that the initial momentum of each ball will be \(
\overrightarrow{p_i}=m \vec{v}
\)
\(
\overrightarrow{p_i}=(0.05)(5)=0.25 \mathrm{kgms}^{-1}=0.25 \mathrm{~N-s} \dots(i)
\)
After the collision, on rebounding, the direction of the velocity of each ball is reversed; hence, the final momentum of each ball will be
\(
\begin{aligned}
& \vec{p}=m(-\vec{v}) \\
& =-0.25 \mathrm{~N-s}
\end{aligned}
\)
Hence, opt (d) is verified.
Thus, it is clear that impulse imparted to each ball is equal to change in momentum of each ball
\(
=p_f-p_i
\)
\(
\begin{aligned}
& =-0.25-(0.25) \\
& =-0.50 \mathrm{~kg} \mathrm{~ms}^{-1} \\
& =-0.50 \mathrm{~N-s}
\end{aligned}
\)
Here opt (c) is also verified.

Hint

(a, c) Recall the concept of the resultant of two vectors, when they are perpendicular
\(
R=\sqrt{A^2+B^2+2 A B \cos \theta}
\)
As they are perpendicular, \(\cos 90^{\circ}=0\)
So resultant will be \(R=\sqrt{A^2+B^2}\)
As shown in the diagram
According to the problem, mass \(=m=10 \mathrm{~kg}\)
\(
F_1=6 \mathrm{~N}, F_2=8 \mathrm{~N}
\)
Net force \(=F=\sqrt{F_1^2+F_2^2}=\sqrt{36+64}=10 \mathrm{~N}\)
\(
a=\frac{F}{m}=\frac{10}{10}=1 \mathrm{~m} / \mathrm{s}^2 \text { along } F \text {. }
\)
Let \(\theta_1\) be the angle between \(F\) and \(F_1\).
\(
\begin{aligned}
& \tan \theta_1=\frac{8}{6}=\frac{4}{3} \\
& \theta_1=\tan ^{-1}(4 / 3) \text { w.r.t } F_1=6 \mathrm{~N}
\end{aligned}
\)
Let \(\theta_2\) be angle between \(F\) and \(F_2\).
\(
\begin{aligned}
& \tan \theta_2=\frac{6}{8}=\frac{3}{4} \\
& \theta_2=\tan ^{-1}\left(\frac{3}{4}\right) \text { w.r.t } F_2=8 \mathrm{~N}
\end{aligned}
\)