1 Marks Question

Question 11 Mark

Shows the x-t plot of one-dimensional motion of a particle. Is it correct to say from the graph that the particle moves in a straight line for t < 0 and on a parabolic path for t > 0? If not, suggest a suitable physical context for this graph.

Answer

No, because the x-t graph does not represent the trajectory of the path followed by a particles. From the graph, it is noted that at t = 0, x = 0.

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Question 21 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 31 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, $a=g=9.8 \mathrm{~m} / \mathrm{s}^2$
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:
$\mathrm{s}=\mathrm{ut}+(1 / 2) \mathrm{at}^2$
$90=0+(1 / 2) \times 9.8 \mathrm{t}^2$
$\mathrm{t}=\sqrt{18.38}=4.29 \mathrm{~s}$
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 $\mathrm{v}=\mathrm{at}$, 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, $u_r=v(1-1 / 10)=9 \mathrm{v} / 10=9 \times 42.04 / 10=37.84 \mathrm{~m} / \mathrm{s}$
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 41 Mark

Draw position-time graph for a body at rest.

Answer

In the above x-t graph, body is at rest at position $x_o$.

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Question 51 Mark

Is it possible to have negative value in speed and displacement?

Answer

Speed is always +ve. Displacement can have -ve, 0 or +ve values.

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Question 61 Mark

Why the speed of an object can never be negative?

Answer

Speed is distance covered per unit time. Since distance cannot be negative therefore speed cannot be negative.

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Question 71 Mark

Write an expression for distance covered in $n^{th}$ second for a uniformly accelerated motion.

Answer

If a is the uniform acceleration, then $\text{s(n}^{\text{th}})=\text{u}+\frac{1}{2}\text{a}(2\text{n}-1)$ where, s ($n^{th}$) is the distance covered in $n^{th}$ second, u is the initial velocity.

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Question 81 Mark

The length covered by a body is found to be directly proportional to the square of time. What is the nature of acceleration?

Answer

$\text{x}\propto\text{t}_2\therefore\text{v}\propto\text{t}$ and $\text{a}\propto\text{t}^0$ (i. e.) acceleration is constant.

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Question 91 Mark

Draw v-t graph for non-uniform accelerated motion.

Answer

Here, acceleration is increasing.

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Question 101 Mark

A car moving at a speed of 10m/s is accelerated at the rate of $2m/s^2$. Find out the velocity after 6 sec.

Answer

Velocity after 6 seconds $=10+2 \times 6=22 \mathrm{~ms}^{-1}$.

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Question 111 Mark

Two particles A and B start from rest and move for equal time on a straight line. The particle A has an acceleration a for the first half of the total time and 2a for the second half. The particle B has an acceleration 2a for the first half and a for the second half. Which particle has covered larger distance?

Answer

Second particle

Area under (v, t) graph gives displacement. For graph (2) area is more.

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Question 121 Mark

Speed of a particle cannot be negative. Why?

Answer

Speed is the distance travelled in unit time and distance cannot be negative. negative.

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Question 131 Mark

Draw position-time graph for a non-uniform motion, when body starts from origin with increasing velocity.

Answer

In the above x-t graph, the slope of x-t graph is increasing, it means that the velocity is increasing.

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Question 141 Mark

Why does the earth impart the same acceleration to all bodies?

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Question 151 Mark

It is a common observation that rain clouds can be at about a kilometre altitude above the ground. Rate of change of momentum is force. Estimate how much force such a drop would exert on you.

Answer

Key concept: This problem can be solved by kinematic equations of motion and Newton’s second law that $\text{F}_\text{ext}=\frac{\text{dp}}{\text{dt}}$ will be used, where dp is change in momentum over time dt. Force exerted by a rain drop is $\text{F}=\frac{\text{Change in momentum}}{\text{Time}}=\frac{\text{p}-0}{\text{t}}$ $=\frac{4.7\times10^{-3}}{2.8\times10^{-5}}=168\text{N}$

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Question 161 Mark

Is it possible that a body could have constant speed but varying velocity?

Answer

Yes, a body could have constant speed but varying velocity if only the direction of motion changes.

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Question 171 Mark

Which speed is measured by speedometer of your scooter?

Answer

Instantaneous speed of the scooter is measured by the speedometer.

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Question 181 Mark

What can you say about the nature of acceleration, associated with a mass whose y-t graph is shown?

Answer

Slope covers the same magnitude. So acceleration is constant.

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Question 191 Mark

Does the displacement of an object depend on the choice of the position of origin of the coordinate system?

Answer

No, the displacement of the object does not depend on the choice of the position of the origin.

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Question 201 Mark

When a body accelerates by $\alpha \text{ t},$ what is the velocity after time 't', when it starts from rest?

Answer

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

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Question 211 Mark

Write the expression for distance covered in $n^{th}$ the second by a uniformly accelerated body.

Answer

If a is the uniform acceleration then, $\text{s}=\text{u}+\frac{1}{2}\text{a}(2\text{n}-1)$

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Question 221 Mark

Consider that the acceleration of a moving body varies with time. What does the area under acceleration-time graph for any time interval represent?

Answer

The area under acceleration-time graph for any time interval represents the change of velocity of the body during that time interval.

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Question 231 Mark

What is the significance of the slope of x-t graph.

Answer

Slope of x-t graph provides velocity of motion. The nature of motion is identified by the shape of the graph.

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Question 241 Mark

Find the acceleration and velocity of a ball at the instant it reaches its highest point if it was thrown up with velocity v.

Answer

Acceleration is $9.8m/s^2$ (downwards) and velocity is zero at the highest point.

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Question 251 Mark

When a ball hits a wall with a velocity of 50m/s and bounces back with the same speed, what is the change in velocity of the ball?

Answer

$\Delta \text{p}=-2\text{mv}=-100\text{m}$ $\therefore\text{m}\Delta\text{v}'=-100\text{m}$ $\Rightarrow \text{v}'=-100\text{m/s}$

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Question 261 Mark

What is the ratio of the time taken to go up and come down by a body thrown vertically up?

Answer

1 : 1 in the absence of air resistance.

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Question 271 Mark

Two masses in the ratio 1 : 2 are thrown vertically up with the same speed. What is the effect on the time by the mass?

Answer

Mass does not influence time.

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Question 281 Mark

What is the condition for an object to be considered as a point object?

Answer

An object can be considered as a point object, if the distance travelled by it is very large than its size.

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Question 291 Mark

Can Earth be regarded as a 'point object' if only the orbital motion of Earth around the sun is considered?

Answer

Yes. This is because the size of the Earth is very small as compared to the size of the orbit of the Earth around the sun.

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Question 301 Mark

A monkey climbs up a slippery pole for 3 seconds and subsequently slips for 3 seconds. Its velocity at time t is given by v(t) = 2t(3 - t); 0 < t < 3 and v(t) = -(t - 3)(6 - t) for 3 < t < 6s in m/s. It repeats this cycle till it reaches the height of 20m. At what time is its acceleration maximum in magnitude?

Answer

Given velocity $\text{v}(\text{t})=2\text{t}(3-\text{t})=6\text{t}-2\text{t}^2$ In a periodic motion when velocity is zero acceleration will be maximum putting v = 0 in Eq. (i). $0=6\text{t}-2\text{t}^2\Rightarrow0=\text{t}(6-2\text{t})$ $=\text{t}\times2(3-\text{t})=0\Rightarrow\text{t}=0\ \ \text{or}\ \ 3\text{s}$

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Question 311 Mark

Can a body have constant speed but a varying velocity?

Answer

Yes, it is possible only if direction changes.

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Question 321 Mark

Do the following two graphs represent same type of motion? Name the motion.

Answer

Both the graphs represent the non-uniform type of motion.

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Question 331 Mark

A ball is thrown straight up. What is its velocity and acceleration at the top?

Answer

v = 0, a = g.

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Question 341 Mark

Give an example of a body possessing zero velocity and still accelerating.

Answer

A body thrown up vertically has zero velocity at the top-most point, but has acceleration of g.

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Question 351 Mark

The displacement-time graph for two particles X and Y are straight lines making angles of 30° and 60° with the time axis. What is the ratio of the velocities of Y and X?

Answer

$\frac{\text{v}_\text{Y}}{\text{v}_\text{X}}=\frac{\tan60^\circ}{\tan30^\circ}$ $=\frac{\sqrt{3}}{\frac{1}{\sqrt{3}}}=3:1$

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Question 361 Mark

For which condition, the magnitude of average velocity is equal to the average speed for a particular motion?

Answer

When a particle is moving along a straight line with the fixed direction.

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Question 371 Mark

The position coordinate of a moving particle is given by $\mathrm{x}=6+18 \mathrm{t}+9 \mathrm{t}^2$, where x is in metres and t in seconds. What is the velocity at $\mathrm{t}=2 \mathrm{~s}$ ?

Answer

Given, $\text{x}=6+18\text{t}+9\text{t}^2$ $\text{v}_\text{t}=\frac{\text{dx}}{\text{dt}}=18+18\text{t}$ At t = 2, $\text{v}_2=18+18\times2$ $=54\text{m/s}$

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Question 381 Mark

Two particles A and B are moving along the same straight line. Bis ahead of A. Velocities remaining unchanged, what would be effect on the magnitude of relative velocity if A ahead of B?

Answer

There will be no effect on the magnitude of relative velocity.

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Question 391 Mark

At which point on its path a projectile has the smallest speed?

Answer

Highest.

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Question 401 Mark

What is the ratio of SI to CGS unit of acceleration?

Answer

Ratio of SI to CGS unit of acceleration is $10^2$.

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Question 411 Mark

When a body accelerates by $\beta \text{t},$ what is the velocity after time t, when it starts from rest?

Answer

Given, acceleration $\text{a}=\beta \text{}t$ It can be written as $\int\text{dv}=\int \beta\text{tdt}$ On intergrating, we get $\text{v}=\frac{\beta\text{t}^2}{2}+\text{C}=\frac{\beta\text{t}^2}{2}$ [$\because$ C = 0]

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Question 421 Mark

A mass is dropped from certain height. At the same time another equal mass is thrown with a horizontal velocity of u m/s. Which one of the two will reach the ground first?

Answer

Both of them will reach the ground at the same time.

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Question 431 Mark

Answer

No, because the x-t graph does not represent the trajectory of the path followed by a particles. From the graph, it is noted that at t = 0, x = 0.

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Question 441 Mark

The position coordinate of a moving particle is given by $x=6+18 t+9 t^2$ ( $x$ in metres and $t$ in seconds). What is its velocity at $\mathrm{t}=2 \mathrm{sec}$ ?

Answer

$\mathrm{x}=6+18 \mathrm{t}+9 \mathrm{t}^2, \mathrm{v}=18+18 \mathrm{t} \mathrm{~v}_{\mathrm{t}}=2=18+18 \times 2=54 \mathrm{~ms}^{-1}$

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Question 451 Mark

Write the expression for distance covered in $n^{th}$ second by a uniformly accelerated body.

Answer

If a is the uniform acceleration, then $\text{s}=\text{u}+\frac{\text{a}}{2}(2\text{n}-1),$ where u is the initial velocity.

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Question 461 Mark

The displacement-time graph of a particle is parallel to time-axis, what is the velocity of the particle?

Answer

The velocity of the particle is zero because the slope of (x-t) graph is zero.