MCQ Questions for Class 11 Engineering Physics Motion In A Straight Line Quiz 2

A car initially travelling north at 5 m/s has a constant acceleration of

$2 m/s^2$ northward. How far does the car travel in the first 10 s?

0%
60 m
0%
50 m
0%
100 m
0%
150 m
0%
250 m

Explanation

Given,

Initial velocity, $u=5\,m{{s}^{-1}}$

Acceleration, $a=2\,m{{s}^{-2}}$

Apply kinematic equation of motion

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

$s=5\times 10+\dfrac{1}{2}\times 2\times 10^2=150\,m$

Hence in first 10 second card travel $150\,m$

A ball with a mass of 0.5 Kg is thrown vertically upward with a speed of 15 m/s. What are its speed and direction two seconds later ?

0%
10 m/s upwards
0%
5 m/s upwards
0%
zero
0%
5 m/s downward
0%
10 m/s downward

Explanation

Given,

Initial velocity, $u=15m{{s}^{-1}}$

Acceleration, $a=-10\,m{{s}^{-2}}$

Apply kinematic equation of motion

$v=u+at$

$v=15-10\times 2=-5\,m{{s}^{-1}}$

Hence, velocity after $2\,\sec$ is $5\,m{{s}^{-1}}$ downward.

An object is thrown from height of h, its covers height h in time T. After time T/2 where will be the object.

0%
At height h/2 m from ground
0%
At height h/4 m from ground
0%
At height 3h/2 m from ground
0%
At height 3h/4 m from ground

Explanation

Given,

Object is free fall from height $h$

Initial velocity, $u=0$

Apply kinematic equation of motion

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

$h=\dfrac{1}{2}g{{t}^{2}}\,\,........\,(1)$

At half time

$s=u\dfrac{t}{2}+\dfrac{1}{2}g{{\left( \dfrac{t}{2} \right)}^{2}}=\dfrac{1}{4}\left( \dfrac{1}{2}g{{t}^{2}} \right)=\dfrac{h}{4}$

Height from ground, $h-\dfrac{h}{4}=\dfrac{3h}{4}$ meter

Hence, at half time height from ground is $\dfrac{3h}{4}$ m

A man throws balls into the air one after another. He always throws a ball when the previous one thrown has just reached the highest point. The height to which each ball rises, if he throws 5 balls per second is

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

0%
0.31 m
0%
0.20 m
0%
0.42 m
0%
0.53 m

Explanation

suppose $n - balls \to$

$final\,velocity\,,\,$ $v = 0$

$u = 0$

$T = {\dfrac{1} {n}}\sec$

$0 = u - g{\dfrac{1}{n}}$

${\dfrac{g} {n}} = u$

${v^2} = {u^2} - 2gh$

$0 = {\dfrac{{g^2}} {{n^2}}} - 2gh$

${\dfrac{{g^2}} {{n^2}}} = 2gh$

$h = {\dfrac{g} {2{n^2}}}$

$n = 5$

$= {\dfrac{10} {2 \times 25}}$

$= {\dfrac{1}{5}}m$

A ball is released from certain height h reaches ground in time T. Where will it be from the ground at time

$\frac { 3T }{ 4 } ?$

0%
$\frac { 9h }{ 16 }$
0%
$\frac { 7h }{ 16 }$
0%
$\frac { 3h }{ 4 }$
0%
$\frac { 27h }{ 64 }$

Explanation

Given that,

Height $=h$

Time $t=\dfrac{3T}{4}$

Since, the ball is released from rest, its initial velocity is zero. When released from height h it reaches the ground in time T.

If we apply the second equation of motion

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

$h=\dfrac{g{{T}^{2}}}{2}$

${{T}^{2}}=\dfrac{2h}{g}$

Now, we apply the same equation to calculate the height covered in time 3T/4

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

$s=g\times \dfrac{{{\left( \dfrac{3T}{4} \right)}^{2}}}{2}$

$s=\left( \dfrac{9g}{32} \right)\times {{T}^{2}}$

$s=\left( \dfrac{9g}{32} \right)\times \dfrac{2h}{g}$

$s=\dfrac{9h}{16}$

Now, the distance from the ground

$s=h-\dfrac{9h}{16}$

$s=\dfrac{7h}{16}$

Hence, the distance from the ground is $\dfrac{7h}{16}$

A man slides down a snow covered hill along a curved path and falls

$20$ m below his initial position. The velocity in m/sec with which he finally strikes the ground is?

$(g=10 m/sec^2)$ .

0%
$20$
0%
$400$
0%
$200$
0%
$40$

Explanation

The equation of motion is

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

${v }^{ 2 }=0+2gh$

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

$v=\sqrt { 2\times 10\times 20 }$

$v=20\dfrac { m }{ s }$

If we use plus and minus signs to indicate the direction of velocity and acceleration in one dimension, in which of the following situation does the object speed up?

0%
Negative velocity and negative acceleration
0%
Positive velocity and negative acceleration
0%
Positive velocity and zero acceleration
0%
Negative velocity and positive acceleration
0%
none of the above.

Explanation

$\vec{v}=\vec{u}+\vec{a}t$

$or\,\,\vec{v}=-\vec{u}-\vec{a}t$

Direction of initial velocity and direction of acceleration should be same, either negative or positive

Hence, at negative velocity and negative acceleration object will speed up

An object is thrown from the height of 125 cm take g = 10 m/s. Find the ratio of distance covered by object in the 1 st and last 1 sec

0%
1:9
0%
4:9
0%
4:4
0%
2:9

Explanation

Given,

Initial velocity, $u=0$

Time period of free fall from height $125\,m$

$t=\sqrt{\dfrac{2h}{g}}=\sqrt{\dfrac{2\times 125}{10}}=5\,\sec$

Apply kinematic equation of motion

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

Ratio of displacement for ${{1}^{st}}$ sec and last $1\,\sec$

$\dfrac{{{s}_{1}}-{{s}_{0}}}{{{s}_{4}}-{{s}_{5}}}=\dfrac{\dfrac{1}{2}g\left( {{1}^{2}}+0 \right)}{\dfrac{1}{2}g\left( {{5}^{2}}+{{4}^{2}} \right)}=\dfrac{1}{9}$

Hence , ratio is $1:9$

The rate of change of velocity is:

0%
Force
0%
Momentum
0%
Acceleration
0%
Displacement

Two cars are travelling towards each other on a straight road at velocities

$15\ m/s$ and

$16\ m/s$ respectively. When they are

$150\ m$ apart, both the drivers apply the brakes and the cars decelerate at

$3\ m/s^{2}$ and

$4\ m/s^{2}$ until they stop. Separation between the cars when they come to rest is :

0%
$86.5 m$
0%
$89.5 m$
0%
$85.5 m$
0%
$80.5 m$

Explanation

We know,

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

Given,

Initial velocity of car 1,

$u_1=15\ m/s$ , Acceleration,

$a=3\ m/s^2$

Initial velocity of car 2,

$u_2=16\ m/s$ , Acceleration,

$a=4\ m/s^2$

For car 1,

Final Velocity,

$v_1 = 0\ m/s$

$-15^2 = -2(3) \times s$

$\therefore s_1= 37.5$ m

For car 2,

Final Velocity,

$v_2 = 0\ m/s$

$-16^2 = -2(4) \times s$

$\therefore s_2= 32$ m

Total displacement,

$s_1+s_2 = 37.5+32 = 69.5$ m

$\therefore$ separation between them,

$= 150-69.5 =80.5$ m

A body leaving a certain point

$O$ moves with an acceleration which is constant. At the end of the first second after it left

$O$ , its velocity is

$1.5m/\sec$ . At the end of the sixth second after it left

$O$ the body stops momentarily and begins to move backwards. The distance traveled by the body before it stops momentarily is

0%
$9 m$
0%
$18 m$
0%
$54 m$
0%
$36 m$

Explanation

$v=u+at$

$1.5=u+5a...........(i)$

$0=u+6a.............(ii)$

From (i) and (ii)

$0=1.5+a(6-5)$

$a=-1.5m/s^2$

Om substituting

$a=-1.5m/s^2$ in equation (ii)

$0=u+6(-1.5)$

$0=u-9$

$u=9m/s$

The distance traveled by the body in its forward journey is

$s_1$ can be calculated as follows,

$s_1=ut+\frac{1}{2}at^2$

$s_1=9\times6+\frac{1}{2}\times(-1.5)\times6^2$

$s_1=54+(-1.5\times18)$

$s_1=27m$

The distance traveled by the body in its forward journey

$(s_1)$ = the distance traveled by the body in its backward journey

$(s_2)=27m$

So, the total distance traveled by the body in forward journey and its backward journey

$=s_1+s_2=27+27=54m$

The average velocity of a freely falling body is numerically equal to half of the acceleration due to gravity. The velocity of the body as it reaches the ground is numerically equal to:

0%
$g$
0%
$\dfrac{g}{2}$
0%
$\dfrac{g}{\sqrt{2}}$
0%
$\sqrt{2}g$

Explanation

We have average velocity

$= \dfrac{g}{2}$
Since it is a free fall, average velocity,

$v=\dfrac{S}{t}= \dfrac{\dfrac{1}{2}gt^2}{t} = \dfrac{g}{2}$ , where

$t$ is the time of descent and initial velocity is zero.

We know that for a free fall, velocity with which it reaches the ground:

$V = gt$ , where

$t$ is time of descent .

Substituting the same in the above equation for average velocity:

$\dfrac{V}{2} = \dfrac{g}{2}$

$\therefore V=g$

A driver takes

$0.20s$ to apply the brakes after he sees a need for it. This is called the reaction time of the driver. If he is driving a car at a speed of

$54 km/h$ and the brakes causes a deceleration of

$6.0 \ { m }/{ { s }^{ 2 } }$ , find the distance traveled by the car after he sees the need to put the brakes on.

0%
$18.63 m$
0%
$20 m$
0%
$26.85 m$
0%
$27.67 m$

For a body moving with uniform acceleration

$a$ , initial and final velocities in a time interval

$t$ are

$u$ and

$v$ respectively. Then, its average velocity in the time interval

$t$ is :

0%
$(v+\dfrac{at}{2})$
0%
$(v-\dfrac{at}{2})$
0%
$(v-at)$
0%
$(u+\dfrac{at}{2})$

Explanation

$\textbf{Hint}$ : Use the equations of motion

$\textbf{Step 1}$ : Given a body is moving with uniform acceleration 'a', initial velocity is u and final velocity is v, time is t.

We have to find the average velocity.

From the equations of motion we know

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

and

$v=u+at$

If an object velocity is increasing at a constant rate then it is called as uniform acceleration.

$\textbf{Step 2}$ :

We know Average velocity can be calculated using the formula

$Average \quad velocity= \dfrac{Total \quad displacement}{Total \quad Time}=\dfrac{ut+\dfrac{1}{2}at^{2}}{t}$

$Average \quad velocity=u+\dfrac{1}{2}at$

From equations of motion

$v=u+at$

Using this equation we can write

$u=v-at$

Substitute

$u=v-at$ in the above equation obtained for Average velocity

we get

$Average \quad velocity=v-at+\dfrac{1}{2}at=v-\dfrac{1}{2}at$

So both option B and D are correct.

A ball dropped from one metre above the top of a window, crosses the window in

$t_{1}\;s$ . If the same ball is dropped from

$2\;m$ above the top of the same window, time taken by it to cross the window is

$t_{2}\; s$ . Then

0%
$t_{2} = t_{1}$
0%
$t_{2} = 2t_{1}$
0%
$t_{2} > t_{1}$
0%
$t_{2} < t_{1}$

Explanation

Solution 1

The ball which is dropped from greater height above the top of the window will have higher speed while crossing the window, by 3rd equation of motion:

$v^2=u^2+2aS$ , where

$u=0$

The ball with higher speed will take less time to cross the window.

$\because$

$H_2 > H_1$

$\implies$

$v_2 > v_1$

$\therefore$

$t_2 < t_1$

Hence Option D is correct

Solution 2

Statement D is correct
Situation 1:
We want time to cross x.

$v^{2}=u^{2}+2as$

$u=0, a=-g, s=-1$

$v^{2}=2g\Rightarrow v_{1}=\sqrt{2g}$

$(1+x)=\dfrac{1}{2} gt^{2}\Rightarrow T_{2}=\sqrt{\dfrac{2(1+x)}{g}}$

$1=\dfrac{1}{2}gt^{2}=T_{1}=\sqrt{\dfrac{2}{g}}$
So time taken to cross

$x=t_{1}=T_{2}-T_{1} =\sqrt{\dfrac{2}{g}}(\sqrt{1+x}-1)$
Situation 2:

$(2+x)=\dfrac{1}{2}gt^{2}\Rightarrow T_{2}=\sqrt{\dfrac{2(2+x)}{g}}$

$2=\dfrac{1}{2}gt^{2}\Rightarrow T_{1}=\sqrt{\dfrac{2\times 2}{g}}$
So time taken to cross

$x=t_{2}=T_{2}-T_{1}=\sqrt{\dfrac{2}{g}}(\sqrt{2+x}-\sqrt{2})$
Clearly

$t_{2}< t_{1}$

A train accelerates from rest at a constant rate

$a_{1}$ for distance

$S_{1}$ and time

$t_{1}$ . After that it decelerates to rest at a constant rate

$a_{2}$ for distance

$S_{2}$ at time

$t_{2}$ . Then the correct relation among the following is

0%
$\displaystyle \frac{S_{1}}{S_{2}} = \frac{a_{1}}{a_{2}} = \frac{t_{1}}{t_{2}}$
0%
$\displaystyle \frac{S_{1}}{S_{2}} = \frac{a_{2}}{a_{1}} = \frac{t_{1}}{t_{2}}$
0%
$\displaystyle \frac{S_{1}}{S_{2}} = \frac{a_{1}}{a_{2}} = \frac{t_{2}}{t_{1}}$
0%
$\displaystyle \frac{S_{1}}{S_{2}} = \frac{a_{2}}{a_{1}} = \frac{t_{2}}{t_{1}}$

Explanation

Given that the particle was at rest at

$t\ =\ 0$

$s_1 = u_1t + \frac {1}{2} a_1t_1^2 .... (1)$

$\Rightarrow v_1 = u_1 + a_1t_1 .... (2)$

$v_1 \rightarrow$ final speed after

$t_1$

Now, final speed after this is

$v_2$
And

$v_1 = v_2$

$\Rightarrow v_2 = u_2 - a_2t_2$

$\Rightarrow v_1 = a_2t_2 .... (3)$

$\Rightarrow v_2^2 = u_2^2 - 2a_2s_2$

$\Rightarrow v_1^2 = 2a_2s_2$

From (2),

$a_1^2t_1^2 = 2a_2s_2$

$\Rightarrow s_2 = \dfrac {a_1^2t_1^2}{2a_2}$

$\Rightarrow s_2 = \dfrac {a_1 (2s_a)}{2\frac {a}{2}a_2}$

$\Rightarrow \dfrac {eq^2 (2)}{eq^n (3)} = \dfrac {a_1t_1}{a_2t_2} = 1$

$\Rightarrow \dfrac {a_1}{a_2} = \dfrac {t_2}{t_1}$

$\Rightarrow \dfrac {a_1}{a_2} = \dfrac {t_2}{t_1} = \dfrac {s_2}{s_1}$

A proton in a uniform electric field moves along a straight line with constant acceleration starting from rest. If it attains a velocity

$4\times 10^3$

$km/s$ at a distance of

$2\;cm$ , the time required to reach the given velocity is :

0%
$10^{-3}\;s$
0%
$10^{-6}\;s$
0%
$10^{-8}\;s$
0%
$10^{-5}\;s$

Explanation

$v^2 = u^2 + 2as$

$u = 0, s = 2 \times 10^{-2} mt$

$a = \dfrac {v^2}{2s} = \dfrac {(4 \times 10^3 \times 10^3)^2}{2 \times 2 \times 10^{-2}} m/s^2$

$a = 4 \times 10^{14}$

Now,

$v = u + at$

$\dfrac {v}{a} = t$

$t = \dfrac{4 \times 10^6}{4 \times 10^{14}} = 10^{-8}s$

From a building two balls A & B are thrown such that A is thrown upwards and B downwards (both vertically with same speed). If

$V_{A}$ and

$V _{B}$ are their respective velocities on reaching the ground then :

0%
$V_{B} < V_{A}$
0%
$V_{A} = V_{B}$
0%
$V_{A} < V_{B}$
0%
Their velocities depends on their masses

Explanation

For A:

$s_A = vt_1 - \dfrac {1}{2} gt_1^2$

$s_B = -vt_2 - \dfrac {1}{2} gt_2^2$

We have,

$s_A = s_B$

So, on solving we get:

$t_2 - t_1 = \dfrac {2v}{g}$ ...........(1)

Also,

$v_A = u_A - gt_1$

$v_B = u_B - gt_2$

Using

$(1)$ , we get:

$\dfrac {v_A}{v_B} = \dfrac {v - gt_1}{-v -gt_2}$

$= \dfrac {v - gt_1}{-v -g (t_1 + \dfrac {2v}{g})} = 1$

$\Rightarrow v_A = v_B$

It takes one minute for a passenger standing on an escalator to reach the top. If the escalator does not move it takes him 3 minutes to walk up. How long will it take for the passenger to arrive at the top if he walks up the moving escalator?

0%
$30\ sec$
0%
$45\ sec$
0%
$40\ sec$
0%
$35\ sec$

Explanation

Let the distance to the top be d.

therefore the speed of the man with escalator not moving

$=(\dfrac{d}{1}) m/min$

therefore the speed of the escalator with the man at rest

$=(\dfrac{d}{3})$ m/min

when both the main and the escalator are moving, effective speed

$= (\dfrac{d}{3} +d)$ m/min

$= \dfrac{4d}{3}$ m/min

Therefore time taken when both men and escalator are moving

$= \dfrac{distance}{speed} = \dfrac{d}{\dfrac{4d}{3}}$

$\dfrac{3}{4} min = 45$ seconds

Hence (B) is correct answer

From an elevated point P, a stone is projected vertically upwards. When it reaches a distance

$d$ below P, its velocity is doubled. The greatest height reached by it above P is :

0%
$d/3$
0%
$3d$
0%
$2d$
0%
$d/2$

Explanation

$v -u =gt$

$2u-u =gt$

$\Rightarrow t = \dfrac{u}{g}$ where, t is the time to go from u (downward) to 2u (downward).

$t_{total} = \dfrac{2u}{g} + \dfrac{u}{g}= \dfrac{3u}{g}$ since

$\dfrac{u}{g}$ is time to go up and same time to come down.

$(2u)^2 - u^2 = 2g d$

$\Rightarrow d = \dfrac{3u^2}{2g}$
Maximum height,

$H = \dfrac {u^2}{2g} = \dfrac {d}{3}$

A bus starts from rest and moves with a uniform acceleration of

$1\;ms^{-2}$ . A boy

$10\;m$ behind the bus at the start runs at a constant speed and catches the bus in

$10\;s$ . Speed of the boy is :

0%
$10 ms^{-1}$
0%
$1 ms^{-1}$
0%
$6 ms^{-1}$
0%
$4 ms^{-1}$

Explanation

Solution 2

Distance travelled by bus in $10s \Rightarrow \dfrac {1}{2} \times 1 \times 10^2 = 50 m$
D $=$ Distance boy has to cover $= 10 + 50 = 60 m$
Speed of boy $=\dfrac {D}{T} = \dfrac {60}{10} = 6\ m/s$

A ball dropped from a height of

$10$ m, rebounds to a height of

$2.5$ m. If the ball is in contact with the floor for

$0.01$ second, its acceleration during contact is (

$g = 9.8$

$ms^{-2}$ )

0%
$20$ $ms^{-2}$
0%
$21$ $ms^{-2}$
0%
$210$ $ms^{-2}$
0%
$2100$ $ms^{-2}$

Explanation

When ball is hitting the ground,
It's velocity,

$v$ is

$= \sqrt {2gh} = \sqrt {2 \times 9.8 \times 10}$
After rebounding up its velocity be

$u$ .

$u^2 = 2gh \Rightarrow \sqrt {2 \times g \times 2.5}$
So,

$v = u + at$

$\Rightarrow \sqrt {2 \times 9.8 \times 10} = - \sqrt {2 \times g \times 2.5} + a (0.01)$

$\Rightarrow \dfrac {\sqrt {2 \times 9.8}}{0.01} (\sqrt {10} + \sqrt {2.5}) = a$

$\Rightarrow a = 2100\ m/s^2$

A car moving with a constant acceleration covers the distance between two points

$180\;m$ apart in

$6\;s$ . If its speed as it passes the second point is

$45\;ms^{-1}$ , its speed at the first point is

0%
10 $ms^{-1}$
0%
15 $ms^{-1}$
0%
30 $ms^{-1}$
0%
45 $ms^{-1}$

Explanation

$180 = u (6) + \frac {1}{2} a (6)^2$

$180 = 6u + 18a$

$30 = u +3a .................. (1)$

$45 = u + at$

$45 = u + a (6) ................. (2)$

$eq^n (2) - eq^n (1)$

$\Rightarrow 15 = 3a \Rightarrow a = 5$

$u = 15 m/s$

A sharp stone of mass

$2$ kg falls from a height of

$10$ m on sand and buries into the sand. It comes to rest in a time of

$0.029$ second. The depth through which it buries into sand is

0%
0.2 m
0%
0.15 m
0%
0.25 m
0%
0.30 m

Explanation

Velocity attained just before contact:

$v = \sqrt {2gh} = \sqrt {2 \times 9.8 \times 10} = \sqrt {196} = 14\ m/s$
$u_1 = v$
$v_1 = u_1 + at$
$0 = 14 + a (0.029)$
$a = -(\dfrac {14}{0.029})$

$s = ut + \dfrac {1}{2} at^2 = 14 (0.029) + \dfrac {1}{2} (\dfrac {-14}{0.029})(0.029)^2$
$s = 0.029 (14 + \dfrac {1}{2} (-14))$

$= 0.029 \times 7$
$= 0.203\ m$
$= 0.2\ m$

A bus is moving along a straight road with a uniform acceleration. It passes through two points A and B separated by a certain distance with velocity of

$30$ kmph and

$40$ kmph respectively.Velocity of the car exactly midway between A and B is

0%
38.3 kmph
0%
35.35 kmph
0%
33.3 kmph
0%
None

Explanation

$v = u + at$

$40 = 30 + at$

$at = 10$

$v^2 = u^2 + 2as$

$40^2 = 30^2 + 2as$

$2as = 700$

Now at midway

$s^1 = \dfrac {s}{2}$

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

$v^2 = 30^2 + \dfrac {700}{2} = 900 + 350$

$v^2 = 1250$

$v = \sqrt {1250}=35.35\ kmph$

A body travels

$200\;cm$ in the first two seconds with cosntant speed and

$220\;cm$ in the next

$4$ seconds with constant deceleration. The speed of the body at the end of the

$7^{th}$ second is

0%
$22.5 cm/s$
0%
$12.5 cm/s$
0%
$57.5 cm/s$
0%
$0 cm/s$

Explanation

Let the initial velocity of the body be

$U$ .

In the first

$2$ seconds,

$200 = 2U \implies U = 100\ cm/s$
In the next

$4$ seconds,

$220 = 4U - \dfrac{1}{2} 16a = 4U - 8a = 4(100) - 8a \rightarrow a= 22.5\ cm/s^2$

Velocity of the body at the end of

$7^{th}$ second is

$V= U - at = 100 - ( 22.5) ( 7) = - 57.5\ cm/s$

While moving with uniform acceleration, a body has covered

$550 \;m$ in

$10$ second and attained a velocity of

$105 \;ms^{-1}$ . Its initial velocity

$u$ and acceleration

$a$ respectively are

0%
$10 ms^{-1}, 5 ms^{-2}$
0%
$10 ms^{-1}, -5 ms^{-2}$
0%
$5 ms^{-1}, 10 ms^{-2}$
0%
$10 ms^{-1}, 0 ms^{-2}$

Explanation

$v = u +at$

$105 = u + 10a \ \ \ ..... (1)$

$550 = u (10) + \dfrac {1}{2} \times a \times 10^2$

$\Rightarrow 550 = 10u + 50 a \Rightarrow 55 = u + 5a \ \ \ ..... (2)$

$eq^n (1) - eq^n (2)$

$\Rightarrow 50 = 5a$

$a = 10\ m/s^2$

$u = 5\ m/s$

The average velocity of a body moving with uniform acceleration after travelling a distance of

$3.06\ m$ is

$0.34\ {m/s}$ . The change in velocity of the body is

$0.18\ {m/s}$ . During this time, its acceleration is

0%
$0.01\ {m/s}^{2}$
0%
$0.02\ {m/s}^{2}$
0%
$0.03\ {m/s}^{2}$
0%
$0.04\ {m/s}^{2}$

Explanation

$V_{avg} = \dfrac {D}{t} \Rightarrow 0.34 = \dfrac {3.06}{t}$

$\Rightarrow t = \dfrac {3.06}{0.34}$ =

$9\ sec$

We have,

$v = u + at$

$v - u = at$

$0.18 = a \times 9$

$\Rightarrow a = \dfrac {0.18}{9} = 0.02 \ m/s^2$

A body travels

$200\;m$ in the first two second and

$220\;m$ in the next four second. The velocity at the end of the seventh second from the start will be

0%
10 $ms^{-1}$
0%
15 $ms^{-1}$
0%
220 $ms^{-1}$
0%
5 $ms^{-1}$

Explanation

$200 = u(2) + \dfrac {1}{2} \times a (2)^2 = 2u + 2a = 200$

$\Rightarrow u + a =100 - (1)$

Distance in next

$4\ sec$

$\Rightarrow s_6 - s_2$

$\Rightarrow 220 = (u (6)] + \dfrac {1}{2} a (6)^2) - (u (2) + \dfrac {1}{2} a (2)^2)$

$420 = 6u + 18a - (2)$

$420 = 6 (100 - a) + 18a (from 1)$

$\Rightarrow -180 = 12a \Rightarrow a = - 15 m/s^2 , u = 115 m/s$

$v\ (at \;end \;of \;7 \;sec) = u + at$

$= 115 + (-15 \times 7) = 10 m/s$

A ball after having fallen from rest under the influence of gravity for

$6s$ , crashes through a horizontal glass plate, thereby losing two-third of its velocity. Then it reaches the ground in

$2s$ , height of the plate above the ground is

0%
$19.6m$
0%
$39.2m$
0%
$58.8m$
0%
$78.4m$

Explanation

$v_B = v_A + gt = 0 + g(6) = 6g$
Now,

$\dfrac {2}{3}^{rd}$ of the velocity is lost.
Hence,

$\dfrac {1}{3} \times 6g = 2g$ velocity is left.

$h = ut + \dfrac {1}{2} gt = (2g) (2) + \dfrac {1}{2} g (2)^2 = 6g$

$= 58.8\ m$

A ball is thrown straight upward with a speed

$v$ from a point

$h$ meters above the ground. The time taken for the ball to strike the ground is

0%
$\dfrac{v}{g}\left [1+\sqrt{1+\dfrac{2hg}{v^{2}}} \right ]$
0%
$\dfrac{v}{g}\left [1-\sqrt{1-\dfrac{2hg}{v^{2}}} \right ]$
0%
$\dfrac{v}{g}\left [1-\sqrt{1+\dfrac{2hg}{v^{2}}} \right ]$
0%
$\dfrac{v}{g}\left [{2+\dfrac{2hg}{v^{2}}} \right ]$

Explanation

$s = vt - \dfrac {1}{2} gt^2$
Now, at

$C$ (maximum height point),

$h^` = \dfrac {v^2}{2g}$
So, for journey from

$C$ to

$B$ ,
time

$= \sqrt {\dfrac {2(h+h^`)}{g}} = \sqrt {\dfrac {2(h + \dfrac {v^2}{2g})}{g}} = t_2$

$\Rightarrow$ total time

$= t_1 + t_2 = \dfrac {v}{g} (1 + \sqrt {1 + \dfrac {2hg}{v^2}})$

A balloon starts rising from the ground with an acceleration of

$1.25 \;ms^{-2}$ , After

$8$ seconds, a stone is released from the balloon, The stone will

$(g=10\;ms^{-2})$

0%
cover a distance of 40 m in reaching the ground.
0%
have a displaceent of 50 m
0%
reach the ground in 4 second
0%
begin to move down after being released

Explanation

$s = \dfrac {1}{2} at^2$

$a = 1.25\ m/s^2 \Rightarrow s = \dfrac {1}{2} \times 1.25 \times 8^2 = 40\ m$

After

$8\ sec$ speed of balloon:

$v = u - gt$

$v = -(1.25) \times 8 = - 10\ m/s$

Now, when stone is released distance to cover is

$40\ m$ .

$v^2 = u^2 - 2gs$

$v^2 = (100)^2 + 2 \times 10 \times 40 \Rightarrow v = 30 m/s$

$v = u - gt \Rightarrow - 30 = +10 - 10t$

$t = 4\ s$

A ball is dropped from the top of a building. The ball takes

$0.5$ s to fall past the

$3$ m length of a window at certain distance from the top of the building.
(

$g=10\;m/s^{2}$ )
How far is the top of the window from the point at which the ball was dropped ?

0%
$0.5$ m
0%
$0.5225$ m
0%
$0.6125$ m
0%
$0.8$ m

Explanation

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

$3 = u (0.5) + \dfrac {1}{2} \times 10 (0.5)^2$
$3 = (0.5)u + 5(0.25)$

$\dfrac {1.75}{0.5} = u$

$\Rightarrow u = 3.5\ m/s$

$u^2 = u_1^2 + 2gs$

$\dfrac {(3.5)^2}{2g} = s$

$\Rightarrow s = \dfrac {3.5 \times 3.5}{2g}$ $\Rightarrow \ s = 0.6125\ m$

A stone is dropped from the

$16^{th}$ storey of a multistoried building reaches the ground in

$4$ seconds. The no. of storeys travelled by the stone in

$2^{nd}$ second is

0%
4
0%
3
0%
5
0%
2

Explanation

Let height of each storey

$= x$

$16x = ut + \dfrac {1}{2} gt^2 = 0 + \dfrac{1}{2} gt^2$
As

$t=4 s$ ,

$16x = \dfrac{16g}{2}$

$x=\dfrac{g}{2}$

$2^{nd}$ second,

$S_2 - S_1 = \dfrac{1}{2} g( 2^2 -1^2 ) = \dfrac {3g}{2}$
So,

$\dfrac {3g}{2} = 3x$ hence stone traveled 3 storeys.

A ball is dropped from the top of a building. The ball takes

$0.5$ s to fall past the

$3$ m length of a window at certain distance from the top of the building. Time of travel above the window is (Take

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

0%
0.5 s
0%
0.25 s
0%
0.35 s
0%
0.75 s

Explanation

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

$3 = u (0.5) + \dfrac {1}{2} \times 10 (0.5)^2$

$3 = (0.5)u + 5(0.25)$

$\dfrac {1.75}{0.5} = u$

$\Rightarrow u = 3.5\ m/s$

$v_1 = u = 3.5 m/s$

$v_1 = u_1 + gt$

$3.5 = 10 (t) \Rightarrow t = 0.35\ sec$

A ball is dropped from the top of a building. The ball takes

$0.5s$ to fall past the window of

$3m$ in length at certain distance from the top of the building. (

$g=10m/s^{2}$ ) How fast was the ball going as it passed the bottom of the window?

0%
3.5 $ms^{-1}$
0%
8.5 $ms^{-1}$
0%
5 $ms^{-1}$
0%
12 $ms^{-1}$

Explanation

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

$3 = u (0.5) + \dfrac {1}{2} \times 10 (0.5)^2$

$3 = (0.5) u + 5 (0.25)$

$\dfrac {1.75}{0.5} = u$

$\Rightarrow u = 3.5\ m/s$
$v = u + gt$
$v = (3.5) + 10 (0.5) = 8.5\ m/s$

A body falls from a height of

$200$ m. If gravitational attraction ceases after

$2$ s, further time taken by it to reach the ground is (

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

0%
$5\ s$
0%
$9\ s$
0%
$13\ s$
0%
$17\ s$

Explanation

Distance travelled in

$2\ sec$ .

$s = ut + \dfrac {1}{2}gt^2 = \dfrac {1}{2}g(2)^2 = 2g = 20\ mt$

Remaining

$= 180\ m$

$v = u + gt = 0 + 10 \times 2 = 20 m/s$

So, after this velocity will be constant (20 m/s) downwards (since no

$g$ is acting)
Hence, time taken

$= \dfrac {180}{20} = 9\ sec$

A body thrown vertically up with a velocity

$u$ reaches the maximum height

$h$ after

$T$ seconds.Which of the following statements is correct?

0%
At a height $\frac{h}{2}$ from the ground its velocity is $\frac{u}{2}$
0%
At a time T its velocity is $u$
0%
At a time 2T its velocity is $-u$
0%
At a time 2T its velocity is $\frac{u}{2}$

Explanation

At height

$h/2$ ,

$V^2 -u^2 = -2g(\dfrac{h}{2})$

$V^2 = u^2 - gh$

Since,

$0=u^2-2gh$

$V^2 = u^2 - u^2/2$

$V=u/\sqrt 2$

$V = u - gt$

At maximum height,

$V = 0$
At time T, its velocity is 0

$T = \dfrac {u}{g} .................... (1)$
at

$2T, V = u -gt$

$V = u - g(2T) = u - g\times 2 (\dfrac {u}{g})$ (from 1)

$\Rightarrow V = -u$

A juggler throws up balls at regular intervals of time. Each ball takes

$2\ s$ to reach the highest position. If the first ball is in the highest position by the time the fifth one starts, then the separation between the first and the second balls is

0%
1.225m
0%
2.45m
0%
4.9m
0%
3.8m

Explanation

Let the interval between each balls be

$\Delta t$ ,

$\Rightarrow$ difference between

$1^{st}$ and

$5^{th}$ ball

$= 4 \Delta t$

$\Rightarrow 4 \Delta t = 2 \Rightarrow \Delta t = \frac {1}{2} : Also, 4 \Delta t = \dfrac {u}{g}$

$u = 4 \Delta t g - (1)$

So, separation between

$1^{st}$ and

$2^{nd}$ ball,

$s_1 = u (2) - \dfrac {1}{2}g (2)^2$

$s_2 = u (2 - \Delta t) - \dfrac {1}{2} g (2 - \Delta t)^2$

So,

$s_1 - s_2 = u (\Delta t) - \dfrac {1}{2}g (4 \Delta t - \Delta t^2)$
From (1)

$= 4 (\Delta t)^2g - \dfrac {g}{2} (4 \Delta t - \Delta t^2)$

$= 4 (\dfrac {1}{2})^2 g - \dfrac {g}{2} (\dfrac {4}{2} - \dfrac {1}{4}) = g - \dfrac {g}{2} (\dfrac {7}{4})$

$= 1.225\ mt$

A bag is dropped from a helicopter rising vertically at a constant speed of

$2$

$m/s$ . The distance between the two after

$2$ s is (Given

$g=9.8m/s^2$ )

0%
$4.9\ m$
0%
$19.6\ m$
0%
$29.4\ m$
0%
$39.2\ m$

Explanation

for helicopter

$a=0,\ t=2\mathrm{s}\mathrm{e}\mathrm{c},\ u=2m/s,\ h_{1}=ut$
for body

$h_{2}=-ut+\displaystyle \frac{1}{2}gt^{2}$
distances between them

$=h_{1}-h_{2}$

So,

Distance travelled by pocket in

$2$ sec,

$s_1 = 2(2) - \dfrac {1}{2}g(2)^2 = 4 - 5(2)^2 = - 15.6\ m$

Distance travelled by helicopter in

$2$ sec

$s_2 = 2(2) = 4\ m$

Hence, distance between the two

$\Rightarrow s_2 - s_1 = 4 - (-15.6)$

$= 19.6\ m$

A body throws balls vertically upwards. He throws one, while the previous one is at its highest point. Maximum height reached by a ball if he throws one ball each per second at uniform speed is

0%
$19.6m$
0%
$9.8m$
0%
$4.9m$
0%
$2.45m$

Explanation

Since, he throws one ball each per second, so time taken by ball to reach the maximum height

$= 1 \ sec$

$\Rightarrow h = \dfrac {u^2}{2g}$

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

At,

$t = 1\ sec$

$v = u - gt = u - \dfrac {g}{2} = 9 - \dfrac {g}{2}$

$t = 1\ sec \Rightarrow u = g - (1) = \dfrac {g}{2} = 4.9\ m$

A body thrown up with a velocity of 98 m/s reaches a point P in its path 7 second after projection. Since its projection it comes back to the same position after:

0%
13s
0%
14s
0%
6s
0%
22s

Explanation

Here,

$u=98m/s$ ,

${t}_{1}=7s$ . When body reaches maximum height, its velocity becomes

${v}_{1}=0$ .
Maximum height reached can be obtained using:

${v}^{2}={u}^{2}+2ah$

$h_{max}=\dfrac{{u}^{2}}{2g}$
Now considering initial velocity as

${v}_{1}=0$ , displacement as

$s=\dfrac{{u}^{2}}{2g}$ ,

We can find time for the body to come back from the maximum height using:

$s=ut+\dfrac { 1 }{ 2 } a{ t }^{ 2 }\\ \dfrac { { u }^{ 2 } }{ 2g } =\dfrac { 1 }{ 2 } g{ t }^{ 2 }\\ t=\dfrac { u }{ g } \\$
Thus, total time,

$t=\dfrac{2u}{g}=20\ s$
Now the time it takes to reach point P from ground and to reach from point P to ground is same i.e.

$7s$ each time.

Thus, it takes

$20s - 7s = 13s$ to come back to the same point P since its projection.

A particle is projected vertically up and another is let fall to meet at the same instant. If they have velocities equal in magnitude when they meet, the distances travelled by them are in the ratio

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

Explanation

$\text{Let}\ A\ \& B\ \text{meet at P.}$

$\text{For}\ A \Rightarrow v = gt .................. (1)$

$\text{For}\ B \Rightarrow v^{'}= u - gt ....................(2)$

$\text{The magnitude of } v\ \text{and}\ v^{'}\ \text{to be same}$

$|v| = |v^{'}|$

$\Rightarrow u - gt = gt$

$u = 2gt \Rightarrow t = \dfrac {u}{2g}$

$\text{So, dist. travelled by B:}$

$\Rightarrow S = ut - \dfrac {1}{2}gt^2$

$S= u\left(\dfrac {u}{2g}\right) - \dfrac {1}{2}g\left(\dfrac {u}{2g}\right)^2$

$S=\dfrac{u^2}{2g}-\dfrac{u^2}{8g}= \dfrac {3}{8}\left(\dfrac {u^2}{g}\right)$

$\text{Dist. travelled by}\ B\Rightarrow S^{'} = \dfrac {1}{2}gt^2 = \dfrac {1}{2}g\left(\dfrac {u}{2g}\right)^2 = \dfrac {u^2}{8g}$
So,

$\dfrac {S}{S^{'}} = \dfrac {3}{1}$

A ball of mass

$100$ gm is projected vertically upwards from the ground with a velocity of

$49$

$m/s$ . At the same time another identical ball is dropped from a height of

$98$ m. After some time the two bodies collide. When they collide, their velocities are (Given

$g=9.8\; m/s^2$ )

0%
29.4 m/s upwards, 29.4 m/s downwards
0%
29.4 m/s upwards, 19.6 m/s downwards
0%
19.6 m/s upwards, 19.6 m/s downwards
0%
None

Explanation

Equations to use :

time of meet

$t=\frac{h}{u}$

$v_{1}=gt$ ( for freely falling body)

$v_{2}=u-gt$ ( for body projected up)

For A:

$s_A = 49t - \dfrac {1}{2}gt^2 - (1)$
For B:

$s_B = \dfrac {1}{2}gt^2 - (2)$

$s_A + s_B = 98 = 49t \ \ \ \ (from (1) \& (2))$

$\Rightarrow t = 2 sec$

So,

$v_A = u_A - gt = 49 - 9.8 \times 2 = 29.4\ m/s$ upwards

$v_B = gt = 9.8 \times 2 = 19.6\ m/s$ downwards

A body projected up with a speed

$u$ took

$T$ seconds to reach the maximum height

$H$ . Pick out the correct statement

0%
It reaches $\dfrac{H}{2}$ in $\dfrac {T}{2}$ s
0%
It acquires velocity $\dfrac{u}{2}$ in $\dfrac{T}{2}$ s
0%
Its velocity is $\dfrac{u}{2}$ at $\dfrac{H}{2}$
0%
Same velocity at $2T$ s

Explanation

$v = u - gT$

$\Rightarrow T = \dfrac {u}{g} \ \ \ .... (1)$

$H = uT - \dfrac {1}{2}gT^2$
In

$\dfrac {T}{2} sec \Rightarrow h^1 = u (\dfrac {T}{2}) - \dfrac {1}{2}g \dfrac {T^2}{4}$

Hence

$h^1 \neq \dfrac {H}{2}$

$v = u -gt$
At,

$t = \dfrac {T}{2}$

$v = u - g\dfrac {T}{2} = u - \dfrac {u}{2} = \dfrac {u}{2} \ \ \ (from (1) T = \dfrac {u}{g})$

$v^2 = u^2 - 2gH$

$\dfrac {H}{2}, v^2 = u^2 - 2g\dfrac {H}{2} = u^2 - gH \neq \dfrac {u^2}{4} \Rightarrow v \neq \dfrac {u}{2}$

Hence, option

$B$ is correct.

From the top of a tower

$36$ m high, a body is dropped and at the same time another body is projected vertically upward from the ground. If they meet midway, find the initial velocity of the projected body and its velocity when the two bodies come together

0%
$5\sqrt{6}m, 0\ m/s$
0%
$\dfrac{42}{\sqrt{5}}m,0\ m/s$
0%
$6\sqrt{6}m , 2\ m/s$
0%
$8\sqrt{6}m, 4.5\ m/s$

Explanation

Equation to use :

for first body

$x=\displaystyle \frac{1}{2}gt^{2}$
for second body

$(h-x)=ut-\displaystyle \frac{1}{2}gt^{2}$

Now solving,

For A

$\Rightarrow 18 = \dfrac {1}{2}gt^2$

$\sqrt {\dfrac {2 \times 18}{g}} = t_1$

For B

$\Rightarrow 18 = ut - \dfrac {1}{2}gt^2 (t = t_1)$

$\Rightarrow 18 = u \sqrt {\dfrac {36}{g}} - \dfrac {g}{2} \times \dfrac {36}{g} \Rightarrow 36 = u \sqrt {\dfrac {36}{g}}$

$\Rightarrow u = 18.78 = \dfrac {42}{\sqrt 5} m/s$

When they come in contact:
Velocity of B is,

$v_B = u - gt$

$= \dfrac {42}{55} - 9.8 \times 1.916$

$v_B = 0\ m/s$

A rocket is fired and ascends with constant vertical acceleration of

$10m/s^{2}$ for 1 minute. Its fuel is exhausted and it continues as a free particle.The maximum altitude reached is (

$g=10m/s^{2}$ )

0%
18 km
0%
36 km
0%
72 km
0%
108km

Explanation

For

$1$ minute, distance,

$d$ travelled is:

$d = ht^0 + \dfrac {1}{2} at^2 = \dfrac {1}{2} \times 10 \times (60)^2 = \dfrac {36000}{2} = 18000\ mt$

$v = u + at = 10 \times 60 = 600\ m/s$

Now, for max

$m$ height:

$v_1^2 = 4_1^2 - 2gs (v_1 = v)$

$\Rightarrow s = \dfrac {(600)^2}{2g} = \dfrac {600 \times 600}{2 \times 10} = 18000 \ mt$

$\Rightarrow$ Total attitude

$= d + s = 18000 + 18000$

$= 36000\ m$

$= 36\ Km$

A stone projected vertically up from the ground reaches a height

$y$ in its path at

$t_{1}$ seconds and after further

$t_{2}$ seconds reaches the ground. The height

$y$ is equal to

0%
$\frac{1}{2}g(t_{1}+t_{2})$
0%
$\frac{1}{2}g(t_{1}+t_{2})^{2}$
0%
$\frac{1}{2}gt_{1}t_{2}$
0%
$g t_{1}t_{2}$

Explanation

$y = ut_1 - \dfrac {1}{2}gt_1^2$

$\dfrac {t_1 + t_2}{\Downarrow} = \dfrac {2u}{g}$
(total time of journey)
So,

$u = g\dfrac {(t_1 + t_2)}{2}$

$\Rightarrow y = g\dfrac {(t_1 + t_2)}{2}t_1 - \dfrac {1}{2}gt_1^2 = \dfrac {1}{2}gt_1t_2$

A paper weight is dropped from the roof of a block of multistorey flats, each storey being

$3$ meters high. It passes the ceiling of the

$20^{th}$ storey at

$30$

$m/s$ . If

$g = 10 m/ s^{2}$ , how many storey does the flat have?

0%
25
0%
30
0%
35
0%
40

Explanation

$v^2 = u^2 + 2gs$

$\Rightarrow (30)^2 = 0^2 + 2 \times 10 \times s$

$s = \dfrac {900}{20} = 45\ m$

$\Rightarrow$ It has travelled

$\dfrac {45}{3} = 15$ storeys
And it is

$21^{st}$ storey from bottom.

Hence, total storeys

$= 20 + 15 = 35$ storeys.

A body is thrown vertically up to reach its maximum height in

$t$ seconds. The total time from the time of projection to reach a point at half of its maximum height while returning ( in seconds ) is

0%
$\sqrt{2}t$
0%
$\left [ 1+\dfrac{1}{\sqrt{2}} \right ]t$
0%
$\dfrac{3t}{2}$
0%
$\dfrac{t}{\sqrt{2}}$

Explanation

$max^m$ height

$v =o$ ,

$v = u - gt, \Rightarrow t = \dfrac {u}{g} = \sqrt {\dfrac {2H}{g}} - (1)$
Now situation is,
We want time to reach C.
Time from

$A to B = t$
Time from

$B to C \Rightarrow v_c^2 = v_B^2 + 2g \dfrac {H}{2}$

$\dfrac {H}{2} = + \dfrac {1}{2} gt_1^2$

$v_C = \sqrt {gH}$

$t_1 = \sqrt {\dfrac {H}{g}} \Rightarrow t_1 = \dfrac {t}{\sqrt 2} (from (1))$
So, total time

$= t + t_1 = t + \dfrac {t}{\sqrt 2} = t (1 + \dfrac {1}{\sqrt 2})$

CBSE Questions for Class 11 Engineering Physics Motion In A Straight Line Quiz 2 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Physics Motion In A Straight Line Quiz 2 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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