Motion in A Straight Line Test

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Motion in A Straight Line Test - 13

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Question 1
1 / -0

Two points $$A$$ and $$B$$ move from rest along a straight line with same acceleration $$f$$ and $$f'$$ respectively. If $$A$$ takes $$m\ \text{sec}$$ more than $$B$$ and describes $$n$$ units more than $$B$$ in acquiring the same speed then

A
$$(\mathrm{f}-\mathrm{f}')\mathrm{m}^{2}=\mathrm{f}\mathrm{f}'\mathrm{n}$$

B
$$(\mathrm{f}+\mathrm{f}')\mathrm{m}^{2}=\mathrm{f}\mathrm{f}'\mathrm{n}$$

C
$$\displaystyle \frac{1}{2}(\mathrm{f}+\mathrm{f}')\mathrm{m}=\mathrm{f}\mathrm{f}'\mathrm{n}^{2}$$

D
$$(f^{'}-f)n=\frac{1}{2}ff^{'}m^{2}$$

Solution

$$\mathrm{v}^{2}=2\mathrm{f}(\mathrm{d}+\mathrm{n})=2\mathrm{f}'\mathrm{d}$$
$$\mathrm{v}=\mathrm{f}'(\mathrm{t})=(\mathrm{m}+\mathrm{t})\mathrm{f}$$

eliminate $$\mathrm{d}$$ and $$\mathrm{m}$$ we get $$(f^{'}-f)n=\frac{1}{2}ff^{'}m^{2}$$
Question 2
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A lizard, at an initial distance of $$21$$ cm behind an insect, moves from rest with an
acceleration of $$2 cm/s^{2}$$ and pursues the insect which is crawling uniformly along a
straight line at a speed of $$20$$ cm/s. Then the lizard will catch the insect after

A
$$20$$ s

B
$$1$$ s

C
$$21$$ s

D
$$24$$ s

Solution

Let t be the time in which the lizard will catch the insect.

Distance traveled by lizard $$=$$ 21 + distance traveled by insect.

$$ut+\frac { 1 }{ 2 } a{ t }^{ 2 }=21+vt$$

$$0+\frac { 1 }{ 2 } 2{ t }^{ 2 }=21+20t$$

$${ t }^{ 2 }-20t-21=0$$

solving this $$t= 21 , -1 $$

t cannot be -ve.

$$t=21 sec.$$
Question 3
1 / -0

Two stones are thrown up simultaneously from the edge of a cliff $$240\ m$$ high with initial speed $$10\ m/s$$ and $$40\ m/s$$ respectively. Which of the following graph best represents the time variation of relative position of the second stone with respect to the first ?
(Assume stones do not rebound after hitting the ground and neglect air resistance, take $$g = 10\ m /s^2$$)
(The figures are schematic and not drawn to scale.)

A

B

C

D

Solution

For the first ball,
$$-240 = 10t -\frac{1}{2}gt^2$$
$$ \therefore 5t^2-10t -240$$
$$\therefore t^2-2t-48 =0$$
$$\Rightarrow t=8,-6$$
The 1st particle will reach the ground in 8 secs.
Upto 8 secs, the relative velocity between the particles is 30m/sec and the relative acceleration is zero..
For the 2nd particle,
$$-240 = 40t -5t^2$$
$$ \Rightarrow t^2 -8t -48 =0$$
$$ t =12\: secs$$
The second particle will strike the ground in 12 secs.
Option(A): Linear behaviour after 8 secs, hence can be eliminated.
Option(B) shows increase until 12 secs , hence is not correct. after 8 secs, separation will decrease.
Option(c) is the correct one.
Question 4
1 / -0

STATEMENT-1: For an observer looking out through the window of a fast moving train, the nearby objects appear to move in the opposite direction to the train, while the distant objects appear to be stationary.

STATEMENT-2: If the observer and the object are moving at velocities $$\vec{\mathrm{V}}_{1}$$ and $$\vec{\mathrm{V}}_{2}$$ respectively with reference to a laboratory frame, the velocity of the object with respect to the observer is $$\vec{\mathrm{V}}_{2}-\vec{\mathrm{V}}_{1}$$.

A
STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1

B
STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT-1

C
STATEMENT -1 is True, STATEMENT-2 is False

D
STATEMENT -1 is False, STATEMENT-2 is True

Solution

Both the statements are true,but correct explanation of statement one is
we,know that $${\theta}=\frac{arc}{radius}$$
so value of $${\theta}$$ will be more for nearby object and radius will be less wile value of $${\theta}$$ will be less for distant object and radius will be more.
that is the reason behind this phenomenon and near one will move faster and distant object will move relatively less(i.e both are moving w.r.t any frame of refrence)
Question 5
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A toy car with charge q moves on a frictionless horizontal plane surface under the influence of a uniform electric field $$\overrightarrow{E}$$. Due to the force $$q\overrightarrow{E}$$, its velocity increases from $$0$$ to $$6\ m/s$$ in one-second duration. At that instant, the direction of the field is reversed. The car continues to move for two more seconds under the influence of this field. The average velocity and the average speed of the toy car between $$0$$ to $$3$$ seconds are respectively:

A
$$1$$ m/s, $$3.5$$ m/s

B
$$2$$ m/s, $$4$$ m/s

C
$$1.5$$ m/s, $$3$$ m/s

D
$$1$$ m/s, $$3$$ m/s

Solution

Acceleration toward +ve
x-direction in first case is a
$$v-u=at$$
$$a=\frac{6}{1}=6m/s^{2}$$
when field reserved acceleration changed to
$$a=-6m/s^{2}$$
$$x=\frac{1}{2}at_{1}^{2}=\frac{1}{2}*6*1=3m$$
$$y=vt+\frac{1}{2}at_{2}^{2}=6*2+\frac{1}{2}(-6)*2^{2}$$
$$y=12-12=0$$
$$x_{2}=$$ distance traveled is 1 sec
$$x_{2}=6*1-\frac{1}{2}(6)*1$$
$$=3m$$
total distance cover $$=x_{1}+2x_{2}=9m$$
Total dispalcement $$=3m$$

Average velocity = $$\dfrac{displacement \ cover}{time \ taken}=\dfrac{3}{3}=1 \ m/s$$

Average speed $$=\dfrac{distance \ cover}{time \ taken}=\dfrac{9}{3}=3 \ m/s$$
Question 6
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A ball is dropped from a bridge $$122.5 m$$ high. After the first ball has fallen for $$2 seconds$$, a second ball is thrown straight down after it, what must be the initial velocity of the second ball be, so that both the balls hit the surface of water at the same time?

A
$$26.1 {m}/{s}$$

B
$$9.8 {m}/{s}$$

C
$$55.5 {m}/{s}$$

D
$$49 {m}/{s}$$

Solution

Time taken by the first object to reach the ground $$= t$$, so

$$122.5 = ut + \displaystyle\frac{1}{2} g{t}^{2}$$

$$122.5 = \displaystyle\frac{1}{2} \times 10 \times {t}^{2}$$

$$\Rightarrow t = 5 sec$$ (approx)

Time to be taken by the second ball to reach the ground $$= 5 2 = 3 sec$$.

If $$u$$ be its initial velocity then,

$$122.5 = u \times 3 + \displaystyle\frac{1}{2} g{t}^{2} = 3u + \displaystyle\frac{1}{2} \times 10 \times 9$$

$$3u = 122.5 -45 = 77.5$$

$$u = 26$$ (approx)
Question 7
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If a car at rest accelerated uniformly to a speed of $$144 {km}/{hour}$$ in $$20 second$$ it covers a distance:

A
$$400 m$$

B
$$1440 m$$

C
$$2880 m$$

D
$$25 m$$

Solution

$$u = 0, v = 144   {km}/{hour}= 144 \times \displaystyle\frac{5}{18} {m}/{sec} = 40 {m}/{sec}$$
$$v = u + at$$
$$\Rightarrow   a = \displaystyle\frac{v-u}{t} = \displaystyle\frac{40-0}{20} = 2 {m}/{{sec}^{2}}$$
$$\therefore     s = ut + \displaystyle\frac{1}{2} a{t}^{2}$$
$$= \displaystyle\frac{1}{2} \times 2 \times {\left(20\right)}^{2} = 400   m$$
Question 8
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A particle of mass $$2\ m$$ is projected at an angle $$45^{\circ}$$ with the horizontal with a velocity of $$20\sqrt {2}m/ s$$. After $$1s$$, explosion takes place and the particle is broken into the two equal pieces. As a result of explosion, one part comes to rest. The maximum height from the ground attained by the other part is

A
$$50\ m$$

B
$$25\ m$$

C
$$40\ m$$

D
$$35\ m$$

Solution

Given: Initial velocity $$u_{0} = 20\sqrt {2}m/s$$; angle of projection $$\theta = 45^{\circ}$$
Therefore horizontal and vertical components of initial velocity are
$$u_{x} = 20\sqrt {2}\cos 45^{\circ} = 20\ m/s$$
and $$u_{y} = 20\sqrt {2}\sin 45^{\circ} = 20\ m/s$$
After $$1s$$, horizontal component remains unchanged while the vertical component becomes
$$v_{y} = u_{y} - gt$$
Due to explosion, one part comes to rest.
Hence, from the conservation of linear momentum, vertical component of second part will become $$v_{y}' = 20 m/s$$.
Therefore, maximum height attained by the second part will be
$$H = h_{1} + h_{2}$$
Using, $$h = ut + \dfrac {1}{2} at^{2}$$
$$\Rightarrow h_{1} = (20\times 1) - \dfrac {1}{2} \times 10\times (1)^{2} = 15\ m$$
$$a = g = 10\ m/s^{2}$$
$$h_{2} = \dfrac {v'^{2}_{y}}{2g} = \dfrac {(20)^{2}}{2\times 10} = 20 m$$
$$H = 20 + 15 = 35\ m$$.
Question 9
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A ball is thrown vertically upwards with a given velocity such that it rises for T seconds (T > 1). What is the distance traversed by the ball during the last one second of ascent (in meters) ? (Acceleration due to gravity is g $$m/s^{2}$$ ) .

A
$$v+g\dfrac{(1-2T)}{2}$$

B
$$ vT + \frac {1}{2} g [T^2 - (T-1)^2] $$

C
$$\dfrac {g}{2}$$

D
$$\frac {1}{2} g [T^2 - (T-1)^2] $$

Solution
Question 10
1 / -0

A body starts from rest and acquires a velocity $$10m s^{-1}$$ in 2 s. Find the acceleration.

A
$$10 m s^{-2}$$

B
$$20 m s^{-2}$$

C
$$5 m s^{-2}$$

D
$$40 m s^{-2}$$

Solution

Acceleration = Change in velocity per unit time.

Therefore, Acceleration $$a = \dfrac{10 m/s}{2 s} = 5 m/s^2 $$