Motion in A Straight Line Test

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Motion in A Straight Line Test - 44

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Question 1
1 / -0

Three particles $A, B$ and $C$ are thrown from the top of a tower with the same speed. $A$ is thrown straight up, $B$ is thrown straight down and $C$ is thrown horizontally. They hit the ground with speed $v_{A}, v_{B}$ and $v_{C}$ respectively.

A
$v_{A} = v_{B} = v_{C}$

B
$v_{A} > v_{B} > v_{C}$

C
$v_{A} = v_{B} > v_{C}$

D
$v_{A} > v_{B} = v_{C}$

Solution

For both $A$ and $B, u_{vertical}$ is the same, because when a ball is projected upwards with an initial velocity, it will come back with the same velocity. At $Q$ both have the same initial velocity. For $C$ , its vertical velocity is zero but it has a horizontal velocity $u_{C} = u_{A} = u_{B}$ . It falls through the height freely. But $v_{A}$ and $v_{B}$ will be greater than the final velocity $v_{C}$ , because $v_{A} = u_{A} + gt; v_{B} = u_{B} + gt$ . But for $C, u_{C} + (gt)$ are not in the same direction. It is less, but $v_{A} = v_{B} > v_{C}$ .

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Question 2
1 / -0

A car and bike start racing in a straight line. The distance of finish line from starting line is 100m. The minimum acceleration of car to win, if it accelerates uniformly starting from rest and the bike moves with a constant velocity of 10 m/s, is

A
$0.5 m/s^2$

B
$2 m/s^2$

C
$1m/s^2$

D
$3 m/s^2$

Solution

For bike :
Speed   $v = 10 \ m/s$
Distance to be covered   $d = 100 \ m$
Time taken $t = \dfrac{d}{v} = \dfrac{100}{10} = 10 \ s$
For car to win the race :
Time $t = 10 \ s$
Initial velocity $u = 0$
Using    $S = ut+\dfrac{1}{2}at^2$
Or   $100 = 0+\dfrac{1}{2}a(10)^2$
$\implies \ a = 2 \ m/s^2$

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Question 3
1 / -0

A body is thrown horizontally from the top of a tower of height $5\ m$ . It touches the ground at a distance of $10\ m$ from the foot of the tower. The initial velocity of the body is $\left( g=10\ m{ s }^{ -2 } \right)$

A
$2.5\ { ms }^{ -1 }$

B
$5\ { ms }^{ -1 }$

C
$10\ { ms }^{ -1 }$

D
$20\ { ms }^{ -1 }$

Solution

Height of the tower $H= 5 \ m$
Time of flight $T = \sqrt{\dfrac{2H}{g}} = \sqrt{\dfrac{2\times 5}{10}} = 1 \ s$
Range of projectile is $R=10\ m$
Initial velocity of projectile $u = \dfrac{R}{T} = \dfrac{10}{1} = 10 \ m/s$

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Question 4
1 / -0

The deceleration of a car traveling on a straight highway is a function of its instantaneous velocity $v$ given by $w=a\sqrt { v }$ , where $a$ is a constant. If the initial velocity of the car is $v_o$ , the distance the car will travel and the time it takes before it stops are

A
$\cfrac { 2 }{ 3 } m v_o^{3/2},\cfrac { 1 }{ 2 } v_o^{3/2} s$

B
$\cfrac { 3 }{ 2a } m v_o^{1/2},\cfrac { 1 }{ 2a } v_o^{1/2}s$

C
$\cfrac { 3a }{ 2 } m v_o^{1/2},\cfrac { a }{ 2 } v_o^{3/2}s\quad$

D
$\cfrac { 2 }{ 3a } m v_o^{3/2},\cfrac { 2 }{ a } v_o^{1/2} s\quad$

Solution

Deceleration of the car is given as   $w = a\sqrt{v}$
Using $w = \dfrac{-vdv}{ds}$
$\therefore$ $\int^o_{v_o} v^{1/2}dv =- a\int^S_o ds$
We get   $\dfrac{2}{3} v_{o}^{3/2} = aS$
$\implies \ S = \dfrac{2}{3a}v_o^{3/2}$
Using $w =- \dfrac{dv}{dt}$
$\therefore$ $-\dfrac{dv}{dt} = a\sqrt{v}$
Integrating    $\int^o_{v_o} v^{-1/2} dv = -a\int_o^t dt$
we get,   $2\sqrt{v_o} = at$
$\implies \ t = \dfrac{2\sqrt{v_o}}{a}$

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Question 5
1 / -0

A $20\ kg$ bullet pierces through a plate of mass $M_{1} = 1\ kg$ and then comes to rest inside a second plate of mass $M_{2}= 2.98\ kg$ as shown in the figure. It is found that the two plates initially at rest and now move with equal velocities. Find the percentage loss in the initial velocity of the bullet when it is between $M_{1}$ and $M_{2}$ . (Neglect any loss of material of the plates due to the action of bullet).

A
$6$ %

B
$25$ %

C
$50$ %

D
$72.5$ %

Solution

Let initial velocity of bullet $={ V }_{ 1 }㎧$
Velocity with which each plate moves $={ V }_{ 2 }㎧$
According to the law of conservation of momentum the initial momentum of the bullet is equal to the sum of the final momentum of the second plate including the bullet
$\therefore m{ V }_{ 1 }={ M }_{ 1 }{ V }_{ 2 }+\left( { M }_{ 2 }+m \right) { V }_{ 2 }$
$20{ V }_{ 1 }=1\times { V }_{ 2 }+\left( 2.98+20 \right) { V }_{ 2 }$
$20{ V }_{ 1 }={ V }_{ 2 }+22.98{ V }_{ 2 }$
${ V }_{ 1 }=\cfrac { 23.98 }{ 20 } { V }_{ 2 }$
${ V }_{ 1 }=1.199{ V }_{ 2 }\rightarrow 1$
Let the velocity of bullet when it comes out of first plate $={ V }_{ 3 }$
The momentum of the bullet the first and second, a plate is equal to the sum of the momentum of the second plate and the bullet.
$20{ V }_{ 3 }=\left( 20+2.980{ V }_{ 2 } \right)$
$20{ V }_{ 3 }=22.98{ V }_{ 2 }$
$\therefore { V }_{ 3 }=1.149{ V }_{ 2 }\rightarrow 2$
Loss Percentage in the initial velocity of the bullet when it is moving between ${ m }_{ 1 }\& { m }_{ 2 }$ is expressed as the following
Loss % $=\cfrac { { V }_{ 1 }-{ V }_{ 3 } }{ { V }_{ 1 } } \times 100$
Loss % $=\cfrac { 1.199{ V }_{ 2 }-1.149{ V }_{ 2 } }{ 1.199{ V }_{ 2 } } \times 100$
Loss % $=\cfrac { 0.05 }{ 1.199 } \times 100$
$=5.995$ %

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Question 6
1 / -0

A body falling from a high Minaret travels $40m$ in the last $2$ seconds of its fall to ground. Height of Minaret in metres is
(take $g=10m/{ s }^{ 2 }$ )

A
$60$

B
$45$

C
$80$

D
$50$

Solution

Taking the height of minaret is $H$ and time taken by body to fall from top to bottom be $T$ .
$\therefore H=(1/2)gT^2$ .....(1)
In last two second body travels a distance of 40m, henc ein $(T-2)sec$ body will travel $(H-40)m$ .
$(H-40) = (1/2) g (T-2)^2$ .......(2)
$\therefore$ solving (1) and (2),
$T=3sec , H=45m$
Hence option $B$ is correct.

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Question 7
1 / -0

The speed of a car is reduced from 90 km/hr to 36 km/hr in 5 s. What is the distance travelled by the car during this time interval.

A
87.5 m

B
90 m

C
50 m

D
100 m

Solution

$velocity (u) = 90km/h$ .

$= 25m/s$

$final velocity (v) = 36 km/h$

$= 10m/s$

$time(t) = 5 s$ .

acceleration (a)= $\dfrac{v-u}{t}$

= $\dfrac{10-25}{5}$

= $\dfrac{-15}{5}$

= $-3m/s^2$

let,

Using the third equation of motion

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

$2 (-3)(s) = (10)^2 -(25)^2$

$-6s = 100 - 625$

$- 6s = - 525$

$6s = 525$

$s = \dfrac{525}{6}$

$= 87.5m$

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Question 8
1 / -0

A particle is found to be at rest when seen from a frame $S_{1}$ and moving with a constant velocity when seen from another frame $S_{2}$ . Mark out the possible options.

A
Both the frames are inertal

B
$S_{1}$ is inertial and $S_{2}$ is noninertial

C
$S_{1}$ is noninertial and $S_{2}$ is inerital

D
none of these

Solution

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Question 9
1 / -0

A particle is thrown upwards from ground. It experiences a constant resistance force which can produce retardation $2{ m/s }^{ 2 }$ . The ratio of time of ascent to the time of descent is (g= $10{ m/s }^{ 2 }$ )

A
1:1

B
$\sqrt { \dfrac { 2 }{ 3 } }$

C
$\dfrac { 2 }{ 3 }$

D
$\sqrt { \dfrac { 3 }{ 2 } }$

Solution

Let the maximum height attained by the particle be $H$ .
Time taken by it to reach maximum height   $T = \sqrt{\dfrac{2H}{g_{eff}}}$
During upward motion :
The particle experiences acceleration due to gravity $10 \ m/s^2$ in downward direction and also, resistive retardation $2 \ m/s^2$ in downward direction.
So, effective acceleration $g_{eff} = 10+2 = 12 \ m/s^2$
Thus, time of ascent   $t_a = \sqrt{\dfrac{2H}{12}}$
During downward motion :
The particle experiences acceleration due to gravity $10 \ m/s^2$ in downward direction and also, resistive retardation $2 \ m/s^2$ in upward direction.
So, effective acceleration $g_{eff} = 10-2 = 8 \ m/s^2$
Thus, time of descent   $t_d = \sqrt{\dfrac{2H}{8}}$
Thus   $\dfrac{t_a}{t_d} = \sqrt{\dfrac{2}{3}}$
Correct answer is option B.

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Question 10
1 / -0

A and B start walking towards each other from the opposite ends of a 15 km long straight road, at a speed of 5 km/hr and 7 km/hr respectively. How far apart will they be after one hour?

A
2 km

B
3 km

C
5 km

D
7 km

Solution

$x_A=v_At=5\times 1=5km$
$x_B=v_Bt=7\times 1=7km$
Total distance covered $=12km$
Distance between $A$ and $B$ at time $t=1h$
$15-12=3km$

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Motion in A Straight Line Test - 44

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Motion in A Straight Line Test - 44

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Question 1

vA=vB=vCv_{A} = v_{B} = v_{C}vA​=vB​=vC​

vA>vB>vCv_{A} > v_{B} > v_{C}vA​>vB​>vC​

vA=vB>vCv_{A} = v_{B} > v_{C}vA​=vB​>vC​

vA>vB=vCv_{A} > v_{B} = v_{C}vA​>vB​=vC​

Solution

Question 2

0.5m/s20.5 m/s^20.5m/s2

2m/s22 m/s^22m/s2

1m/s21m/s^21m/s2

3m/s23 m/s^23m/s2

Solution

Question 3

2.5 ms−12.5\ { ms }^{ -1 }2.5 ms−1

5 ms−15\ { ms }^{ -1 }5 ms−1

10 ms−110\ { ms }^{ -1 }10 ms−1

20 ms−120\ { ms }^{ -1 }20 ms−1

Solution

Question 4

23mvo3/2,12vo3/2s\cfrac { 2 }{ 3 } m v_o^{3/2},\cfrac { 1 }{ 2 } v_o^{3/2} s32​mvo3/2​,21​vo3/2​s

32amvo1/2,12avo1/2s\cfrac { 3 }{ 2a } m v_o^{1/2},\cfrac { 1 }{ 2a } v_o^{1/2}s2a3​mvo1/2​,2a1​vo1/2​s

3a2mvo1/2,a2vo3/2s\cfrac { 3a }{ 2 } m v_o^{1/2},\cfrac { a }{ 2 } v_o^{3/2}s\quad 23a​mvo1/2​,2a​vo3/2​s

23amvo3/2,2avo1/2s\cfrac { 2 }{ 3a } m v_o^{3/2},\cfrac { 2 }{ a } v_o^{1/2} s\quad 3a2​mvo3/2​,a2​vo1/2​s

Solution

Question 5

666%

252525%

505050%

72.572.572.5%

Solution

Question 6

606060

454545

808080

505050

Solution

Question 7

87.5 m

90 m

50 m

100 m

Solution

Question 8

Both the frames are inertal

S1S_{1}S1​ is inertial and S2S_{2}S2​ is noninertial

S1S_{1}S1​ is noninertial and S2S_{2}S2​ is inerital

none of these

Solution

Question 9

1:1

23\sqrt { \dfrac { 2 }{ 3 } }32​​

23\dfrac { 2 }{ 3 }32​

32\sqrt { \dfrac { 3 }{ 2 } }23​​

Solution

Question 10

2 km

3 km

5 km

7 km

Solution

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$v_{A} = v_{B} = v_{C}$

$v_{A} > v_{B} > v_{C}$

$v_{A} = v_{B} > v_{C}$

$v_{A} > v_{B} = v_{C}$

$0.5 m/s^2$

$2 m/s^2$

$1m/s^2$

$3 m/s^2$

$2.5\ { ms }^{ -1 }$

$5\ { ms }^{ -1 }$

$10\ { ms }^{ -1 }$

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

$\cfrac { 2 }{ 3 } m v_o^{3/2},\cfrac { 1 }{ 2 } v_o^{3/2} s$

$\cfrac { 3 }{ 2a } m v_o^{1/2},\cfrac { 1 }{ 2a } v_o^{1/2}s$

$\cfrac { 3a }{ 2 } m v_o^{1/2},\cfrac { a }{ 2 } v_o^{3/2}s\quad$

$\cfrac { 2 }{ 3a } m v_o^{3/2},\cfrac { 2 }{ a } v_o^{1/2} s\quad$

$6$ %

$25$ %

$50$ %

$72.5$ %

$60$

$45$

$80$

$50$

$S_{1}$ is inertial and $S_{2}$ is noninertial

$S_{1}$ is noninertial and $S_{2}$ is inerital

$\sqrt { \dfrac { 2 }{ 3 } }$

$\dfrac { 2 }{ 3 }$

$\sqrt { \dfrac { 3 }{ 2 } }$