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Question 511 Mark

When a particle moves with constant velocity, its average velocity, its instantaneous velocity and its speed are all equal. Comment on this statement.

Answer

Its average velocity and instantaneous velocity will be equal but we cannot compare these with average speed as one is vector and other is scalar.

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Question 521 Mark

A ball dropped from height h reaches the ground in t s. After what time the ball was passing through a point at a height $\frac{\text{h}}{2}$?

Answer

$\text{t}=\sqrt{2\frac{\text{h}}{\text{g}}}$ $\text{t}'=\sqrt{\frac{1}{\text{g}}.\Big(\frac{\text{h}}{2}\Big)}=\sqrt{\frac{\text{h}}{\text{g}}}$ $\Rightarrow \text{t}'=\frac{\text{t}}{2}$

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Question 531 Mark

Suppose two trains A and B are moving with uniform velocities along parallel tracks in the same direction and the velocities of A and B be 60km/h in East and 65km/h in East. Find the relative velocity of B w.r.t. A.

Answer

Relative velocity of Bw.r.t. $A, V_{A B}=V_A-V_B$

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Question 541 Mark

The velocity of a particle is towards west at an instant. Its acceleration is not towards west, not towards east, not towards north and not towards south. Give an example of this type of motion.

Answer

Projectile at highest point because at the highest point vertical velocity become zero.

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Question 551 Mark

Look at the graphs carefully and state, with reasons, which of these cannot possibly represent one-dimensional motion of a particle.

Answer

The given v-t graph, shown in (d), does not represent one-dimensional motion of the particle. This is because the total path length travelled by the particle cannot decrease with time.

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Question 561 Mark

For which condition, the average velocity will be equal to the instantaneous velocity?

Answer

When a body moves with a uniform velocity, then $V_{av} = V_{ist}$

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Question 571 Mark

What is the shape of displacement-time graph for uniform linear motion?

Answer

A straight line inclined to the time axis.

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Question 581 Mark

Express an acceleration of $10m/ s^2$ in $km/ h^2$.

Answer

Acceleration$=\frac{10\text{m}}{(1\text{s)}^2}=\frac{10\times10^{-3}}{\Big[\frac{1}{60\times60}\text{h}\Big]^2}$ $=(3600)^2\times10^{-2}\text{km}/\text{ h}^2$ $=1.29\times10^5\text{km}/\text{ h}^2$

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Question 591 Mark

What will happen to a hydrogen balloon released on the moon?

Answer

The balloon will fall with an acceleration of $g/6ms^{-2}$.

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Question 601 Mark

For which condition, the distance and the magnitude of displacement of an object have the same values?

Answer

The distance and the magnitude of displacement of an object have the same values, when the body is moving along a straight line path in a fixed direction.

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Question 611 Mark

Why does a parachute descend slowly?

Answer

Due to increased surface area of a parachute, the air resistance is more, so it descends slowly.

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Question 621 Mark

If position of a particle at instant t is given by $x = 2t^3$, find the acceleration of the particle.

Answer

Given, $x = 2t^3$, velocity, $\text{v}=\frac{\text{dx}}{\text{dt}}=\frac{\text{d}(2\text{t}^3)}{\text{dt}}=6\text{t}^2$ $\therefore$ Accceleration, $\text{a}=\frac{\text{dv}}{\text{dt}}=\frac{\text{d}(6\text{t}^2)}{\text{dt}}=12\text{t}$

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Question 631 Mark

Give an example of uniformly accelerated linear motion.

Answer

Motion of a body under gravity.

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Question 641 Mark

Draw velocity-time graph for an object, starting from rest. Acceleration is constant and remains positive.

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Question 651 Mark

Can a moving body have relative velocity zero with respect to another body? Give an example.

Answer

Yes, two cyclists moving with same velocity in the same direction.

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Question 661 Mark

The position x of a body is given by $\text{x}=\text{A}\sin(\omega\text{t}).$ Find the time at which the displacement is maximum.

Answer

The value of position x will be maximum, when the value of sin (t) is maximum, for this $\sin(\omega\text{t})=1=\sin\frac{\pi}{2}$ $\omega\text{t}=\frac{\pi}{2}$ $\Rightarrow \text{t}=\Big(\frac{\pi}{2\omega}\Big)$

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Question 671 Mark

A man is standing on top of a building $100m$ high. He throws two balls vertically, one at $t = 0$ and other after a time interval $($less than $2$ seconds$)$. The later ball is thrown at a velocity of half the first. The vertical gap between first and second ball is $+15m$ at $t = 2s$. The gap is found to remain constant. Calculate the velocity with which the balls were thrown and the exact time interval between their throw.

Answer

We solve this problem by using kinematic equations with proper sign convention and to calculate time interval we will take We solve this problem by using kinematic equations with proper sign convention and to calculate time interval we will take difference of displacements.

Let the speeds of the two balls $(1$ and $2)$ be $v_1$ and $v_2$ where:
if $\text{v}_1=2\text{v},\text{v}_2=\text{v}$ if $y_1$ and $y_2$ and the displacement covered by the balls $1$ and $2$, respectively,
before coming to rest, then, $\text{y}_1=\frac{\text{v}_1^2}{2\text{g}}=\frac{4\text{v}^2}{2\text{g}}$ and $\text{y}_2=\frac{\text{v}_2^2}{2\text{g}}=\frac{\text{v}^2}{2\text{g}}$
Since $\text{y}_1-\text{y}_2=15\text{m},\frac{4\text{v}^2}{2\text{g}}-\frac{\text{v}^2}{2\text{g}}=15\text{m or }\frac{3\text{v}^2}{2\text{g}}=15\text{m}$ or $\text{v}^2=\sqrt{5\text{m}\times(2\times10)}\text{m/s}^2$ or $\text{v}=10\text{m/s}$
Clearly, $v_1 = 20m/s$ and $v_2 = 10m/s$ as $\text{y}_1=\frac{\text{v}_1^2}{2\text{g}}=\frac{(20\text{m})^2}{2\times10\text{m}}=20\text{m}$ .
$\text{y}_2=\text{y}_1-15\text{m}=5\text{m}$ If $t_2$ is the time taken by the ball $2$ to cover a displacement of $5m$, then from
$\text{y}_2=\text{v}_2\text{t}-\frac{1}{2}\text{gt}^2_2$
$5=10\text{t}_2-5\text{t}^2_2\ \text{or}\ \text{t}^2_2-2\text{t}_2+1=0$
where $\text{t}_2=1\text{s}$ Since $t_1 ($time taken by ball $1$ to cover distance of $20m)$ is $2s$,
time interval between the two throws $=\text{t}_1-\text{t}_2=2\text{s}-1\text{s}=1\text{s}$
Important note: We should be very careful when we are applying the equation of rectilinear motion.
These equations are applicable only in case of constant acceleration. Some important observations for motion under gravity:

The motion is independent of the mass of the body, as in any equation of motion, mass is not involved.

That is why a heavy and light body when released from the same height, reach the ground simultaneously and with same velocity, i.e., $\text{t}=\sqrt{\frac{2\text{h}}{\text{g}}}$ and $\text{v}=\sqrt{2\text{gh}}.$

In case of motion under gravity time taken to go up is equal to the time taken to fall down through the same distance.

Time of descent $(t_1)$ = time of ascent $(\text{t}_2)=\frac{\text{u}}{\text{g}}$
Total time of flight $\text{T}=\text{t}_\text{x}+\text{t}_2=\frac{2\text{u}}{\text{g}}$

In case of motion under gravity, the speed with which a body is projected up is equal to the speed with which it comes back to the point of projection.

As well as the magnitude of velocity at any point on the path is same whether the body is moving in upwards or downward direction.

A body is thrown vertically upwards. If air resistance is to be taken into account, then the time of ascent is less than the time of descent $t_2> t_1$

Let u be the initial velocity of body, then time of ascent $\text{t}_1=\frac{\text{u}}{\text{g}+\text{a}}$ and $\text{h}=\frac{\text{u}^2}{2(\text{g}+\text{a})}$
where $g$ is acceleration due to gravity and $a$ is retardation by air resistance and for upward motion both will act vertically downward.
For downward ‘motion $a$ and $g$ will act in opposite direction because a always act in direction opposite to motion and $g$ always act vertically downward.
So, $\text{h}=\frac{1}{2}(\text{g}-\text{a})\text{t}^2_2$
$\Rightarrow\frac{\text{u}^2}{2(\text{g}+\text{a})}=\frac{1}{2}(\text{g}-\text{a})\text{t}^2_2$
$\Rightarrow\text{t}_2=\frac{\text{u}}{\sqrt{(\text{g}+\text{a})(\text{g}-\text{a})}}$
Comparing $t_1$ and $t_2$ we can say that $\text{t}_2 > \text{t}_1,$ since $(\text{g}+\text{a})>(\text{g}-\text{a})$

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Question 681 Mark

When a body accelerates by at, what is the velocity after time ‘t', when it starts from rest?

Answer

$\text{a}=\alpha\text{t}\text{ i.e. },\int\text{dv}=\text{a}\int\text{t dt}$ $\Rightarrow \text{v}=\frac{\alpha\text{t}^2}{2}$

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Question 691 Mark

A truck and a car with the same kinetic energy are brought to rest by the application of brakes which provide equal retarding forces. Which of them will come to rest in a shorter distance?

Answer

Both truck and car will stop at the same distance.

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MCQ 701 Mark

The position of a particle moving in the $X-Y$ plane at any time $t$ is given by; $x=\left(3 t^2-6 t\right)$ meters; $y=\left(t^2-2 t\right)$ meters. Select the correct statement:

A
Acceleration is zero at $t = 0$
B
Velocity is zero at $t = 0$
✓
Velocity is zero at $t = 1$ second
D
Velocity and acceleration of the particle are never zero

Answer

Correct option: C.

Velocity is zero at $t = 1$ second

$\text{x}=3\text{t}^2-6\text{t},$ So $($Velocity$)_x=\frac{\text{dx}}{\text{dt}}=6\text{t}-6$
$($Acceleration$)_x=\frac{\text{d}^2\text{y}}{\text{dt}^2}=6\text{t};\text{y}=\text{t}^2-2\text{t};$
so $($Velocity$)_y=\frac{\text{d}^2\text{y}}{\text{dt}^2}=2;$ At time $t = 1,$
$\frac{\text{dx}}{\text{dt}}=6\times1-6=0$ and $\frac{\text{dy}}{\text{dt}}=2\times1-2=0$

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Question 711 Mark

Define one dimensional motion.

Answer

A particle moving along a straight line or a path is said to undergo one dimensional motion.

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Question 721 Mark

Why is it not necessary for a body following another to stop, to avoid collision?

Answer

If the relative velocity becomes zero, the collision can be avoided.

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Question 731 Mark

Displacement-time (S-T) graph of any object is shown in figure. Draw velocity-time graph for this motion.

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Question 741 Mark

A car starts accelerating from rest for sometime, maintains the velocity for sometime and then comes to rest with uniform deceleration. Draw V-t graph.

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Question 751 Mark

State the condition when the magnitude of velocity and speed of an object are equal.

Answer

When a body moves in a straight line.

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Question 761 Mark

What does the slope of velocity-time graph represent?

Answer

The slope of velocity-time graph represents acceleration.

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Question 771 Mark

Define displacement of a particle.

Answer

The change in the position co-ordinates of a particle over a given period of time is called the displacement of the particle.

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Question 781 Mark

Define uniformly accelerated motion.

Answer

A body has uniformly accelerated motion if it moves with constant acceleration.

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Question 791 Mark

A ball is dropped from a height of $90m$ on a floor. At each collision with the floor, the ball loses one tenth of its speed. Plot the speed$-$time graph of its motion between $t = 0$ to $12s$.

Answer

Downward motion of Ball

Ball is dropped from a height, $s=90 \mathrm{~m}$
Initial velocity of the ball, $\mathrm{u}=0$
Acceleration, $\mathrm{a}=\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^2$
Final velocity of the ball $=\mathrm{v}$
From second equation of motion, time $(t)$ taken by the ball to hit the ground can be obtained as:
$s=u t+(1 / 2) \mathrm{at}^2$
$90=0+(1 / 2) \times 9.8 \mathrm{t}^2$
$\mathrm{t}=\sqrt{18.38}=4.29 \mathrm{~s}$
From first equation of motion, final velocity is given as:
$v=u+a t$
$=0+9.8 \times 4.29=42.04 \mathrm{~m} / \mathrm{s}$
Here we see that $v=a t$, so velocity varies linearly with downward motion of ball.
This part of motion is represented by first line $(1)$ in the below graph

First rebound

Now Rebound velocity of the ball, $\mathrm{u}_{\mathrm{r}}=\mathrm{v}(1-1 / 10)=9 \mathrm{v} / 10=9 \times 42.04 / 10=37.84 \mathrm{~m} / \mathrm{s}$
This is represented the line $(2)$ in the graph. We are assumed negligible time of collision between ball and floor

Upward motion of ball

Time $\left(t_1\right)$ taken by the ball to reach maximum height is obtained with the help of first equation of motion as:
$\mathrm{v}=\mathrm{u}+\mathrm{at}_1$
$0=37.84+(-9.8) \mathrm{t}_1$
$\mathrm{t}_1=-37.84 /-9.8=3.86 \mathrm{~s}$
Total time taken by the ball $=\mathrm{t}+\mathrm{t}_1=4.29+3.86=8.15 \mathrm{~s}$
This is represented by line $(3)$ in the graph

Second descent of ball

As the time of ascent is equal to the time of descent, the ball takes $3.86 s$ to strike back on the floor for the second time.
The velocity with which the ball rebounds from the floor $=9 \times 37.84 / 10=34.05 \mathrm{~m} / \mathrm{s}$
Total time taken by the ball for second rebound $=8.15+3.86=12.01 \mathrm{~s}$
This is represented by line $(4)$ in the graph

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Question 801 Mark

A car travels at a speed of 60km/hr due north and the other at a speed of 60km/hr due east. Are the velocities equal? If no, which one is greater? If you find any of the questions irrelevant, explain.

Answer

There are equal because they have different directions. There magnitude can be compared.

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Question 811 Mark

If in case of a motion, displacement is directly proportional to the square of time elapsed, what do you think about its acceleration i.e., constant or variable? Explain why?

Answer

Acceleration is constant, since the equation of motion canbe applied only then.

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Question 821 Mark

Constant acceleration means that x-t graph will have constant slope? Yes/ No

Answer

Acceleration means that velocity is non-uniform. So, x-t graph will be curved.

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Question 831 Mark

Define non-uniformly accelerated motion.

Answer

A body has non-uniformly accelerated motion if it moves with variable acceleration.

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Question 841 Mark

Reaction time : When a situation demands our immediate action, it takes some time before we really respond. Reaction time is the time a person takes to observe, think and act. For example, if a person is driving and suddenly a boy appears on the road, then the time elapsed before he slams the brakes of the car is the reaction time. Reaction time depends on complexity of the situation and on an individual.
You can measure your reaction time by a simple experiment. Take a ruler and ask your friend to drop it vertically through the gap between your thumb and forefinger (Fig. 2.8). After you catch it, find the distance $d$ travelled by the ruler. In a particular case, $d$ was found to be $21.0 cm$. Estimate reaction time.

Answer

The ruler drops under free fall. Therefore, $v_o=0$, and $a=-g=-9.8 m s ^{-2}$.
The distance travelled $d$ and the reaction time $t_r$ are related by
$d=-\frac{1}{2} g t_r^2$
$t_r=\sqrt{\frac{2 d}{g}} s$
Given $d=21.0 \ cm$ and $g=9.8 m s ^{-2}$ the reaction time is
$t_r=\sqrt{\frac{2 \times 0.21}{9.8}} s \equiv 0.2 s .$

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Physics 1 Marks Question