MCQ Questions for Class 11 Engineering Physics Motion In A Straight Line Quiz 5

How much time elapses during this interval?

0%
5 s
0%
20 s
0%
10 s
0%
1 s

Explanation

Answer is C.

We select positive direction for our coordinate system to be the direction of the velocity and choose the origin so that

$x_i\, =\, 0$ when the braking begins; Then the,initial velocity is

$u_x\, =\, + 25$ m/s at t = 0, and the final velocity and position are

$v_x\, =\, + 15$ m/s and x = 200 m at time t.
Since the acceleration is constant, the average velocity in the interval can be found from the average of the initial and final velocities.
Therefore,

$v_{av.x}\, =\, \displaystyle \frac{1}{2}(u_x\, +\, v_x)\, =\, \displaystyle \frac{1}{2}(15\, +\, 25)\, =\, 20m/s$ .
The average velocity can also be expressed as

$V_{av.x}\, =\, \displaystyle \frac{\Delta x}{\Delta t}$ . With

$\Delta x$ = 200 m
and

$\Delta t\, =\, t\, -\, 0$ , we can solve for t:

$t\, =\, \displaystyle \frac{\Delta x}{v_av.x}\, =\, \displaystyle \frac{200}{20} \, =\, 10s$ .
Hence, the time elapses during this interval is 10 s.

A ball is thrown vertically upwards with a certain initial velocity. Assume that there is no resistance due to air. Among the graphs below, the graph that is not an appropriate representation of the motion of the ball is:-

0%
A
0%
B
0%
C
0%
D

Explanation

A ball is thrown vertically upward.

Let initial velocity is

$u$ .

Velocity at any time

$t$

$\Rightarrow v= u-gt$ , ........1

Height at any time

$t$

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

Velocity at any height

$h$

$\Rightarrow v^2=u^2-2gh$ ....3

Kinetic Energy=

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

$\dfrac{1}{2}m (u-gt)^2$ (using equation 1)

$\therefore$ Kinetic energy is varies parabolically with time , Kinetic Energy will zero at time

$t= \dfrac{u}{g}$ , So option (A) is correct.

Potential Energy=

$mgh$ =

$mg(ut-\dfrac{1}{2}gt^2)$ , (Using equation 2)

$\therefore$ Potential Energy is varies parabolically with time , So Option (B) is correct.

Kinetic Energy =

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

$\frac{1}{2} m(u^2-gh)$ , (Using equation 3)

$\therefore$ Kinetic Energy is varies linearly with height

$h$ , So option (C) is wrong.

Potential Energy =

$mgh$ =

$mg\dfrac{u^2-v^2}{g}$ =

$\frac{}{}m(u^2-v^2)$ , (Using equation 3)

$\therefore$ Potential Energy is varies parabolically with speed. So option (D) is correct.

Consider two observers moving with respect to each other at a speed

$v$ along a straight line. They observe a block of mass

$m$ moving a distance

$l$ on a rough surface. The following quantities will be same as observed by the two observers.

0%
Kinetic energy of the block at time t
0%
Work done by friction
0%
Total work done on the block
0%
Acceleration of the block

Explanation

acceleration of the block is same in both frames as they are only moving in different velocities (but constant).

KE is dependent of

$v_{rel}$ . Work done on block by forces changes as position vector changes differently in each frame.

An observer finds the magnitudes of the acceleration of two bodies to be the same. This necessary implies that the two bodies.

0%
are at rest with respect to each other
0%
are at rest or move with constant velocities with respect to each other
0%
are accelerated with respect to each other
0%
may be at rest, moving with constant velocities or accelerated with respect to each other

Explanation

Case 1 :

Both the bodies are moving with same velocity. Thus they are rest w.r.t each other but they have same acceleration.

Case 2 :

Both the bodies are moving with different constant velocities. Hence both are moving with constant w.r.t to each other and yet they can have the same acceleration.

Case 3 :

Both are moving with different velocities but in opposite direction with same acceleration.

Acceleration of 1 w.r.t 2,

$a_{12} = a - (-a) =2a$

Hence both are accelerating w.r.t each other, although having same acceleration.

Option D is correct.

A car of mas sm starts moving so that its velocity varies according to the law

$v = \beta\, \sqrt{s}$ where

$\beta$ is a constant, ands is the distance covered. The total work performed by all the forces which are acting on the car during the first t seconds after the beginning of motion is

0%
$m\beta^{4}t^{2}/8$
0%
$m\beta^{2}t^{4}/8$
0%
$m\beta^{4}t^{2}/4$
0%
$m\beta^{2}t^{4}/4$

Explanation

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

$\beta^2s=0+2as$

$a=\dfrac{\beta^2}{2}$

$s=ut+0.5at^2$

$s=0+0.5(\beta^2/2)t^2$
Work=mas
W=

$m(\beta^2/2)(0.5(\beta^2/2)t^2)$
W=

$\dfrac{m\beta^4 t^2}{8}$

A block of mass m is suspended by a light thread from an elevator.The elevator is accelerating upward with uniform acceleration a. The work done by tension on the block during t seconds is (1 = 0) :

0%
$\displaystyle \frac{m}{2} (g + a) at^{2}$
0%
$\displaystyle \frac{m}{2} (g - a) at^{2}$
0%
$\displaystyle \frac{m}{2} gat^{2}$
0%
0

Explanation

T acting on string due to mass= +mg
T acting on string due to upward accelration= +ma
Total T= m(g+a)

$S=ut+ 0.5at^2= 0.5at^2$

$W= (0.5at^2)(m(g+a))$

$W=\dfrac{m(g+a)at^2}{2}$

The speed of the particle moving along with straight line become half after every next second. The initial speed is

$V_0$ . The total distance travelled by the particle will be

0%
$V_0$
0%
$2V_0$
0%
$\infty$
0%
None of the above

Explanation

Distance = $V_0 \times 1+\dfrac{V_0}{2}\times 1+\dfrac {V_0}{4}\times 1+.....$

$= V_0 \times \left (1+\dfrac{1}{2}+\dfrac{1}{4}+...... \right ) = 2V_0$

A car is moving with a uniform velocity of 40 km h

$^{-1}$ in straight line. Its acceleration after 1 hour is:

0%
40 km h $^{-1}$
0%
20 km h $^{-1}$
0%
30 km h $^{-1}$
0%
zero

A truck of mass

$2800 kg$ is moving with a speed of

$15\:m/s$ . A frictional retarding force of

$1200 N$ are acting on it, then in

$10 s$ it shall travel a distance of:

0%
$156 m$
0%
$122.8 m$
0%
$162.5 m$
0%
$118 m$

Explanation

Acceleration of the truck:

$a=\displaystyle \frac{1200-500}{2800}=\frac{1}{4}m/s^{2}$

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

$=\displaystyle 15\times 10+\frac{1}{2}\times \frac{1}{4}\times 100=162.5\:m$ .

The dispIacement of a body is given by

$2s\, =\, gt^2$ where g is a constant. The velocity of the body at any time t is:

0%
$gt$
0%
$\dfrac{gt }{ 2}$
0%
$\dfrac{gt^2} {2}$
0%
$\dfrac{gt^3}{ 6}$

Explanation

Given that:

$2s = gt^{2}$

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

Comparing this with the equation:

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

we get,

$u = 0\ \& \ a=g$

$v = u + at = 0 + gt$

$A$ and

$B$ are arguing about uniform acceleration.

$A$ states that acceleration means "the longer you go."

$B$ states that acceleration means "the further you go." Who is right?

0%
$A$
0%
$B$
0%
Both $A$ and $B$
0%
None of these

A ball is thrown upwards in a train which is accelerating. The ball will fall ______________.

0%
ahead of the thrower
0%
behind the thrower
0%
in the hands of thrower
0%
data inadequate

A body under uniformly accelerated motion covers

$20 m$ and

$60 m$ in first two seconds of time from the beginning. The body starts with an initial velocity

0%
Zero
0%
$5{m}/{s}$
0%
$10{m}/{s}$
0%
$15{m}/{s}$

Explanation

Here

$t = 1, {S}_{1} = 20$ and

$t = 2$ and

${S}_{2} = 60 + 20 = 80$

$\because {S}_{1} = ut + \displaystyle\frac{1}{2}a{t}^{2}$

$\therefore 20 = u + \displaystyle\frac{1}{2}a$
or,

$2u + a = 40$ .....(1)
and also

${S}_{2} = ut + \displaystyle\frac{1}{2}a{t}^{2}$

$\therefore 80 = 2u + \displaystyle\frac{1}{2}a \times 4 = 2u + 2a$
or,

$2u + a = 40$ ......(2)
From equations (1) and (2) we get

$u = 0$ .

A truck increases its speed from

$10km/h$ to

$50 km/h$ in

$20$ seconds. Its acceleration is:

0%
$0.55 m/s^2$
0%
$2.55 m/s^2$
0%
$0.75 m/s^2$
0%
$8.65 m/s^2$

Explanation

Given: Initial speed,

$u=10\ km/h=10\times \dfrac{5}{18}m/s$

Final speed,

$v=50\ km/h=50\times \dfrac{5}{18}m/s$

Using equation of motion,

Acceleration,

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

$=\dfrac{\dfrac{5}{18}\times(50-10)}{20}$

$=\dfrac{5}{18}\times 2$

$=0.55\ m/s^2$

Which one of the following represents the displacement-time graph of two objects a and b moving with zero relative speed ?

0%
0%
0%
0%
None of these

Explanation

For zero relative velocity the velocity of both body must be same.

$v=v_a-v_b=0$

The slope of displacement vs time graph must be same hence option B is correct.

A ball is thrown vertically upwards with a velocity

$u$ . What is the maximum height to which it will rise before falling back?

0%
${u^2/2g}$
0%
$g/u$
0%
$2g/u$
0%
$u/2g$

Explanation

Answer is A.

If we know the initial velocity, from the equation,

${ v }^{ 2 }-{ u }^{ 2 }=2as$ with s = h and v = 0 at the top of the path, initial velocity is given as 'u' and a = -g then:
h =

$\frac { { u }^{ 2 } }{ 2g }$ .
Therefore, the maximum height reached =

$\frac { { u }^{ 2 } }{ 2g }$ .

A ball is dropped on the floor from a height of 10 m. It rebounds to a height of 2.5 m.If the ball is in contact with the floor for 0.01 sec, then average acceleration during contact is

0%
$\displaystyle 2100{ m }/{ { s }^{ 2 } }$
0%
$\displaystyle 1400{ m }/{ { s }^{ 2 } }$
0%
$\displaystyle 700{ m }/{ { s }^{ 2 } }$
0%
$\displaystyle 400{ m }/{ { s }^{ 2 } }$

Explanation

Let u be the velocity with which the ball hits the ground, then

$\displaystyle { u }^{ 2 }=2gh$

$\displaystyle =2\times 9.8\times 10=196$

$\displaystyle \therefore \quad u=14 m/sec$
If v be the velocity with which it rebounds, then

$\displaystyle { v }^{ 2 }=2\times 9.8\times 2.5=49$

$\displaystyle \Rightarrow v=7m/sec$

$\displaystyle \therefore \quad \Delta v=\left( v-u \right)$

$\displaystyle =\left( 7m/sec \right) -\left( -14m/sec \right)$

$\displaystyle =21m/sec$

$\displaystyle \therefore \quad a=\frac { \Delta v }{ \Delta t } =\frac { 21 }{ 0.01 }$

$\displaystyle =2100{ m }/{ { s }^{ 2 } }$

If a car at rest accelerates uniformly and attains a speed of 72 km/hr in 10 s, then it covers a distance of:

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

Explanation

Given

u=0

v=72 km/h=20 m/s

t=2s

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

$a=2$

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

putting values

$s=100m$

A rocket is launched to travel vertically upward with a constant velocity of 20 m/s. After travelling 35 s the rocket develops a snag and its fuel supply is cut off. The rocket then travels like a free body, the height achieved it is

0%
600 m
0%
720 m
0%
800 m
0%
820 m

Explanation

For upward motion of rocket with uniform velocity

$u=20ms^{-1} ; t=35s ;$

$h=ut=20\times35=700m$
For upward motion after fuel cut off

$u=20ms^{-1} ;v=0 ; g=-10ms^{-2}$
using formula

$v=u+gt$

$0=20-10\times t$

$t=2s$
from formula

$v^2=u^2+2gh$
putting values,we get,

$0=20^2-2\times10\times h$

$0=400-20h$

$h=\frac{400}{20}=20m$
Total height achieved

$=700+20=720m$
hence,option b is correct.

A block is released from rest at the top of a frictionless inclined plane 16 m long. It reaches the bottom 4 sec later. A second block is projected up the plane from the bottom at the instant the block is released in such a way that it returns to the bottom simultaneously with first block. The acceleration of each block on the incline is

0%
$\displaystyle 1{ m }/{ { s }^{ 2 } }$
0%
$\displaystyle 2{ m }/{ { s }^{ 2 } }$
0%
$\displaystyle 4{ m }/{ { s }^{ 2 } }$
0%
$\displaystyle 9.8{ m }/{ { s }^{ 2 } }$

Explanation

We know,

$\displaystyle s=\frac { 1 }{ 2 } { at }^{ 2 }$

$\displaystyle \therefore \quad 16=\frac { 1 }{ 2 } \times a\times { \left( 4 \right) }^{ 2 }\quad \Rightarrow a=2{ m }/{ { s }^{ 2 } }$

A stone dropped from the top of a building taken 5 s to reach the ground. If it is stopped momentarily 4 s after it is dropped and then released again, how much time would it take from the moment it is released again to reach the ground?

$\displaystyle (Take\quad g = 10 {ms} ^{-2})$

0%
1 s
0%
3 s
0%
4 s
0%
5 s

Explanation

First Case-

Let the distance covered by the stone in 5 s is

$h$ .

Using

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

$\Rightarrow h= 0\times 5+\frac{1}{2} \times 10\times 5^2$ (

$\because u=0$ )

$\Rightarrow h= 125 m$ ,

Second Case-

Let the distance covered by the stone in 4 s is

$h_1$ .

Using

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

$\Rightarrow h= 0\times 4+\frac{1}{2} \times 10\times 4^2$ (

$\because u=0$ )

$\Rightarrow h_1=80 m$

At this moment stone is stopped and then released,

Let it take

$t$ seconds to cover the remaining distance.

Remaining distance =

$h-h_1$ = 125-80=45

$m$ ,

Using

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

$\Rightarrow 45= 0\times t+\frac{1}{2} \times 10\times t^2$ (

$\because u=0$ )

$\Rightarrow t=3 s$

A body, when dropped from a certain height, falls to the ground in 4 s. What is the time taken by the body to cover the last 100 cm?

$\displaystyle (Take\quad g={ 10\quad ms }^{ -2 })$

0%
$0.65 s$
0%
$0.15 s$
0%
$0.125 s$
0%
$0.025 s$

Explanation

Let Height

$h$ from a body is dropped.

Using

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

$\Rightarrow -h= 0\times 4 -\frac{1}{2}g\times 4^2$

$\Rightarrow h= 80 m$ ,

Let time

$t$ to cover the 79

$m$ .

Using

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

$\Rightarrow -79= 0\times t -\frac{1}{2}g\times t^2$

$\Rightarrow t=\sqrt{\frac{79}{5}}$ = 3.975

$s$

$\therefore$ Required time=

$4-3.975= 0.025$

$s$

The distance travelled by a body starting with a velocity of

$\displaystyle 20\ { ms }^{ -1 }$ , and moving with an acceleration of

$\displaystyle 2\ { ms }^{ -2 }$ , in the

$\displaystyle { 8 }^{ th }$ second is ____ m.

0%
35
0%
20
0%
120
0%
10

Explanation

Initial velocity

$u = 20 \ m/s$
Acceleration

$a = 2 \ m/s^2$
Distance covered in

$8^{th}$ second

$S_t = u+\dfrac{a}{2}(2t-1)$
where

$t = 8 \ s$
Thus,

$S_8 = 20+\dfrac{2}{2}(2\times 8-1) = 35 \ m$

An automobile travelling with a speed of

$\displaystyle 72 km {h}^ {-1}$ , can be stopped within a distance of 30 m, by applying brakes. What will be the stopping distance, if the automobile speed is increased to

$\displaystyle \sqrt { 3 }$ times and the same braking force is applied?

0%
30 m
0%
90 m
0%
60 m
0%
120 m

Explanation

Given Initial Velocity

$u=72 kmh^{-1}$ = 20

$ms^{-1}$ , Distance

$s= 30 m$ , Final Velocity

$v=0$ ,

Let acceleration is

$a$ .

Using

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

$\Rightarrow 0= 20^2 +2\times a\times 30$

$\Rightarrow a= -\dfrac{20}{3}$ ,

Now initial velocity is increases by

$\sqrt3$ times and the same braking force is applied.

Let distance covered is

$s$ .

Using

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

$\Rightarrow 0= (20\sqrt 3)^2+2\times (-\dfrac{20}{3}) \times s$

$\Rightarrow s=90 m$

Distance covered is 90 m.

A stone thrown vertically passes a certain point P at the end of 2 seconds and 8 seconds respectively. Find the maximum height reached by the stone .

$\displaystyle (Take\quad g ={ 10\quad ms }^{ -2 })$

0%
625 m
0%
125 m
0%
225 m
0%
350 m

Explanation

Let initial velocity

$u$ and point 'P' is at height

$h$ .

At the end of 2 seconds,

Using

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

$\Rightarrow h= u\times 2 - \frac{1}{2}g\times 2^2$

$\Rightarrow h= 2u-20$ .......1

At the end of 8 seconds,

$\Rightarrow h= u\times 8- \frac{1}{2}g\times 8^2$

$\Rightarrow h= 8u-320$ ....2

Solving equations 1 and 2,

$\Rightarrow u=50 {m}/{s}$ ,

At maximum height , velocity

$v=0$ ,

Using

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

$\Rightarrow 0= 50^2-2\times 10\times s$

$\Rightarrow s=125m$ ,

$\therefore$ Maximum height

$= 125 m$

A stone is released from a hot air balloon which is rising steadily with a velocity of

$\displaystyle { 4\quad ms }^{ -1 }$ . The velocity of the stone at the end of 3 s after it is released is _____

$\displaystyle { ms }^{ -1 }$

0%
29.4
0%
25.4
0%
32.5
0%
62.7

Explanation

Let the upward direction to be negative.

Initial velocity of stone

$u = -4 \ m/s$

Acceleration of the stone

$a = g = 9.8 \ m/s^2$

Velocity of stone

$v = u+at$

$\therefore$

$v = -4+9.8\times 3 = 25.4 \ m/s$

A body starts moving with an initial velocity of

$\displaystyle 5\ { ms }^{ -1 }$ and an acceleration of

$\displaystyle 1\ { ms }^{ -2 }$ . The distance travelled by it in the 5th second is

0%
$9.5$ m
0%
$22.5$ m
0%
$50$ cm
0%
$10$ cm

Explanation

Initial velocity

$u = 5 \ m/s$

Acceleration

$a = 1 \ m/s^2$

Distance covered in

$5th$ second,

$S_5 = u+\dfrac{1}{2}a(2t-1)$

where

$t = 5 \ s$

$\therefore$

$S_5 = 5+\dfrac{1}{2}(2\times 5-1) = 9.5 \ m$

A hollow iron ball (A) and a solid iron ball (B) and cricket ball (C) are dropped from the same height. Which among the three balls reaches the ground first? Assuming there is no resistance due to air.

0%
A
0%
B
0%
C
0%
All the three balls reaches ground simultaneously

The ratio of the time taken by a body moving with uniform acceleration in reaching two points P and Q along a straight line path is 1 :If the body starts from rest and moves linearly, the ratio of the distance between P and Q from the starting point is:

0%
4 : 1
0%
1 : 4
0%
2 : 3
0%
3 : 1

Explanation

The ratio of time taken is 1 :2,

Using

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

The ratio of distance

$\dfrac{s_1}{s_2}= \dfrac{\dfrac{1}{2}gt_1^2}{\dfrac{1}{2}gt_2^2}=\dfrac{t_1^2}{t_2^2}$ , ( u=0 given)

$\Rightarrow \dfrac{s_1}{s_2}=\dfrac{1}{4}$

A body starts from rest and moves with uniform acceleration for 3s. It then decelerates uniformly for 2s. and stops. If the deceleration is

$\displaystyle 3 \ { ms }^{ -2 }$ the maximum velocity of the body is ____

$\displaystyle { ms }^{ -1 }$

0%
zero
0%
2
0%
6
0%
Cannot be determined

Explanation

Velocity will be maximum before the deceleration starts.

Given Final Velocity v=0, acceleration

$a=-3ms^{-2}$ and time

$t=3 s$ ,

Let maximum velocity is

$u_{max}$ ,

Using

$v=u+at$

$\Rightarrow 0=u_{max} -3\times 2$

$\Rightarrow u_{max}=6 ms^{-1}$

A stone projected up vertically with a velocity v, reaches points x, y and z in its path with velocities

$\displaystyle \frac { v }{ \sqrt { 5 } } ,\frac { v }{ \sqrt { 10 } } and\frac { v }{ \sqrt { 15 } }$ respectively. Find the ratio xy : yz.

0%
3:1
0%
4:1
0%
6:1
0%
3:2

Explanation

Let

$S_x$ ,

$S_y$ and

$S_z$ are the height from the ground of the point

$x$ ,

$y$ and

$z$ respectively.

Using

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

$\Rightarrow (\dfrac{v}{\sqrt5})^2=v^2-2gS_x$

$\Rightarrow S_x=\dfrac{v}{25}$ ,

$\Rightarrow (\dfrac{v}{\sqrt10})^2=v^2-2gS_y$

$\Rightarrow S_y=\dfrac{9v}{200}$ ,

$\Rightarrow (\dfrac{v}{\sqrt15})^2=v^2-2gS_z$

$\Rightarrow S_z=\dfrac{14v}{300}$ ,

$xy=S_y-S_x= \dfrac{9v}{200}- \dfrac{v}{25}=\dfrac{v}{200}$

$yz=S_z-S_y= \dfrac{14v}{300}- \dfrac{9v}{200}=\dfrac{v}{600}$

$\therefore$

$xy:yz=\dfrac{v}{200}:\dfrac{v}{600}=3:1$

A bike moving along a straight road covers

$35$ m in the

$4$ th second and

$40$ m in the

$5$ th second. What is its initial velocity and acceleration (if the acceleration is assumed to be uniform )?

0%
$\displaystyle { 17.5\ ms }^{ -1 }$
0%
$\displaystyle { 8\ ms }^{ -1 }$
0%
$\displaystyle { 7.8\ ms }^{ -1 }$
0%
$\displaystyle { 38.5\ ms }^{ -1 }$

Explanation

Distance covered in

$nth$ second is given by

$S_n = u+\dfrac{1}{2}(2n-1)$

Case 1 :

$n = 4$

$S_n = 35$

So,

$35 = u+\dfrac{a}{2}(2\times 4-1)$

$35 = u+3.5 \ n$ ......(1)

Case 2 :

$n = 5$

$S_n = 40$

So,

$40 = u+\dfrac{a}{2}(2\times 5-1)$

$40 = u+4.5 \ n$ ......(2)

Equation (2) - (1), we get

$40-35 = 4.5 a-3.5 a$

$a = 5 \ m/s^2$

Now,

$35 = u+3.5\times 5$

$\implies \ u = 17.5\ m/s$

A body is projected horizontally from a certain height, (h) then time of descent is

0%
$\displaystyle { t }_{ d }=\sqrt { \frac { 2h }{ g } }$
0%
$\displaystyle { t }_{ d }=\sqrt { \frac { h }{ g } }$
0%
$\displaystyle { t }_{ d }=\sqrt { h g}$
0%
$\displaystyle { t }_{ d }=\sqrt { h + g}$

Explanation

When a body is projected horizontally, its initial velocity in the vertical direction is zero i.e.

$u_y = 0$

Displacement of body in vertical direction

$S_y = h$

Acceleration due to gravity in vertical direction

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

Using

$S_y = u_y t +\dfrac{1}{2}a_yt^2$

$\therefore$

$h = 0+\dfrac{1}{2}gt_d^2$

$\implies \ t_d = \sqrt{\dfrac{2h}{g}}$

Find the initial velocity of projection of a ball thrown vertically up if the distance moved by it in 3rd second is twice the distance covered by it in 5th second.

$\displaystyle (Take g = 10\ { ms }^{ 2 })$

0%
$\displaystyle 45\ { ms }^{ -1 }$
0%
$\displaystyle 75\ { ms }^{ -1 }$
0%
$\displaystyle 20\ { ms }^{ -1 }$
0%
$\displaystyle 85\ { ms }^{ -1 }$

Explanation

Let initial velocity

$u$ ,

Let

$S_1$ be the distance covered in

$n$ seconds and

$S_2$ be the distance covered in

$(n+1)$ .

Let

$S_n$ be the distance covered in

$nth$ seconds.

Using

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

$\Rightarrow S_1= un-\frac{1}{2}gn^2$ ....1

$\Rightarrow S_2= u(n+1)-\frac{1}{2}g(n+1)^2$ ....2

Subtracting equations 1 from 2,

$S_2-S_1=u(n+1)-\frac{1}{2}gn^2-(un-\frac{1}{2}g(n+1)^2)$

$\Rightarrow S_n=S_2-S_1= u-\frac{1}{2}(2n+1)$

$\therefore S_3= u-35$ ......3

$\therefore S_5= u-55$ ......4

Using the condition given in question ,

$S_3=2S_5$

$\Rightarrow u-35=2(u-55)$

$\Rightarrow u=75 ms^{-1}$

A stone is dropped by a person from the top of a tower, which is 200 m tall. At the same time another stone is thrown upwards, with a velocity of

$\displaystyle 50\ { ms }^{ -1 }$ by a person standing at the foot of the tower. Find the time after which two stones meet.

0%
$4 \ s$
0%
$5 \ s$
0%
$8\ s$
0%
$10\ s$

Explanation

Let the two stones meet at a distance of x mfrom the top of the tower, and 't' be the time taken. Let us assume the downward direction as positive.
For the stone that is dropped its initial velocity

$\displaystyle u = 0\quad { ms }^{ -1 }$ ; displacement s = x and acceleration = acceleration due to gravity (g).
Using

$\displaystyle s=\quad ut+\frac { 1 }{ 2 } { at }^{ 2 },\\ we\quad get\quad x\quad =\quad (0)\quad t\quad +\frac { 1 }{ 2 } { gt }^{ 2 }\quad \rightarrow \quad (1)$ .
For the stone that is projected vertically upwards, its initial velocity,

$\displaystyle u =-50\quad { ms }^{ -1 }$ ; displacement s =-(200-x) and acceleration a =g.
using

$\displaystyle s=\quad ut+\frac { 1 }{ 2 } { at }^{ 2 },\\ we\quad get\quad -(200-x)=-50\times t+1/2{ gt }^{ 2 }\\ 200=50t-1/2{ gt }^{ 2 }+x\quad \rightarrow \quad (1)(2)\\ From\quad the\quad equations\quad (1)\quad and\quad (2)\\ we\quad have\quad 200\quad =50t-1/2{ gt }^{ 2 }+1/2{ gt }^{ 2 }\\ \Rightarrow 200=50\therefore \quad t=4s$

A body starts from rest and moves with uniform acceleration for 2s. If then decelerates uniformly for 3s and stops. If deceleration is

$\displaystyle { 4\quad ms }^{ -2 }$ , the acceleration of the body is ____

$\displaystyle { ms }^{ -2 }$

0%
10
0%
8.7
0%
4
0%
6

Explanation

Given Final Velocity v=0, deceleration

$a=4ms^{-2}$ and deceleration time

$t=3 s$ ,

Let

$u$ be the velocity before the deceleration is started.

Using

$v=u+at$

$\Rightarrow 0=u -4\times 3$

$\Rightarrow u=12, ms^{-1}$ ,

Now , During acceleration, Initial velocity

$u_1$ =0, time

$t= 2s$ , Final velocity

$u=12ms^{-1}$ ,

Let acceleration is

$a$ .

Using

$v=u+at$

$\Rightarrow 12=0+a\times 2$

$\Rightarrow a= 6 ms^{-2}$

Two cars arrive at certain point with velocity of

$\displaystyle { 30\ ms }^{ -1 } , { 25\ ms }^{ -1 }$ and travel in a straight line with uniform acceleration

$\displaystyle { 0.25\ ms }^{ -2 }\ and\ { 0.5\ ms }^{ -2 }$ respectively.
(A) Find the distance at which they meet again.
(B) Also determine the time after which the final velocity one of the cars is equal to the initial velocity of the other.

0%
$1400 \ m$ , $10\ s$
0%
$2800 \ m$ , $20\ s$
0%
$1400 \ m$ , $30\ s$
0%
$1000 \ m$ , $10\ s$

Explanation

Solution: (A) Both cars travels for same time (t)
First car :

$\displaystyle u= { 30\quad ms }^{ -1 }; a= { 0.25\quad ms }^{ -2 }$

$\displaystyle { S }_{ 1 }=(30\times t)+\frac { 1 }{ 2 } (0.25\times { t }^{ 2 })$
Second car :

$\displaystyle u= { 25\quad ms }^{ -1 }; a= { 0.5\quad ms }^{ -2 }$

$\displaystyle { S }_{ 2 }=(25\times t)+\frac { 0.5 }{ 2 }{ t }^{ 2 })$

$displaystyle But\quad { S }_{ 1 }={ S }_{ 2 }\\ \therefore 30t+\frac { 1 }{ 2 } { 0.25 }\quad t^{ 2 }\\ =25t+\frac { 1 }{ 2 } { 0.5t }^{ 2 }=10t={ 0.25t }^{ 2 }\\ t=\frac { 10 }{ 0.25 } =40s\\ Length\quad of\quad the\quad path\quad travelled\\ (S)=\quad (30\times 40)+\frac { 0.25 }{ 2 } \times { (40 })^{ 2 }\\ =1200+200=1400m$
(B)

$\displaystyle v=30\quad { ms }^{ -1 },\quad u=25\quad { ms }^{ -1 }\\ a=0.5\quad { ms }^{ -2 },\quad t=\quad ?\\ v=u+at\Rightarrow 30=25+0.5\times t\\ \Rightarrow 5=0.5t\Rightarrow t=10s$

When a body is dropped from a tower, it covers 75% of total height of the tower in the last second of its fall. What is the total height of the tower?

0%
$4.9$ m
0%
$9.8$ m
0%
$5.6$ m
0%
$6.0$ m

Explanation

Let h be the height of the tower the initial velocity u=0

$\displaystyle h=\frac { 1 }{ 2 } { gt }^{ 2 }\\ { S }_{ n }=\frac { g }{ 2 } (2t-1)=0.75h\\ =\frac { g }{ 2 } (2t-1)=0.75(\frac { 1 }{ 2 } { gt }^{ 2 })\quad (from\quad 1))\\ 2t-1=\frac { 75{ t }^{ 2 } }{ 100 } =\frac { 3 }{ 4 } { t }^{ 2 }\\ 8t-4={ 3t }^{ 2 }\\ { 3t }^{ 2 }-8t+4=0\\ t=\frac { 8\pm \sqrt { 64-4\times 3\times 4 } }{ 2\times 3 } =\frac { 8\pm \sqrt { 16 } }{ 12 } =\frac { 8\pm 4 }{ 12 } \\ t=1,-2/3\Rightarrow t=1s\\ h=\frac { 1 }{ 2 } \times { gt }^{ 2 }=\frac { 1 }{ 2 } \times 9.8\times 1=4.9\quad m$

A freely falling body crosses points P, Q and R with velocity V,

$2$ V and

$3$ V respectively. Find the ratio of the distance PQ to QR.

0%
$5 : 3$
0%
$3 : 5$
0%
$1 : 2$
0%
$2 : 1$

Explanation

Let the acceleration be

$a$ .

For P to Q :

Initial speed

$u = V$

Final speed

$v = 2V$

Using

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

$\therefore$

$2aS_{PQ} = (2V)^2-V^2$

We get

$S_{PQ} = \dfrac{3V^2}{2a}$

For Q to R :

Initial speed

$u = 2V$

Final speed

$v = 3V$

Using

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

$\therefore$

$2aS_{QR} = (3V)^2-(2V)^2$

We get

$S_{QR} = \dfrac{5V^2}{2a}$

So.

$S_{QR}:S_{PQ} = 3:5$

A car moves with a constant velocity of

$\displaystyle { 10\ ms }^{ -1 }$ for 10 s along a straight road. Then it moves with uniform acceleration of

$\displaystyle { 2\ ms }^{ -2 }$ for

$5$ seconds. Find the total displacement at the end of the 15 s of its motion.

0%
$175$ m
0%
$125$ m
0%
$150$ m
0%
$105$ m

Explanation

For constant velocity :

Constant velocity

$u = 10 \ m/s$

Time

$t = 10 \ s$

So, displacement in 10 seconds

$S_1 = ut = 10\times 10 = 100 \ m$

For accelerated journey :

Initial speed

$u = 10 \ m/s$

Acceleration

$a = 2 \ m/s^2$

Time

$t = 5\ s$

Displacement in 5 seconds,

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

$\implies \ S_2 = 10\times 5+\dfrac{1}{2}\times 2\times 5^2 = 75 \ m$

Total displacement in 15 seconds

$S_{total} = 100+75 = 175 \ m$

A body is dropped from a height of

$2$ m. It penetrates into the sand on the ground through a distance of

$10$ cm before coming to rest. What is the retardation of the body in sand?

0%
$\displaystyle -9.8\ {ms} ^{-1}$
0%
$\displaystyle 196\ {ms} ^{-2}$
0%
$\displaystyle -196\ {ms} ^{-2}$
0%
$\displaystyle 9.8\ {ms} ^{-2}$

Explanation

Initial velocity of the body

$u =0$

Velocity of the body just before reaching the sand

$V = \sqrt{2gH}$

$\implies \ V = \sqrt{2\times 9.8\times 2} = \sqrt{4\times 9.8} \ m/s$

Final velocity in the sand

$V_f = 0$

Distance covered in sand

$S = 10 \ cm = 0.1 \ m$

Using

$V_f^2 - V^2 = 2aS$

$0 - (\sqrt{4\times 9.8})^2 = 2\times a\times 0.1$

$\implies \ a = -196 \ m/s^2$

Retardation is

$196 \ m/s^2$

A body falls from a height of

$45$ m above the ground. Find the time taken by the body to reach the ground.

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

0%
$5$ s
0%
$2$ s
0%
$3$ s
0%
$6$ s

Explanation

Initial speed

$u = 0 \ m/s$

Acceleration

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

Height

$h = 45 \ m$

Using

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

$45 = 0+\dfrac{1}{2}(10)t^2$

$\implies \ t = 3 \ s$

Two stones A and B are dropped from the top of two different towers such that they travel

$44.1$ m and

$63.7$ m in the last second of their motion respectively. Find the ratio of the heights of the two towers from where the stones were dropped.

0%
$16 : 25$
0%
$7 : 9$
0%
$5 : 7$
0%
$25 : 49$

Explanation

A stone is dropped from the tower, Let it takes

$t$ seconds to reach the ground.

Using

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

Distance travelled in

$t$ seconds

$\Rightarrow -S_t=0\times t-\frac{1}{2}gt^2$

$\Rightarrow S_t=\frac{1}{2}gt^2$ ....1

Distance travelled in

$t-1$ seconds

$\Rightarrow -S_{t-1}=0\times t-\frac{1}{2}g(t-1)^2$

$\Rightarrow S_{t-1}=\dfrac{1}{2}g(t-1)^2$ ....2

Subtract equation 2 from 1,

$S_t-S_{t-1}=\dfrac{1}{2}g(2t-1)$

$\therefore$ The distance travelled in last seconds

$S= \dfrac{1}{2}g(2t-1)$ ,

Let stone A takes time

$t_A$ and stone B takes time

$t_B$ .

$\Rightarrow 44.1=\dfrac{1}{2}g(2t_A-1)$

$\Rightarrow t_A=5$ ,

$\Rightarrow 63.7=\dfrac{1}{2}g(2t_B-1)$

$\Rightarrow t_B=7$ ,

Using equation 1,

Distance travelled by stone A

$=S_A= \dfrac{1}{2}g\times 5^2$

Distance travelled by stone B

$=S_B= \dfrac{1}{2}g\times 7^2$

$\therefore$

$\frac{S_A}{S_B}=\dfrac{\dfrac{1}{2}g\times 5^2}{\frac{1}{2}g\times 7^2}=\dfrac{25}{49}$

$\Rightarrow S_A:S_B= 25:49$

An object projected vertically up from the top of the tower took

$5$ s to reach the ground. The average velocity of the object is

$\displaystyle 5 { ms }^{ -1 })$ , find its average speed. (given

$\displaystyle g\ = 10 { ms }^{ -1 })$ .

0%
$\displaystyle 65\ { ms }^{ -1 }$
0%
$\displaystyle 13\ { ms }^{ -1 }$
0%
$\displaystyle 26\ { ms }^{ -1 }$
0%
$\displaystyle 25\ { ms }^{ -1 }$

Explanation

Height of tower= displacement of an object= average velocity

$\times$ time

$=5\times 5=25 m$ ,

Let initial velocity of the object is

$u$ .

Using

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

$\Rightarrow -25= u\times 5-\frac{1}{2}g\times 5^2$

$\Rightarrow u=20 ms^{-1}$ ,

Now from the tower,Let maximum height reached by object is

$h$ .

At maximum height velocity

$v=0$ ,

Using

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

$\Rightarrow 0=20^2-2gh$

$\Rightarrow h=20m$ ,

$\therefore$ Total distance

$= 2h+25= 65m$

$\text{Average speed}= \dfrac{\text{Total Distance}}{\text{Total Time}}$ ,

$\Rightarrow$ Average speed

$=\dfrac{65}{5}=13ms^{-1}$

A ball thrown vertically upwards with speed 'u' from the top of a tower reaches the ground in

$9$ s. Another ball is thrown vertically downwards from the same position with speed 'u', takes

$4$ s to reach the ground. Calculate the value of 'u' .

$\displaystyle (Take \ g= 10 { ms }^{ -1 })$

0%
$\displaystyle 125\ { ms }^{ -1 }$
0%
$\displaystyle 75\ { ms }^{ -1 }$
0%
$\displaystyle 25\ { ms }^{ -1 }$
0%
$\displaystyle 175\ { ms }^{ -1 }$

Explanation

Ball is thrown from the top of tower,

$\Rightarrow height= -h$ ,

For the first case, velocity =

$u$ , Time

$t= 9 s$ ,

Using

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

$\Rightarrow -h= u\times 9- \frac{1}{2} \times 10\times 9^2$

$\Rightarrow -h= 9u-405$ .....1

For the second case, velocity =

$-u$ , Time

$t= 4 s$ ,

Using

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

$\Rightarrow -h= -u\times 4- \frac{1}{2} \times 10\times 4^2$

$\Rightarrow -h= -4u-80$ .....2

Equating equations 1 and 2,

$\Rightarrow 9u-405=-4u-80$

$\Rightarrow u= 25 ms^{-1}$

A ball which is thrown vertically up from the top of the tower reaches the ground in

$12$ s. Another ball thrown vertically downwards from the same position with the same velocity takes

$4$ s to reach the ground. Find the height of the tower.

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

0%
$180$ m
0%
$120$ m
0%
$220$ m
0%
$240$ m

Explanation

Ball is thrown from the top of tower,

$\Rightarrow height= -h$ ,

For the first case, velocity =

$u$ , Time

$t= 12 s$ ,

Using

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

$\Rightarrow -h= u\times 12- \frac{1}{2} \times 10\times 12^2$

$\Rightarrow -h= 12u-720$ .....1

For the second case, velocity =

$-u$ , Time

$t= 4 s$ ,

Using

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

$\Rightarrow -h= -u\times 4- \frac{1}{2} \times 10\times 4^2$

$\Rightarrow -h= -4u-80$ .....2

Equating equations 1 and 2,

$\Rightarrow 12u-720=-4u-80$

$\Rightarrow u= 40 ms^{-1}$

Put the value of

$u=40 ms^{-1}$ in equation 1,

$\Rightarrow -h= 12\times 40-720$

$\Rightarrow h=240m$

An object travels 10 s with uniform acceleration along a straight line path. During this period if the velocity of the object is increased from

$\displaystyle { \ 5\ ms }^{ -1 } to { \ 25\ s }^{ -1 }$ , then find the distance travelled by the body?

0%
$150$ m
0%
$100$ m
0%
$125$ m
0%
$175$ m

Explanation

Initial speed

$u = 5 \ m/s$

Final speed

$v = 25\ m/s$

Time

$t = 10 \ s$

Using

$v = u+at$

$\therefore$

$25 = 5+a\times 10$

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

Using

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

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

A body is dropped from a certain height 'h' meters. Assuming that the gravitational field is nullified, after the body has travelled

$\dfrac{h}{2}$ meters such that g

$= 0,$ discuss the motion of the body. Find an expression for the time taken by the body to reach the ground.

0%
$\displaystyle \dfrac { 3 }{ 2 } \sqrt { \dfrac { h }{ g } }$
0%
$\displaystyle\sqrt { \dfrac { g }{ h } }$
0%
$\displaystyle \sqrt { gh }$
0%
$\displaystyle \dfrac { 1 }{ 2 } \sqrt { gh }$

Explanation

Till

$\dfrac{h}{2}$ meter,

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

$t=\sqrt { \dfrac { 2s }{ g } } =\sqrt { \dfrac { 2\times \dfrac { h }{ 2 } }{ g } } =\sqrt { \dfrac { h }{ g } }$

so to cover

$\dfrac{h}{2}$ meter with a=g,the time taken

${ t }_{ 1 }=\sqrt { \dfrac { h }{ g } } \\$

After that a=0

$v(at\quad t=\sqrt { \dfrac { h }{ g } } )=g{ t }_{ 1 }=g\sqrt { \dfrac { h }{ g } } =\sqrt { gh }$

with this contant speed,particle will travel next

$\dfrac{h}{2}$ meter.

${ t }_{ 2 }=\dfrac { \dfrac { h }{ 2 } }{ \sqrt { gh } } =\dfrac { 1 }{ 2 } \sqrt { \dfrac { h }{ g } } \\ t={ t }_{ 1 }+{ t }_{ 2 }=\sqrt { \dfrac { h }{ g } } +\dfrac { 1 }{ 2 } \sqrt { \dfrac { h }{ g } } =\dfrac { 3 }{ 2 } \sqrt { \dfrac { h }{ g } }$

A ball thrown vertically upwards with speed 'u' from the top of a tower reaches the ground in

$9$ s. Another ball is thrown vertically downwards from the same position with speed 'u'. take

$4$ s to reach the ground. Calculate the value of 'u'.

$\displaystyle (Take \quad g = 10 { ms }^{ -2 })$

0%
$\displaystyle 15\ { ms }^{ -1 }$
0%
$\displaystyle 20\ { ms }^{ -1 }$
0%
$\displaystyle 25\ { ms }^{ -1 }$
0%
$\displaystyle 45\ { ms }^{ -1 }$

Explanation

Ball is thrown from the top of tower,

$\Rightarrow height= -h$ ,

For the first case, velocity =

$u$ , Time

$t= 9 s$ ,

Using

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

$\Rightarrow -h= u\times 9- \frac{1}{2} \times 10\times 9^2$

$\Rightarrow -h= 9u-405$ .....1

For the second case, velocity =

$-u$ , Time

$t= 4 s$ ,

Using

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

$\Rightarrow -h= -u\times 4- \frac{1}{2} \times 10\times 4^2$

$\Rightarrow -h= -4u-80$ .....2

Equating equations 1 and 2,

$\Rightarrow 9u-405=-4u-80$

$\Rightarrow u= 25 ms^{-1}$

A ball is thrown vertically upwards with an initial velocity such that it can reach a maximum height of 15 m, if at the same instance a stone is dropped from the same height of 15 m, find the ratio of distances travelled by them when they cross each other?

0%
$4 : 3$
0%
$2 : 3$
0%
$1 : 2$
0%
$3 : 1$

CBSE Questions for Class 11 Engineering Physics Motion In A Straight Line Quiz 5 - 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 5 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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