Motion in A Straight Line Test

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Motion in A Straight Line Test - 26

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Question 1
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Two stones are dropped down simultaneously from different heights. At the starting time, the distance between them is 30 cm. After 1 s, the distance between the two stones will be $$(g = 10 m s^{-2}).$$

A
$$10 cm$$

B
$$20 cm$$

C
$$30 cm$$

D
$$0 cm$$

Solution

After 2 s or any difference in seconds, separation will be 30 cm only as both bodies covers same distance for same time interval under gravity.
Question 2
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A gun is fired at a target. At the moment of firing, the target is released and allowed to fall freely under gravity. Then the bullet :
(Assume zero air resistance)

A
Misses the target by passing above it

B
Hits the target

C
Misses the target by passing below it

D
May or may not hit

Solution

Initial vertical component of the velocities (in downward direction) of both the bullet and the target are zero. So, the bullet and the target fall down by equal amount and thus the bullet hits the target.
Question 3
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A body A is thrown vertically upwards with such a velocity that it reaches a maximum height of $$h$$ in time $$t$$. Simultaneously another body B is dropped from height $$h$$. It strikes the ground and does not rebound. The velocity of A relative to B v/s time graph is best represented by (upward direction is positive)

A

B

C

D

Solution

For the first half, when ball A goes up
$$V_A=U_A-gT$$
$$V_B=-gT$$
$$V_{AB}=V_A-V_B$$
$$V_{AB}=U$$.......$$(1)$$
For the second half, when ball A comes down
$$V_A=-gT$$
$$V_B=0$$
$$V_{AB}=V_A-V_B$$
$$V_{AB}=-gT$$........$$(2)$$
Hence from the equation $$(1)$$ we observe that relative velocity is independent of $$T$$. Which will be valid till ball A reaches maximum height point. It is given time taken in reaching $$h$$ is $$t$$.
Hence for the time $$t$$, $$V_{AB}$$ will be constant, whereas after time $$t$$, relative velocity changes its direction and increases afterwards.
Question 4
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A body standing on a long railroad car throws a ball straight upwards, the car is moving on the horizontal road with an acceleration $$1\ m{ s }^{ -2 }$$. The vertical velocity given is $$9.8\ m{ s }^{ -1 }$$. How far behind the boy the ball will fall on the railroad car?

A
$$1\ m$$

B
$$\cfrac {3}{2}\ m$$

C
$$\cfrac {7}{4}\ m$$

D
$$2\ m$$

Solution

For vertical direction we see that the relative distance travelled is zero. Thus
$${ s }_{ rel }=ut+\dfrac { 1 }{ 2 } a{ t }^{ 2 }\\ \Rightarrow 0=9.8t-\dfrac { 1 }{ 2 } { 9.8t }^{ 2 }\\ \Rightarrow t=2s.$$

Now for horizontal direction we have
$$s=ut+\dfrac { 1 }{ 2 } a{ t^2 }\\ s=0+\dfrac { 4 }{ 2 } =2m$$
Question 5
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The acceleration of a particle as seen from two frames $$S_{1}$$ and $$S_{2}$$ has equal magnitude $$5\ ms^{-2}$$.

A
The frames must be at rest with respect to each other.

B
The frames may be moving with respect to each other but neither should be accelerated with respect to the other.

C
The acceleration of frame $$S_{2}$$ with respect to $$S_{1}$$ be $$0$$ or $$10\ ms^{-2}$$

D
The acceleration of frame $$S_{2}$$ with respect to $$S_{1}$$ lies between $$0$$ and $$10\ ms^{-2}$$

Solution

Using parallelogram law,
Let,
$$\vec {S_1} \,\,\, \vec {S_2}$$ be the arms of the parallelogram,
Given,
$$\vec {S_1} = \vec {S_2}$$
the angle between these two vectors may lie between $$0^o$$ to $$180^o$$
Let us find the resultant taking $$\theta = 0^o \,\,\, and \,\,\,\, 180^o$$
$$R = \sqrt{{S_1}^2 + {S_2}^2 - 2.{S_1}{S_2}cos\theta}$$
$$\therefore R = \sqrt{5^2 + 5^2 - 2.(5)(5)cos{0^o}} = 0$$
AND
$$R = \sqrt{5^2 + 5^2 - 2.(5)(5)cos{180^o}} = 10$$
$$\therefore $$ The acceleration of frame $$S_{2}$$ with respect to $$S_{1}$$ lies between $$0$$ and $$10ms^{-2}$$
Question 6
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A flower pot falls off a window sill and falls past the window below. It takes $$0.5s$$ to pass through a $$2.0m$$ high window. Find how high is the window sill from the top of the window?

A
$$10cm$$

B
$$7.5cm$$

C
$$11.25cm$$

D
$$15cm$$

Solution

For motion from $$B$$ to $$C$$

$$t=0.5s, s=BC=2\ m$$

Velocity at $$B$$ is $$ { u }_{ B }$$

$$ s={ u }_{ B }t+\dfrac { 1 }{ 2 } g{ t }^{ 2 }\quad \Rightarrow \quad { u }_{ B }=1.5\ m/s$$

For motion from $$A$$ to $$B$$

$$ s=AB=h, { u }_{ A }=0\quad { v }_{ B }{ =u }_{ B }=1.5\ m/s$$

$$ { v }_{ B }^{ 2 }={ u }_{ A }^{ 2 }+2gh\quad $$ or

$$ \left( 1.5 \right)^2 =0+2\left( 10 \right) h\quad or\quad h=0.1125m=11.25cm$$
Question 7
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A particle is projected vertically upward with a speed $$u$$.

A
When it rises to half its maximum height its velocity is $$\dfrac{u}{2}$$

B
Time taken to rise to half its maximum height is half the time taken to reach maximum height

C
Time taken to rise to three fourth its maximum height is half the time taken to reach maximum height

D
The acceleration of the particle when it reaches its maximum height is zero.

Solution

Time taken by particle to reach maximum height,
v = u - gt
0 = u- gt
t = u/g

we know,
H = $$\dfrac { { u }^{ 2 } }{ 2g } $$
$$\dfrac { 3H }{ 4 } $$ = $$\dfrac { 3{ u }^{ 2 } }{ 8g } $$

$$h = ut - \dfrac { 1 }{ 2 } g{ t }^{ 2 }$$
$$\dfrac { 3{ u }^{ 2 } }{ 8g } = ut -\dfrac { 1 }{ 2 } g{ t }^{ 2 }$$
$$g{ t }^{ 2 }$$-2ut+$$\dfrac { 3{ u }^{ 2 } }{ 4g }=0 $$
on solving quadratic, we get
$$t = \dfrac{u}{2g}$$
Question 8
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From the top of a tower, two stones whose masses are in the ratio $$1:2$$ are thrown, one straight up with an initial speed $$u$$ and the second straight down with same speed $$u$$. neglecting air resistance,

A
the heavier stone hits the ground with a higher speed

B
the lighter stone hits the ground with a higher speed

C
both the stones will have same speed when they hit the ground

D
the speed cannot be determined with the given data

Solution

For motion under gravity, speed is independent of masses
The net vertical displacement for both stone are same. Also they are launched with same initial speed $$u$$, therefore both the stone will have same speed when they hit the ground.
Question 9
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A ball is thrown up with certain velocity so that it reaches a height $$h$$. Find the ratio of the times in which it is at $$\dfrac{h}{3}$$.

A
$$\dfrac { \sqrt { 2 } -1 }{ \sqrt { 2 } +1 } $$

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

C
$$\dfrac { \sqrt { 3 } -1 }{ \sqrt { 3 } +1 } $$

D
$$\cfrac {1} {3}$$

Solution

At maximum height, velocity is zero so, $$ 0={ u }^{ 2 }-2gh\quad or\quad u=\sqrt { 2gh } $$
As $$ s=ut+\dfrac { 1 }{ 2 } a{ t }^{ 2 }$$ so $$, \dfrac { h }{ 3 } =\sqrt { 2gh } t-\dfrac { 1 }{ 2 } g{ t }^{ 2 }$$
$$ g{ t }^{ 2 }-2\sqrt { 2gh } t+\dfrac { 2h }{ 3 } =0\quad or\quad t=\dfrac { 2\sqrt { 2gh } \pm \sqrt { 8gh-\dfrac { 8gh }{ 3 } } }{ 2g } $$
$$ \dfrac { { t }_{ 1 } }{ { t }_{ 2 } } =\dfrac { 2\sqrt { 2gh } -\sqrt { 8gh-\dfrac { 8gh }{ 3 } } }{ 2\sqrt { 2gh } +\sqrt { 8gh-\dfrac { 8gh }{ 3 } } } =\dfrac { \sqrt { 3 } -\sqrt { 3-1 } }{ \sqrt { 3 } +\sqrt { 3-1 } } =\dfrac { \sqrt { 3 } -\sqrt { 2 } }{ \sqrt { 3 } +\sqrt { 2 } } $$
Question 10
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Directions For Questions

The velocity of a particle is the rate at which its position varies with time. The position of a particle in a particular reference frame is given by a position vector drawn from the origin of that frame of reference. At $$t={ t }_{ 1 }$$ assume the particle is at $$A$$ and its position in $$xy$$ plane being described by position vector $$\overrightarrow{ r }_1 $$. At a later time $${ t }_{ 2 }$$ the particle reaches at point $$B$$ as shown in figure and its position vector is $$\overrightarrow{r}_2 $$.
The displacement vector $$\Delta\overrightarrow{r}= \overrightarrow{ r }_2 -\overrightarrow{r}_1$$ describes the change in posittion of the particle from $$A$$ to $$B$$ in time $${ t }_{ 2 }-{ t }_{ 1 }$$. Average velocity of the particle during this interval is
$$\overrightarrow{ v }_{av } =\cfrac { displacement\quad (a\ vector) }{ time\quad elapsed\quad (a\ scalar) } $$

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Which of the following statements regarding motion of particle is true?

A
The motion between $$A$$ and $$B$$ is known

B
The motion between $$A$$ and $$B$$ is erratic

C
The motion between $$A$$ and $$B$$ may have been steady or erratic

D
The motion between $$A$$ and $$B$$ is steady

Solution

the motion between A and B is steady if the velocity of the particle is constant and erratic if it is variable. From the question it is not clear what happens with the particle during the motion so we cannot say whether the particle is in erratic or steady motion.