Assertion is incorrect but Reason is correct.

Both Assertion and Reason are correct and Reason is the correct explanation for Assertion.

Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion.

Assertion is correct but Reason is incorrect.

MCQ Questions for Class 9 Physics Motion Quiz 5

State whether given statement is True or False.
A body can have constant velocity and still have varying speed?

0%
True
0%
False

A driver applies brakes when he sees a child on the railway track, the speed of the train reduces from

$54\ kmph$ to

$18\ kmph$ in

$5\ s$ . What is the distance traveled by train during this interval of time?

0%
$52\ m$
0%
$50\ m$
0%
$25\ m$
0%
$80\ m$

Explanation

Initial speed

$u = 54 \ kmph$

$\Rightarrow u= 54\times \dfrac{5}{18}$

$\Rightarrow u = 15 \ ms^{-1}$

Final speed

$v = 18 \ kmph$

$\Rightarrow v =18\times \dfrac{5}{18}$

$\Rightarrow v = 5 \ ms^{-1}$

Time

$t = 5 \ s$

Using

$v = u+at$

$\therefore$

$15 = 5+a\times 5$

$\implies \ a = -2 \ m/s^2$

Using

$2aS = v^2 - u^2$

where

$S$ is the distance travelled.

$\therefore$

$2\times (-2)S = 5^2 - 15^2$

$\Rightarrow \ S = 50 \ m$

The speed of a car reduces from

$15 m s^{-1}$ to

$5 m s^{-1}$ over a displacement of 10 m. What is the uniform acceleration of the car?

0%
$-10 m s^{-2}$
0%
$+10 m s^{-2}$
0%
$2 m s^{-2}$
0%
$0.5 m s^{-2}$

Explanation

$\displaystyle a = \frac{v^2 - u^2}{2s} = \frac{5^2 - 15^2}{2 \times 10}$

$\displaystyle = \frac{-20 \times 10}{20}= - 10 m s^{-2}$

To reach the same height on the moon as on the earth, a body must be projected up with:

0%
Higher velocity on the moon
0%
Lower velocity on the moon
0%
Same velocity on the moon and earth
0%
It depends on the mass of the body

Explanation

$\textbf{Hint:}$ Maximum height attained should be same on earth and moon

$\textbf{Explanation:}$

The maximum height attained is given as,

${h_{ma \times }} = \dfrac{{{u^2}}}{{2g}}$

here $u$ is the initial velocity, and $g$ is the acceleration due to gravity

So, initial velocity is directly proportional to gravity

$u\alpha \sqrt g$

And Acceleration due to gravity of earth is more than moon,

${g_{earth}} = 6{g_{moon}}$

$\therefore {u_{moon}} < {u_{earth}}$

Hence, a body must be projected up with low velocity for moon.

A particle is thrown vertically upwards. Its velocity at one fourth of the maximum height is

$20 \ m s^{-1}$ . Then, the maximum height attained by it is

0%
16 m
0%
10 m
0%
8 m
0%
18 m

Explanation

Let us assume the initial velocity to be

$u$ .

$H$ be the maximum height reached by the ball.

Given:

$v=0$

By third equation of motion,

$v^2-u^2=2as$

$0-u^2=2(-g)H$

$\Rightarrow H=\dfrac{u^2}{2g}$ . . . . (i)

Given, for

$h=\dfrac{1}{4}H$ ,

$v=20 \ ms^{-1}$

$h=\dfrac{1}{4} \times \dfrac{u^2}{2g}=\dfrac{u^2}{8g}$

Putting

$h$ and

$v$ and

$a=-g$ in third equation of motion,

$(20)^2-u^2= 2(-g)\left( \dfrac{u^2}{8g}\right)$

$\Rightarrow u^2- \dfrac{u^2}{4}=400$

$\Rightarrow 3u^2 =1600$

$\Rightarrow u^2 =\dfrac{1600}{3}$

Putting the value of

$u^2$ is eq. (i),

Maximum height

$H=\dfrac{\left( \dfrac{1600}{3}\right)}{2g}$

$H=\dfrac{1600}{3 \times 2 \times 10}=26.66 \ m$

$\displaystyle \dfrac{u^2}{8g} \times 2(+g) = (20)^2 - u^2+ \dfrac{u^2}{4} + u^2 = 400$

$5 u^2 = 1600$

$u \displaystyle = \sqrt{\dfrac{1600}{5}} m s^{-1}$

$\displaystyle h_{max} = \dfrac{u^2}{2g} = \dfrac{1600}{5 \times 2 \times 10} = 16 m$

A stone thrown vertically upwards with an initial velocity u from the top of a tower reaches the ground with a velocity of 3 u. The height of the tower is :

0%
$3 \displaystyle \dfrac{u^2}{g}$
0%
$4 \displaystyle \dfrac{u^2}{g}$
0%
$6 \displaystyle \dfrac{u^2}{g}$
0%
$9 \displaystyle \dfrac{u^2}{g}$

Explanation

$-h = ut - \displaystyle \dfrac{1}{2} gt^2$

$-h = \displaystyle \dfrac{4u^2}{g} - \dfrac{1}{2} g \left ( \dfrac{16u^2}{g^2} \right)$

$-h = \displaystyle \dfrac{4u^2}{g} - \frac{8u^2}{g} = \dfrac{-4 u^2}{g}$

$h = \displaystyle \dfrac{4u^2}{g}$

A ball is thrown vertically upwards. It has a speed of

$10 m s^{-1}$ , when it has reached on half of its maximum height. How high does the ball rise? (Take g

$=$ 10 m s

$^{-2}$ )

0%
10 m
0%
5 m
0%
15 m
0%
20 m

Explanation

Let initial velocity be u

$\displaystyle h_{max} = \dfrac{u^2}{2g} ; \dfrac{h_{max}}{2} = \dfrac{u^2-10^2}{+2g}$

$\displaystyle \dfrac{u^2}{4g} = \dfrac{u^2-10^2}{2g}$

$u^2 = 2u^2-200$

$u^2 =200$

$u = 10 \sqrt{2} m s^{-1}$

$\therefore h_{max} = \displaystyle \dfrac{(10 \sqrt{2})^2}{2 \times 10} = 10 m$

A body having zero speed :

0%
Is always under rest.
0%
Has always zero acceleration.
0%
Has always uniform acceleration.
0%
Always under motion.

A ball is released from the top of height 'h' meters. It takes 't' seconds to reach the ground. Where is the ball at the time

$\displaystyle \dfrac{t}{2} s$ ?

0%
At $\displaystyle \left ( \dfrac{h}{4} \right)$ from the ground
0%
At $\displaystyle \left ( \dfrac{h}{2} \right)$ from the ground
0%
At $\displaystyle \left ( \dfrac{3h}{4} \right)$ from the ground
0%
Depends upon mass and volume of the ball

A body thrown vertically up reaches a maximum height of 50 m. Another body with double the mass is thrown up with double the initial velocity will reach a maximum height of :

0%
100 m
0%
200 m
0%
400 m
0%
50 m

Two balls are dropped from heights

$h_1$ and

$h_2$ respectively. The ratio of their velocities on reaching the ground is __________ .

0%
$\sqrt{h_1} : \sqrt{h_2}$
0%
$h_1 : h_2$
0%
$h^2_1 : h^2_2$
0%
none of these

Explanation

Initial velocity of ball

$u = 0$

Let the final velocity on reaching the ground be

$v$

Using 3rd equation of motion-

$v^2 = u^2+ 2gh$

$\Rightarrow v^2 = 0+ 2gh$

$\Rightarrow \ v = \sqrt{2gh}$

We get

$v \propto \sqrt{h}$

$\Rightarrow \ v_1:v_2 = \sqrt{h_1}:\sqrt{h_2}$

Displacement of a particle moving in a circle at any two different instants of time is zero.

0%
True
0%
False

A body moving in a straight line can have a constant velocity with varying speed.

0%
True
0%
False

The ratio of the heights from which two bodies are dropped is 3 : 5 respectively. The ratio of their final velocities is?

0%
$\sqrt{5} : \sqrt{3}$
0%
$\sqrt{5} : \sqrt{2}$
0%
$\sqrt{3} : \sqrt{5}$
0%
none of the above

Explanation

Initial speed,

$u = 0$ (at rest)

Final velocity,

$v$

Acceleration due to gravity,

$a = g$

Let the height from which it is dropped be

$s=h$ .

Using Newton's third equation of motion,

$2as=v^2 - u^2$

$2gh=v^2$

$\implies v=\sqrt{2gh}$

Let the heights from which two bodies are dropped are

$h_1$ and

$h_2$

Given,

$h_1:h_2=3:5$

Final velocity of

$1^{st}$ body,

$v_1=\sqrt{2gh_1}$ ...............(1)

Final velocity of

$2^{nd}$ body,

$v_2=\sqrt{2gh_2}$ ..............(2)

Diviing equation (1) by (2)

$\dfrac{v_1}{v_2}=\dfrac{\sqrt{2gh_1}}{\sqrt{2gh_2}}$

$\dfrac{v_1}{v_2}=\sqrt {\dfrac{h_1}{h_2}}$

$\dfrac{v_1}{v_2}=\sqrt {\dfrac{3}{5}}$

$v_1:v_2=\sqrt3:\sqrt5$

A freely falling body travels with uniform acceleration .

0%
True
0%
False

Explanation

When a body is in free fall motion, acceleration due to gravity

$(g = 10m/s^2)$ always acts in the downward direction. So, the given statement is true that a freely falling body travels with uniform acceleration.

A car attains a velocity of 10 m s

$^{-1}$ in 5 s. If initially, it had been at rest, its acceleration must be ___________ .

0%
50m s $^{-2}$
0%
2 m s $^{-2}$
0%
5 m s $^{-2}$
0%
0.6 m s $^{-2}$

Explanation

Initial velocity of the car

$u = 0$

Final velocity

$v = 10 \ m/s$

Time taken

$t = 5 \ s$

Using

$v = u+at$

$\therefore$

$10 = 0+a\times 5$

$\implies \ a = 2 \ m/s^2$

If a body starts from rest and moves with uniform acceleration, then the displacement of the body is directly proportional to the cube of the time.

0%
True
0%
False

Explanation

Initial velocity of the body

$u = 0$

Let the constant acceleration be

$a$ .

Displacement of the body

$S = ut+\dfrac{1}{2}at^2$

$\therefore$

$S = 0+\dfrac{1}{2}at^2$

We get

$S = \dfrac{at^2}{2}$

Thus, displacement of the body is directly proportional to the square of the time.

A body falling for 2 s, covers a distance equal to that covered in the next second. State whether true or false

0%
True
0%
False

Explanation

Initial speed of the body

$u = 0$

Acceleration due to gravity

$g = 10 \ m/s^2$

For

$t = 2 \ s$ :

Distance covered in 2 s,

$S_2 = ut+\dfrac{1}{2}gt^2$

where

$t =2 \ s$

$\therefore$

$S_2 = 0+\dfrac{1}{2}\times 10\times 2^2 = 20 \ m$

For

$t = 3 \ s$ :

Distance covered in 3 s,

$S_3 = ut+\dfrac{1}{2}gt^2$

where

$t =3 \ s$

$\therefore$

$S_3 = 0+\dfrac{1}{2}\times 10\times 3^2 = 45 \ m$

So, distance covered in 3rd second

$S = S_3-S_2 = 45-20 = 25 \ m$

Thus given statement is false.

The displacement - time graphs of two bodies

$A$ and

$B$ are

$OP$ and

$OQ$ respectively. Velocities of A and B are related as

0%
$V_A > V_B$
0%
$V_A < V_B$
0%
$V_A = V_B$
0%
Cannot Predict

Explanation

Velocity is given by the slope of displacement time graph, therefore, graph making more angle with the x-axis has higher velocity, therefore we have

$V_A > V_B$

A point moves rectilinearly in one direction. Above figure shows the distance

$s$ traversed by the point as a function of the time

$t$ . Using the plot, find the maximum velocity.

0%
$1.5\:m/s$
0%
$2.5\:m/s$
0%
$2 \:m/s$
0%
$5\:m/s$

Explanation

Since velocity is the slope of distance versus time graph, thus maximum velocity means maximum slope or the point of inflation.

From the graph, it can be clearly seen that maximum slope occurs in the straight line region between

$10$ seconds and

$14$ seconds, which is given by

$\quad v = \dfrac { 14-4 }{ 14-10}$ or

$\quad v = \dfrac { 10 }{ 4} = 2.5$ m/s

A car is moving in a straight line. The position time graph

$(x-t)$ is as shown in the given figure. Find the average speed of the car for the section- CD.

0%
$1\ ms^{-1}$
0%
$2.66\ ms^{-1}$
0%
$3\ ms^{-1}$
0%
$0.89\ ms^{-1}$

Explanation

$\underline{\text{For the section-CD}}$
Total distance travelled is =

$(4-0)=4\ m$

Total time taken =

$(6-4.5)=1.5\ sec$

$\text{Average Speed}=\dfrac {\text{Total distance}}{\text{Total time}}$

$\text{Average speed}=\dfrac{4}{1.5}=2.66\ m/s$

0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion.
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion.
0%
Assertion is correct but Reason is incorrect.
0%
Assertion is incorrect but Reason is correct.

Explanation

At point A, sign of slope of

$s-t$ graph is not changing. Therefore, sign of velocity is not changing. As the sign is not changing so the direction of the velocity is also not changing. Hence D is the correct answer.

At which instant do the two cars have the same velocity?

0%
$t_1$
0%
$t_2$
0%
$t_3$
0%
$t_4$

Explanation

Velocity is given by slope of distance-time graph. Thus same velocity means same slope. It can be clearly seen, the two curves have same slope at time

${ t }_{ 2 }$

Which one of the following best describes the motion of car A as shown on the graphs?

0%
Speeding up
0%
Constant velocity
0%
Slowing down
0%
First speeding up, then slowing down

A balloonist is ascending at a velocity of

$12\ m{ s }^{ -1 }$ . A packet is dropped from it when it is at height of

$65 \ m$ from the ground. Time taken by the packet to reach the ground is

0%
$5 \ s$
0%
$-5 \ s$
0%
$7 \ s$
0%
$\cfrac{13}{5} \ s$

Explanation

Using the second equation of motion,

$s=ut+\frac { 1 }{ 2 } a{ t }^{ 2 }$

Here,

$s = 65 \ m,\ \ u = -12 \ m/s$ (since it's moving upwards),

$a=g=10 \ m/s^2$

$65=-12t+\dfrac { 1 }{ 2 } 10{ t }^{ 2 }$

$\Rightarrow 5{ t }^{ 2 }-12t-65=0\\ \Rightarrow t=5\quad ; t=-13/5$

Since, time cannot be negative.
Thus,

$t=5s$

A parachutist after bailing out falls

$50\ m$ without friction. When parachute opens, it decelerates at

$2\ m{ s }^{ -2 }$ . He reaches the ground with speed

$3\ m{ s }^{ -1 }$ . At what height did he bail out?

(

$g=9.81m/{ s }^{ 2 }$ )

0%
$91\ m$
0%
$182\ m$
0%
$293\ m$
0%
$111\ m$

Explanation

$\text{Initial velocity,}\ u=\sqrt{2gh}$

$\text{or,}\ u=\sqrt{2\times 9.8 \times 50}=14\sqrt{5}$

$\text{The velocity at ground,}\ v=3\ \text{m/s}$

$v^2 - u^2 =2as$

$\text{or,}\ {3}^{2}-980=2\times (-2) \times s$

$\text{Thus, }\ s=\dfrac{971}{4},\ \text{which is nearly 243 m.}$

$\text{Initially he has fallen 50 m}$

$\text{Thus total height at which he bailed out} = 243+50=293\ \text{m}$

An object may have
(I) varying speed without having varying velocity.
(II) varying velocity without having varying speed.
(III) non-zero acceleration without having varying velocity.
(IV) non-zero acceleration without having varying speed.

0%
I and II are correct.
0%
II and III are correct.
0%
II and IV are correct.
0%
None of the above.

The muzzle velocity of a certain rifle is

$330\ m{ s }^{ -1 }$ . At the end of one second, a bullet fired straight up into the air will travel a distance of

0%
$(330-4.9)\ m$
0%
$330\ m$
0%
$(330+4.9)\ m$
0%
$(330-9.8)\ m$

Explanation

Given:

The initial velocity

$u=330\ m/s$

Time taken

$t=1\ sec$

Acceleration =

$-g$ (As the bullet is going up)

Using the equation of motion

$s=ut-\dfrac { 1 }{ 2 } g{ t }^{ 2 }$

$\therefore s=330(1)-\dfrac { 1 }{ 2 } 9.8{ (1) }^{ 2 }=(330-4.9)m$

When a ball is

$h$ metre high from a point

$O$ , its velocity is

$v$ . When it is

$h\ m$ below

$O$ , its velocity is

$2v$ . Find the maximum height from

$O$ it will acquire.

0%
$\cfrac { 2h }{ 3 }$
0%
$\cfrac { 5h }{ 3 }$
0%
$\cfrac { 3h }{ 2 }$
0%
$2h$

Explanation

Using third equation of motion

$v^2-u^2=2as$

For motion from Highest to lowest point

$,\ { \left( 2v \right) }^{ 2 }- \left( v \right)^{ 2 }=2g\left( h \right) \quad or\quad \dfrac { { v }^{ 2 } }{ 2g } =\dfrac { 2h }{ 3 }$

We know Maximum height is given by

$H=\frac{v^2}{2g}$
At TOP POINT velocity of ball is

$v$ , so maximum height it will attain

$=\dfrac { { v }^{ 2 } }{ 2g } =\dfrac { 2h }{ 3 }$
From point O,maximum height ball acquire

$=h+\dfrac { 2h }{ 3 } =\dfrac { 5h }{ 3 }$

Which of the following statements are true for a moving body?

0%
If its speed changes, its velocity must change and it must have some acceleration.
0%
If its velocity changes, its speed must change and it must have some directions.
0%
If its velocity changes, its speed may or may not change, and it must have some acceleration.
0%
If its speed changes but direction of motion does not change, its velocity may remain constant.

Explanation

$v=\dfrac { change \ in \ displacement }{ change\ in\ time } \quad and\quad a=\dfrac { change \ in \ velocity }{ change \ in \ time }$

Speed is the magnitude of velocity.
There can be a case where there is change in direction but not in magnitude.

A bullet, moving with a velocity of

$200cm/s$ penetrates a wooden block and comes to rest after traversing

$4cm$ inside it. What velocity is needed for traversing a distance of

$6cm$ in the same block

0%
$104.3cm/s$
0%
$136.2cm/s$
0%
$244.9cm/s$
0%
$272.7cm/s$

Explanation

${ v }^{ 2 }-{ u }^{ 2 }=2as\\ 0-{ (200) }^{ 2 }=2a\times 4\\ \Rightarrow a=5000\quad cm/{ s }^{ 2 }$
Now initial velocity to travel 6 cm,

$0-{ u }^{ 2 }=-2\times 5000\times 6\\ \Rightarrow u=244.9\quad cm/{ s }$

The first stage of the rocket launches a satellite to a height of

$50\ km$ and velocity attained is

$6000\ km{ h }^{ -1 }$ at which point its fuel exhausted. How high the rocket will reach (Take

$g=10\ m/s^2$ and assume

$g$ is constant up to for rocket's entire journey?

0%
$138.9\ km$
0%
$188.9\ km$
0%
$88.9\ km$
0%
$168.9\ km$

When a ball is thrown up vertically with a velocity

${ v }_{ 0 }$ it reaches a height

$h$ . If one wishes to triple the maximum height then the ball be thrown with a velocity

0%
$\sqrt { 3 } { v }_{ 0 }$
0%
$3{ v }_{ 0 }$
0%
$9{ v }_{ 0 }$
0%
$\dfrac{3}{2}{ v }_{ 0 }$

Explanation

Using the third equation of motion,

$v^2 = u^2 - 2as$

$Case I$ : Given ,

$u=v_o$ ,

$v=0$ ,

$s=h$

Substituting in the equation, we get,

$\ 0={ v }_{ o }^{ 2 }-2gh\quad or\quad h=\dfrac { { v }_{ o }^{ 2 } }{ 2g }$

$Case II$ : Given,

$v=0$ ,

$h'=3h$ ,

$u=v_1$

Substituting in the equation, we get,

$\ 0={ v }_{ 1 }^{ 2 }-2g\left( 3h \right) \quad or\quad h=\dfrac { { v }_{ 1 }^{ 2 } }{ 6g }$

On equating both equations, we get,

$\dfrac { { v }_{ 1 }^{ 2 } }{ 6g } =\dfrac { { v }_{ o }^{ 2 } }{ 2g } \quad or\quad { v }_{ 1 }=\sqrt { 3 } { v }_{ o }$

Two balls

$A$ and

$B$ are simultaneously thrown.

$A$ is thrown from the ground level with a velocity of

$20\ m{ s }^{ -1 }$ in the upward direction and

$B$ is thrown from a height of

$40\ m$ in the downward direction with the same velocity. Where will the two balls meet?

0%
$15\ m$
0%
$25\ m$
0%
$35\ m$
0%
$45\ m$

Explanation

Let both the ball meet after time

$t$ and

$x$ be the distance covered by ball

$A$ .
So

$,\ x=20t-\dfrac { 1 }{ 2 } \left( 10 \right) { t }^{ 2 }...\left( 1 \right)$
For ball

$B,\quad \left( 40-x \right) =20t+\dfrac { 1 }{ 2 } \left( 10 \right) { t }^{ 2 }...\left( 2 \right)$
Adding

$(1)\ \&\ (2) ,\quad 40=40t\quad or\quad t=1\ s$
so

$x=20-5=15\ m$

A body moving in circular motion with constant speed has :

0%
constant velocity
0%
constant acceleration
0%
constant kinetic energy
0%
constant displacement

When the speed of a car is

$v$ , the minimum distance over which it can be stopped is

$s$ . If the speed becomes

$nv$ , what will be the minimum distance over which it can be stopped during the same time?

0%
$s/n$
0%
$ns$
0%
$s/ { n }^{ 2 }$
0%
${ n }^{ 2 }s$

Explanation

$0-{ v }^{ 2 }=2as\quad \Rightarrow \quad a=\dfrac { -{ v }^{ 2 } }{ 2s } \\ 0-{ (nv) }^{ 2 }=2ad\quad \Rightarrow \quad -{ (nv) }^{ 2 }=2\times \dfrac { -{ v }^{ 2 } }{ 2s } \times d\\ \Rightarrow d={ n }^{ 2 }s$

The two ends of a train moving with constant acceleration pass a certain point with velocities

$u$ and

$v$ . The velocity with which the middle point of the train passes the same point is

0%
${ \left( u+v \right) }/{ 2 }$
0%
${ \left( { u }^{ 2 }+{ v }^{ 2 } \right) }/{ 2 }$
0%
$\sqrt { { \left( { u }^{ 2 }+{ v }^{ 2 } \right) }/{ 2 } }$
0%
$\sqrt { { u }^{ 2 }+{ v }^{ 2 } }$

Explanation

Let

$s$ be the length of train using

${ v }\displaystyle^{ 2 }={ u }\displaystyle^{ 2 }+2a\times s$

$or\quad { 2as }={ v }^{ 2 }-{ u }^{ 2 }$ ......(1)
Where

$v$ is final velocity after travelling a distance

$s$ with an acceleration

$a$ and

$u$ is initial velocity as per question
Let velocity of middle point of train at same point is

${ v }\displaystyle^{ \prime }$ , then

${ \left( { v }\displaystyle^{ \prime } \right) }\displaystyle^{ 2 }={ u }\displaystyle^{ 2 }+2a\times \left( { s }/{ 2 } \right)$

$={ { u }^{ 2 }+as }$ .....(2)
By equations (1) and (2), we get

${ v }^{ \prime }=\sqrt { \displaystyle\frac { { v }\displaystyle^{ 2 }+{ u }\displaystyle^{ 2 } }{ 2 } }$

A stone is thrown upwards with a velocity

$v$ from the top of a tower. It reaches the ground with a velocity

$3v$ . What is the height of the tower?

0%
$\dfrac{2v^2}{ g }$
0%
$\dfrac{3v^2}{ g }$
0%
$\dfrac{4v^2}{ g }$
0%
$\dfrac{6v^2}{ g }$

Explanation

${ v }^{ 2 }-{ u }^{ 2 }=2as\\ \Rightarrow { (3v) }^{ 2 }-{ v }^{ 2 }=2gh\\ \Rightarrow h=\dfrac { { 4v }^{ 2 } }{ g }$

Two particles are projected vertically upwards with the same velocity on two different planets with accelerations due to gravities

${ g }_{ 1 }$ and

${ g }_{ 2 }$ respectively. If they fall back to their initial points of projection after lapse of times

${ t }_{ 1 }$ and

${ t }_{ 2 }$ , respectively, then

0%
${ t }_{ 1 }{ t }_{ 2 }={ g }_{ 1 }{ g }_{ 2}$
0%
${ t }_{ 1 }{ g }_{ 1 }={ t }_{ 2 }{ g }_{ 2 }$
0%
${ t }_{ 1 }{ g }_{ 2 }={ t }_{ 2 }{ g }_{ 1}$
0%
${ { t }_{ 1 } }^{ 2 }+{ { t }_{ 2 } }^{ 2 }={ g }_{ 1 }+{ g }_{ 2 }$

Explanation

For both particles, net vertical displacement is zero.
Particle 1:

$y=ut+\dfrac { 1 }{ 2 } a{ t }^{ 2 }\quad \Rightarrow 0=u{ t }_{ 1 }-\dfrac { 1 }{ 2 } { g }_{ 1 }{ t }_{ 1 }^{ 2 }\quad or\quad u=\dfrac { 1 }{ 2 } { g }_{ 1 }{ t }_{ 1 }$
Particle 2:

$0=u{ t }_{ 2 }-\dfrac { 1 }{ 2 } { g }_{ 2 }{ t }_{ 2 }^{ 2 }\quad or\quad u={ \dfrac { 1 }{ 2 } { g }_{ 2 } }{ t }_{ 2 }$
thus

$u={ \dfrac { 1 }{ 2 } g }_{ 1 }{ t }_{ 1 }={ \dfrac { 1 }{ 2 } { g }_{ 2 } }{ t }_{ 2 }\quad or\quad { g }_{ 1 }{ t }_{ 1 }={ { g }_{ 2 } }{ t }_{ 2 }$

A stone thrown vertically upwards with a speed of

$5{ m }/{ s }$ attains a height

${ H }_{ 1 }$ . Another stone thrown upwards from the same point with a speed of

$10{ m }/{ s }$ attains a height

${ H }_{ 2 }$ . The correct relation between

${ H }_{ 1 }$ and

${ H }_{ 2 }$ is

0%
${ H }_{ 2 }=4{ H }_{ 1 }$
0%
${ H }_{ 2 }=3{ H }_{ 1 }$
0%
${ H }_{ 1 }=2{ H }_{ 2 }$
0%
${ H }_{ 1 }={ H }_{ 2 }$

Explanation

From third equation of motion,

$v^{2} = u^{2} + 2ah$
In first case initial velocity

$u_{1} = 5\ ms^{-1}$
final velocity

$v_{1} = 0,\ a=g$
So,

$0 = 25 - 2gH_{1} \Rightarrow H_{1} = \dfrac{25}{2g}$
and maximum height obtained is H1, then,

$H_{1} = \dfrac{25}{2g} ...........(1)1$
In second case

$u_{1} = 10 ms^{-1} ,\ v_{1} = 0,\ a = g$
So,

$0 = 100 - 2gH_{2} \Rightarrow H_{2} = \dfrac{100}{2g}$
and maximum height obtained is H_2, then,

$H_2 = \dfrac{100}{2g} .............(2)$
from

$(1)$ and

$(2),$ we get,

$H_2=4H_1$

The displacement of a particle as a function of time is shown in figure. The figure indicates that

0%
the particle starts with a certain velocity, but the motion is retarded and finally the particle stops.
0%
the velocity of the particle is constant throughout.
0%
the acceleration of the particle is positive throughout.
0%
the particle starts with a constant velocity, the motion is accelerated and finally the particle moves with another constant velocity.

Explanation

Initially the slope is decreasing and the slope

$=$ 0.
Thus velocity is decreasing as time increases, i.e. there is retardation and ultimately the particle stops. It must have obviously started with some velocity.

The given figure shows the displacement-time curve of two particles

$P$ and

$Q$ . Which of the following statements is correct?

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Both $P$ and $Q$ move with uniform equal speed
0%
$P$ is accelerated and $Q$ is retarded
0%
Both $P$ and $Q$ move with uniform speeds but the speed of $P$ is more than the speed of $Q$
0%
Both $P$ and $Q$ move with uniform speeds but the speed of $Q$ is more than the speed of $P$

Explanation

As the

$x - t$ graph is a straight line, in either case, the velocity of both the particles is uniform. As the slope of the

$x - t$ graph for P is greater, therefore, the velocity of P is greater than that of Q. (Slope of

$x-t$ graph represents velocity)

Is the policeman right in arresting the driver to break speed limit?

0%
Yes
0%
No
0%
Insufficient data to reply
0%
The policeman demanded bribe

Explanation

Max deceleration

$a = -g$

$\Rightarrow a = -9.8ms^{-2}$

Distance,

$S= 15\ m$
Using third equation of motion:

$2as= v^2 - u^2$

(v=0, final speed as vehicle has stopped)

$\Rightarrow 2 \times (-9.8) \times 15 = -u^2$

$\Rightarrow u = 17.1464 ms^{-1}$

$\Rightarrow u = \dfrac{18}{5} \times 17.1464\ kmph$

$\Rightarrow u = 61.7271\ kmph$

So yes the policeman was right in arresting the driver to break the speed limit.

A ball released from a height falls

$5 m$ in one second. In 4 seconds it falls through (Take

$g=10ms^{-2}$ ):

0%
$20 m$
0%
$1.25 m$
0%
$40 m$
0%
$80 m$

Explanation

Since

$s=ut+\displaystyle\frac { 1 }{ 2 } g{ t }\displaystyle^{ 2 }$ , where

$u$ is initial velocity

$\& a$ is acceleration.
Given:

$u=0 \ \&\ a=g$
So, distance travelled in 4 sec is,

$s=\displaystyle\frac { 1 }{ 2 } \times 10\times 16=80m$

A stone is thrown vertically upwards. When the particle is at one-half its maximum height, its speed is

$10 { m }/{ s }$ , then maximum height attained by particle is

$\left( g=10 { m }/{ { s }^{ 2 } } \right)$ :

0%
$8 m$
0%
$10 m$
0%
$15 m$
0%
$20 m$

Explanation

From third equation of motion

${v}^{2}={u}^{2}-2gh$
Given v = 10m/s at h/2. But v=0, when particle attains maximum height h.
Therefore,

${10}^{2}={u}^{2}- 2g(\dfrac{h}{2})$
Now,

${u}^{2}-0=2gh \quad or \quad {u}^{2}=2gh$

$\Rightarrow 100=2gh-gh=gh$

$\Rightarrow h=10m$

A ball is dropped downwards, after

$1$ sec another ball is dropped downwards from the same point. What is the distance between them after

$3$ sec? (Take

$g=10ms^{-2}$ )

0%
$20m$
0%
$25m$
0%
$50m$
0%
$9.8m$

Explanation

Distance at a given time is given by equation,

$S = ut +\dfrac{1}{2}at^{2}$
For first body,

$u_{1} = 0 ,\ S = S_{1}, a = g ,\ t_{1} = 3 sec$

$\therefore S_1 = \dfrac{1}{2}g\times 9$
For second body,

$u_{2} = 0,\ S = S_{2},\ a = g,\ t_{2} = 2 sec$

$\therefore S_{2} = \dfrac{1}{2}g\times 4$
So difference between them after

$3 sec = S_{1} -S_{2}$

$= \dfrac{1}{2}g\times 5$
If

$\ g = 10\ m/s^{2}$ then

$S_{1}- S_{2} = 25 m$

Two stones are thrown from the top of a tower, one straight down with an initial speed

$u$ and the second straight up with the same speed

$u$ . When the two stones hit the ground, they will have speeds in the ratio

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$2:3$
0%
$2:1$
0%
$1:2$
0%
$1:1$

Explanation

In case of stone thrown downwards, initial velocity,

$u = u$
So, using equation

$v^{2} - u^{2} = 2aS$ ,
here,

$u = u, v = 0, a = a, S = Distance$
So, final equation

$= u^{2} = 2aS ...... (1)$
Again, for stone thrown upwards, initial velocity,

$u = -u$
So, using equation

$v^{2} - u^{2} = 2aS$ ,
here,

$u = -u, v = 0, a = a, S = Distance = S$
So, final equation

$= u^{2} = 2aS .......(2)$
Dividing

$(1)$ by

$2,$ we get ratio of

$1 : 1$
In both the cases

$, ( u$ is positive or negative

$) u^{2}$ is positive.

A body dropped from a height

$h$ with an initial speed zero, strikes the ground with a velocity

$3\ { km }/{ h }$ . Another body of same mass dropped from the same height

$h$ with an initial speed

$u=4\ { km }/{ h}$ . Find the final velocity of second mass, with which it strikes the ground.

0%
$3\ { km }/{ h }$
0%
$4\ { km }/{ h }$
0%
$5\ { km }/{ h }$
0%
$6\ { km }/{ h }$

Similar balls are thrown vertically each with a velocity

$20\ { m }/{ s }$ , one on the surface of earth and the other on the surface of moon. What will be ratio of the maximum heights attained by them?

$($ Acceleration on moon

$=\dfrac{g}{6}\ m / s^2$ approximately, where

$g$ is gravitational acceleration due to earth

$)$

0%
$6$
0%
$\displaystyle\dfrac { 1 }{ 6 }$
0%
$\displaystyle\dfrac { 1 }{ 5 }$
0%
None of these

Explanation

Since

${ v }\displaystyle^{ 2 }={ u }\displaystyle^{ 2 }+2as$ .......(1)

For first case

${ u }_{ 1 }=20 { m }/{ s },{ v }_{ 1 }=0,{ a }_{ 1 }=g=-10\ m/s^2,\ { s }_{ 1 }=?$
From equation (1)

$s _ 1=\dfrac { u_1^2 }{ 2\times g }$

$s _ 1=\dfrac { 400 }{ 2\times 10 } =20\ m$

For second case (at moon)

$u_2=20 { m }/{ s },\ v _2=0,\ { a }_{ 2 }=\displaystyle\dfrac { g }{ 6 }$

Similarlly,

$s _ 2=\dfrac { u_2^2 }{ 2\times g }$

${ s }_{ 2 }=\displaystyle\dfrac { 400 }{ 2\times { 10 }/{ 6 } }$

Now,

$\displaystyle\dfrac { { s }_{ 1 } }{ { s }_{ 2 } } =\displaystyle\dfrac { 1 }{ 6 }$

In uniform circular motion

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both velocity and speed are constant
0%
speed is constant but velocity changes
0%
both speed and velocity change
0%
velocity is constant but speed changes

CBSE Questions for Class 9 Physics Motion Quiz 5 - MCQExams.com

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Practice Class 9 Physics Quiz Questions and Answers

CBSE Questions for Class 9 Physics Motion Quiz 5 - MCQExams.com

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Practice Class 9 Physics Quiz Questions and Answers

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