Exemplar MCQs

Hint

\(70 \mathrm{~km} / \mathrm{h}\) towards south-west
Final velocity, \(\vec{V}_{f}=-50 \hat{i} \mathrm{~km} / \mathrm{h}\), Initial velocity, \(\vec{V}_{i}=-50 \hat{i} \mathrm{~km} / \mathrm{h}\) Change in velocity, \(\Delta \vec{V}=\vec{V}_{f}-\vec{V}_{i}\), \(|\Delta V|=\sqrt{50^{2}+50^{2}+2 \times 50 \times 50 \cos \left(90^{\circ}\right)}=70 \mathrm{~km} / \mathrm{h}\)

Hint

The slope of the \(x\)-t graph gives the velocity. In the graph, the slope is constant from \(t=0\) to \(t=t_{0}\), so the velocity is constant. After \(t=t_{0}\), the displacement is zero; i.e., the particle stops.

Hint

Let \(t\) be the time related to each velocity. Total distance travelled \(=v_{1} t+v_{2} t\). Total time taken \(=2 t\). Hence average velocity \(=(v_{1} t+v_{2} t) / 2 t=\left(v_{1}+v_{2}\right) / 2\).

Hint

Average velocity \(=\frac{\text { Total distance }}{\text { Total time }}\)
\(
\begin{aligned}
&=\frac{X+X}{\frac{X}{V_{1}}+\frac{X}{V_{2}}} \\
&\Rightarrow v=\frac{2}{\frac{1}{v_{1}}+\frac{1}{v_{2}}} \\
&\Rightarrow \frac{2}{v}=\frac{1}{v_{1}}+\frac{1}{v_{2}}
\end{aligned}
\)

Hint

g downward is the right answer. Gravity is the only force acting on the stone when it is released. And, we know that gravity is always in the downward direction.

Hint

the ball A thrown upwards: let us say is thrown with speed u upwards.
height it reaches \(=\mathrm{h}\)
then \(\left.v^{2}=u^{2}+2 a s \Rightarrow 0=u^{2}-2 g h \Rightarrow u=\sqrt{(} 2 g h\right)\)
When the ball reaches the same point h, on the way down it will have a speed:
\(
\left.v^{2}=0+2 g h \Rightarrow v=\sqrt{(} 2 g h\right)
\)
So the speeds of \(A\) and \(B\) will be the same when they reach the ground, as they travel the same distance.

Hint

In projectile motion, velocity is perpendicular to acceleration only at the highest point. Here, velocity is along the horizontal direction and acceleration is along the vertically downward direction.

Hint

both will reach simultaneously,because the downward acceleration and the initial velocity in downward direction of the two bullets are the same, they will take the same time to hit the ground and for a half projectile.
Time of flight \(=\mathrm{T}=\sqrt{\frac{2 h}{g}}\).

Hint

\(100 \mathrm{~m}\) is the right answer.
For the same \(u\) range, \(R \propto \sin (2 \theta)\).
\(
\begin{aligned}
&\text { So, } \frac{R_{1}}{R_{2}}=\frac{\sin \left(2 \theta_{1}\right)}{\sin \left(2 \theta_{2}\right)} \\
&\Rightarrow R_{2}=50 \times \frac{\sin (90)}{\sin (30)}=100 \mathrm{~m}
\end{aligned}
\)

Hint

the information is insufficient to decide the relation of \(R_{A}\) with \(R_{B}\). Horizontal range for the projectile, \(R=\frac{u^{2} \sin (2 \alpha)}{g}\) Information of the initial velocity is not given in the question.

Hint

To reach the opposite bank of the river in minimum time, the swimmer should swim at right angles to the direction of the flow of the river. He should swim in a direction due north. Let the width of the river be \(\mathrm{d}\). If the man swims at an angle \(\theta\) with the direction of flow of water, his velocity component perpendicular to the direction of flow of water is \(10 \sin \theta\)
Time taken to reach the opposite bank: \(\mathrm{t}=\frac{\mathrm{d}}{10 \sin \theta}\)
\(\therefore\) t will be minimum when \(\sin \theta\) is maximum i.e \(\theta=90^{\circ}\)

Hint

Let, \(V\) be the velocity of mass in the upwards direction. Therefore, component of velocity along the rope \(=\mathrm{V} \cos \theta\)
Now, \(\mathrm{V} \cos \theta=\mathrm{u}\)
Therefore, \(\mathrm{V}=\frac{\mathrm{u}}{\operatorname{Cos} \theta}\)

Hint

Displacement is zero because the initial and final positions are the same.
Average velocity \(=\frac{\text { Displacement }}{\text { Total time }}=0\)
Distance covered \(=2 \pi r \neq 0\)
Average speed \(=\frac{\text { Distance travelled }}{\text { Total time taken }} \neq 0\)

Hint

We can write the equation as,
\(
X=u(t-2)+\frac{1}{2}(2 a)(t-2)^{2}
\)
clearly, \(\mathrm{X}=0\) at \(\mathrm{t}=2\)
Also, acceleration \(=2\) a when we compare the given equation with
\(
\mathrm{X}=\mathrm{ut}+\frac{1}{2} a \mathrm{t}^{2}
\)

Hint

Since the distance covered by a particle in a given time will never be less than the magnitude of displacement, so average speed is never less than the magnitude of the average velocity. (a) is correct.
\(|\mathrm{dv} / \mathrm{dt}|\) is the magnitude of the acceleration and \(\mathrm{d}|\mathrm{v}| / \mathrm{dt}\) is the rate of change of speed. Consider the case of uniform circular motion. Here speed of the particle is constant, so \(d|v| / d t=0\). But magnitude of the acceleration \(|d v / d t| \neq 0\). (b) is also correct.

As we know, the average velocity of the tip of the minute hand is zero in the time interval of one hour, but its instantaneous velocity is never zero in this interval. So (c) is also correct.

If the average velocity of a particle moving on a straight line is zero in a time interval, it means its displacement is zero. It can only be possible if the particle has returned back to its original position at least once in this time interval. So at the instant when it reverses the direction of its velocity, the instantaneous velocity of the particle will definitely be zero. So (d) is incorrect.

Hint

(b) varying velocity without having varying speed
(d) nonzero acceleration without having varying speed.
Explanation:
Velocity and acceleration are vector quantities that can be changed by changing direction only (keeping magnitude constant).

Hint

If the acceleration and velocity of the particle has opposite sign it has a retarding effect. As is the case of a particle thrown vertically upward. So (a) is true.
(b) is also obviously true.
(c) is not true. Consider the case of a particle thrown vertically upward. At its highest point, the velocity is zero but acceleration is not zero. Acceleration is defined as the rate of change of velocity. If in a time interval velocity is zero, it means in this interval velocity does not change. So acceleration at any instant in this time interval is zero. Hence (d) is true.

Hint

(a) is not true. Consider the case of a particle dropped vertically downward, its velocity at \(t=0\) is zero but acceleration is not zero.
(b) is true because we do not have information about the velocity in the next instants. If in the next instants the velocity still remains zero, the acceleration will be zero.
(c) is true because, if acceleration is zero that means there is no change in velocity. The velocity remains zero in this interval.
(d) is true because if the speed remains zero from \(t=0\) to \(t=10 \mathrm{~s}\), that means the particle is at rest in this interval, so the acceleration is also zero in this interval.

Hint

At any instant, the magnitude of the velocity is also its speed. So (a) is correct. (b) is incorrect because velocity is displacement in unit time while speed is the distance covered in unit time. (c) is incorrect because if speed is zero in a time interval, that means the particle has not moved in this interval and no distance traveled. So average speed has to be zero. (d) is incorrect because if the speed of the particle is never zero, means it has traveled some distance. So average speed in the interval will not be zero.

Hint

(a) is true because in a velocity-time graph the slope (which shows the rate of change of velocity) denotes acceleration. In this graph the slope is constant, so acceleration is constant.
(b) is not true because at \(t=10 \mathrm{~s}\) the velocity changes its sign from positive to negative, means it has turned around.
(c) is not true because displacement in the positive direction of \(v\) is given by the area of the graph above the time axis and displacement in the negative direction of \(v\) is given by the area of the graph below the time axis. In this case, later is greater than the former not equal. So displacement is not zero.
(d) is true because in 0 to \(10 \mathrm{~s}\) speed changes from 10 to \(0 \mathrm{~m} / \mathrm{s}\) and in 10 to \(20 \mathrm{~s}\) speed changes from 0 to \(10 \mathrm{~m} / \mathrm{s}\). So average speed will be the same.

Hint

The slope of the curve at any point gives the velocity of the particle at that point. In the given graph the curve has three troughs and three crests. At these points, the slope of the curve (tangent is horizontal to the time axis) is zero. So at six points velocity is zero, which means it has come to rest. (b) is not true because the slope of the curve is not maximum at t=6 s.  (c) is not true because the gradient of the slope is not positive always between t=0 to t=6 s.  (d) is not true because the final displacement is not less than the initial.

Hint

\(
\begin{aligned}
&\left|\vec{a}_{p s_{1}}\right|=4 \\
&\left|\vec{a}_{p s_{2}}\right|=4 \\
&\vec{a}_{p s 2}=\vec{a}_{p s_{1}}+\vec{a}_{s_{1} s_{2}} \\
&\vec{a}_{s_{1} s_{2}}=\vec{a}_{p s_{2}}-\vec{a}_{p s_{1}} \\
&\vec{a}_{s_{2} s_{1}}=\vec{a}_{p s_{1}}-\vec{a}_{p s_{2}}
\end{aligned}
\)
\(\left|\vec{a}_{p s_{1}}-\vec{a}_{p s_{2}}\right|\) may vary from \(4-4=0\) to \(4+4=8 \mathrm{~m} / \mathrm{s}^{2}\) depending on angle between the acceleration vectors.

Hint

Key concept: Average velocity: It is defined as the ratio of displacement to time taken by the body.
Displacement /Time taken
According to this problem, we need to identify the graph which is having same displacement for two timings. When there are two timings for same displacement, the corresponding velocities should be in opposite directions.

As shown in graph (b), the first slope is decreasing that means the particle is going in one direction and its velocity decreases, becomes zero at the highest point of the curve and then increasing in backward direction. Hence the particle return to its initial position. So, for one value of displacement there are two different points of time and we know that slope of \(x, x\)-t graph gives us the average velocity. Hence, for one time, slope is positive then average velocity is \(\mathrm{A}\) also positive and for other time slope is negative then average velocity is also negative.
As there are opposite velocities in the interval 0 to \(T\), hence average velocity can vanish in (b).

Hint

Key concept: The time rate of change of velocity of an object is called acceleration of the object. It is a vector quantity. Its direction is same as that of change in velocity (Not of the velocity). In the table: Possible ways of velocity change
\(
\begin{array}{|l|l|l|}
\hline \begin{array}{l}
\text { When only direction of } \\
\text { velocity changes }
\end{array} & \begin{array}{l}
\text { When only magnitude of } \\
\text { velocity changes }
\end{array} & \begin{array}{l}
\text { When both magnitude and direction of velocity } \\
\text { change }
\end{array} \\
\hline \begin{array}{l}
\text { Acceleration perpendicular } \\
\text { to velocity }
\end{array} & \begin{array}{l}
\text { Acceleration parallel or } \\
\text { antiparallel to velocity }
\end{array} & \begin{array}{l}
\text { Acceleration has two components-one is } \\
\text { perpendicular to velocity and another parallel or } \\
\text { antiparallel to velocity }
\end{array} \\
\hline \begin{array}{l}
\text { E.g.: Uniform circular } \\
\text { motion }
\end{array} & \begin{array}{l}
\text { E.g.: Motion under } \\
\text { gravity }
\end{array} & \text { E.g: Projectile motion } \\
\hline
\end{array}
\)
Here we will take upward direction positive. As. the lift is coming in downward direction, the displacement will be negative. We have to see whether the motion is accelerating or retarding.
We know that due to downward motion displacement will be negative. When the lift reaches 4 th floor and is about to stop velocity is decreasing with time, hence motion is retarding in nature. Thus, \(x<0 ; a>0\).
Asdisplacementisinnegativedirection, velocity will also be negative, i.e. \(v<0\).
The motion of the lift will be shown like this.

Hint

Key concept: Instantaneous speed: It is the speed of a particle at a particular instant of time. When we say “speed”, it usually means instantaneous speed. The instantaneous speed is the average speed for an infinitesimally small time interval (i.e., \(\Delta \rightarrow 0)\).
Thus, Instantaneous speed \(v=\lim _{\Delta \rightarrow 0} \frac{\Delta s}{\Delta t}=\frac{d s}{d t}\)
As instantaneous speed is less than maximum speed. Then either the velocity is increasing or it is decreasing. For maximum and minimum displacement we have to keep in mind the magnitude and direction of maximum velocity.
As maximum velocity in positive direction is \(\mathrm{v}_0\), magnitude of maximum velocity in opposite direction is also \(v_0\).
Maximum displacement in one direction \(=v_0 T\) Maximum displacement in opposite directions \(=-v_0 T\) Hence, \(v_0 T<x<v_0 T\).
Important point: We should not confuse with the direction of velocities, i.e., in one direction it is taken as positive and in another direction, it is taken as negative.

Hint

(c) Consider the diagram below in which motion is as shown below.

Let the vehicle travels from A to B. Distances, velocities and time taken are shown. To calculate average speed we will calculate total distance covered and will divide by time interval in which it covers that total distance.
Time taken to travel first half distance \(t_1=\frac{L / 2}{v_1}=\frac{L}{2 v_1}\)
Time taken to travel second half distance \(t_2=\frac{L}{2 v_2}\)
Total time \(=t_1+t_2\)
\(
=\frac{L}{2 v_1}+\frac{L}{2 v_2}=\frac{L}{2}\left[\frac{1}{v_1}+\frac{1}{v_2}\right]
\)
We know that
\(
v_{\mathrm{av}}=\text { Average speed }
\)
\(
=\text { total distance/total time }
\)
\(
v_{a v}=\frac{L}{\frac{L}{2}\left[\frac{1}{v_1}+\frac{1}{v_2}\right]}=\frac{2 v_1 v_2^*}{v_1+v_2}
\)
Important point: Students usually thought that \(v_{a v}=\frac{v_1+v_2}{2}\) but it is not the average speed when two equal distances are covered by speed \(v_1\) and \(v_2\). Remember: If \(t_1=t_2=t\), then \(v_{a v}=\frac{v_1+v_2}{2}\). Average speed is equal to the arithmetical mean of individual speeds. (if the particle moves in equal interval of time at different speeds \(v_1\) and \(v_2\).)

And also we should not confuse with distance and displacement. Distance \(\geq\) Displacement.

Hint

Given, \(x=(t-2)^2\)
Velocity, \(v=\frac{d x}{d t}=\frac{d}{d t}(t-2)^2=2(t-2) \mathrm{m} / \mathrm{s}\)
\(
\begin{aligned}
& \text { Acceleration, } a=\frac{d v}{d t}=\frac{d}{d t}[2(t-2)] \\
& =2[1-0] \\
& =2 \mathrm{~m} / \mathrm{s}^2
\end{aligned}
\)
When \(t=0 ; v=-4 \mathrm{~m} / \mathrm{s}\)
\(
\begin{aligned}
& \mathrm{t}=2 \mathrm{~s} ; \mathrm{v}=0 \mathrm{~m} / \mathrm{s} \\
& \mathrm{t}=4 \mathrm{~s} ; \mathrm{v}=4 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)
\(v\)-t graph is shown in the adjacent diagram
Distance travelled \(=\) Area of the graph
\(
\begin{aligned}
& =\text { Area } \mathrm{OAC}+\text { Area } \mathrm{ABD} \\
& =\frac{4 \times 2}{2}+\frac{1}{2} \times 2 \times 4 \\
& =8 \mathrm{~m}
\end{aligned}
\)

Hint

We have to find net velocity with respect to the Earth that will be equal to velocity o fthe girl plus velocity of escalator. Let dispalcemet is \(L\), then
Velocity of girl \(v_g=\frac{L}{t_1}\)
Velocity of escalator \(v_e=\frac{L}{t_2}\)
Net velocity of the girl \(=v_g+v_e=\frac{L}{t_1}+\frac{L}{t_2}\)
It \(t[latex] is total time taken is covering distance [latex]L\), then
\(
\frac{L}{t}=\frac{L}{t_1}+\frac{L}{t_2} \Rightarrow t=\frac{t_1 t_2}{t_1+t_2} .
\)

Hint

When we are calculating the velocity of a displacement time graph we have to take the slope similarly we have to take the slope of the velocity-time graph to calculate acceleration when the slope is constant motion will be uniform. When we are representing motion by a graph it may be displacement time, velocity-time or acceleration time hence, B may represent time for uniform motion velocity-time graph should be a straight line parallel to the time axis. For uniform motion velocity is constant hence, the slope will be positive. Hence quantity A is displacement

For uniformly accelerated motion slope will be positive and A represents velocity.

Hint

\((a, c, e)\)
Key concept: We know that velocity \(v=d x / d t\) and slope of \(x\)-t graph gives
us velocity. This implies slope \(=d x / d t\) for the graph.
As per the diagram, at point \(A\) the graph is parallel to time axis, hence \(d x\)
\(v=d x / d t=0\). As the starting point is A, hence we can say that the particle is starting from rest. Thus option
(a) is correct.
At \(\mathrm{C}\), the graph changes slope, hence velocity also changes. As graph at \(\mathrm{C}\) is almost parallel to time axis, hence we can say that velocity vanishes. Hence option (c) is correct.
As direction of acceleration changes, hence we can say that it may be zero in between.
From the graph it is clear that | slope at \(D|>|\) slope at \(E \mid\)
Hence, the speed at D will be more than at \(E\). Hence option (e) is correct.
Important point: Here, negative slope does not mean less value. It represents the change in the direction of velocity.

Hint

(a, d) Position of the particle is given as a function of time i.e. \(x=t-\operatorname{sint}\) By differentiating this equation w.r.t. time we get velocity of the particle as a function of time.
Velocity \(v=\frac{d x}{d t}=\frac{d}{d t}[t-\sin t]=1-\cos t\)
If we again differentiate this equation w.r.t. time we will get acceleration of the particle as a function of time. 
Acceleration \(a=\frac{d v}{d t}=\frac{d}{d t}[1-\cos t]=\sin t\)
As acceleration \(a>0\) for all \(t>0\)
Hence, \(x(t)>0\) for all \(t>0\)
Velocity \(v=1-\cos t\)
When, \(\cos t=1\), velocity \(v=0\)
\(
\begin{aligned}
v_{\max } & =1-(\cos t)_{\min }=1-(-1)=2 \\
v_{\min } & =1-(\cos t)_{\max }=1-1=0
\end{aligned}
\)
Acceleration \(a=\frac{d v}{d t}=-\sin t\)
When \(t=0 ; x=0, v=0, a=0\)
When \(t=\frac{\pi}{2} ; x=\) positive, \(v=0, a=-1\) (negative)
When \(t=\pi, x=\) positive, \(v=\) positive, \(a=0\)
When \(t=2 \pi ; x=0, v=0, a=0\)
Important points:
(i) When sinusoidal function is involved in an expression we should be careful about sine and cosine functions.
(ii) We should be very careful when calculating maximum and minimum value of velocity because it is in inverse relation with cost in the given expression.

Hint

As shown in the figure above when spring is stretched by length \(x\), restoring force will be \(F=-k x\) (-ve sign shows that the force is always is the direction opposite to displacement \(x\) ). Then the potential energy of the stretched spring \(
=P E=\frac{1}{2} k x^2
\)
The restoring force is central, hence when particle is released it will execute Simple Harmonic Motion about equilibrium position.
Acceleration will be \(a=\frac{F}{m}=\frac{-k x}{m}\)
At equilibrium position, \(x=0 \Rightarrow a=0\)
Hence, when just released \(x=x_{\max }\)
Hence, acceleration is maximum. Thus option (a) is correct.
At equilibrium whole \(\mathrm{PE}\) will be converted to \(\mathrm{KE}\), so \(\mathrm{KE}\) will be maximum and hence, speed will be maximum. Thus option (c) is correct.

Hint

(b, c, d) In this problem, we have to observe the motion from different frames. Here the problem can be solved by the frame of the observer but here we must be clear that we are considering the motion from the ground so we just keep in mind the motion from frame of observer. Compared to the velocity of trains \((10 \mathrm{~m} / \mathrm{s})\) speed of ball is less \((1 \mathrm{~m} / \mathrm{s})\). \((b, c, d)\) In this problem, we have to observe the motion from different frames. Here the problem can be solved by the frame of the observer but here we must be clear that we are considering the motion from the ground so we just keep in mind the motion from frame of observer. Compared to the velocity of trains \((10 \mathrm{~m} / \mathrm{s})\) speed of ball is less \((1 \mathrm{~m} / \mathrm{s})\).
The speed of the ball before collision with side of train is \(10+1=11 \mathrm{~m} / \mathrm{s}\) Speed after collision with side of train \(=10-1=9 \mathrm{~m} / \mathrm{s}\) As speed is changing after travelling \(10 \mathrm{~m}\) and speed is \(1 \mathrm{~m} / \mathrm{s}\), hence time duration of the changing speed is \(10 \mathrm{~s}\).
Since, the collision of the ball is perfectly elastic there is no dissipation of energy, hence total momentum and kinetic energy are conserved.
Since, the train is moving with a constant velocity, hence it will act as an inertial frame of reference as that of Earth and acceleration will be same in both frames.

Remember: We should not confuse with non-inertial and inertial frame of reference. A frame of reference that is not accelerating will be inertial.