Motion in A Straight Line Test

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Motion in A Straight Line Test - 24

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Question 1
1 / -0

An elevator ascends with an upward acceleration of $$0.2\ m/s^{2}$$. At the instant when its upward speed is $$2\ m/s$$, a loose bolt $$5\ m$$ high from the floor drops from the ceiling of the elevator. The time taken by the bolt to strike the floor and the distance it has fallen are (Take $$g=9.8\ m/s^2$$)

A
$$1 s$$, $$1.9 m$$

B
$$1 s$$, $$2.9 m$$

C
$$1 s$$, $$4.9 m$$

D
$$1 s$$, $$3.9 m$$

Solution

$$g=9.8\quad \quad a=0.2\quad \\ { a }_{ rel }=10\quad \\ Now,\quad s=\dfrac { 1 }{ 2 } { a }_{ rel }{ t }^{ 2 }\\ t=\sqrt { \dfrac { 2\times 5 }{ 10 } } =1\quad second.\\ Distance\quad covered\quad by\quad elevator=\dfrac { 1 }{ 2 } \times 0.2\times 1\\ \therefore \quad Displacement\quad of\quad bolt=\quad 5-0.1=4.9\quad meters.$$
Question 2
1 / -0

A boy standing on an open car throws a ball vertically upwards with a velocity of $$9.8\ m/s$$, while moving horizontally with uniform acceleration of $$1\ m/s^{2}$$ starting from rest. The ball will fall behind the boy on the car at a distance of

A
$$1\ m$$

B
$$2\ m$$

C
$$3\ m$$

D
$$4\ m$$

Solution

Initial velocity of the ball
$$u=9.8 m/s$$
$$v=0$$ (at the top)
So, $$v= u-gt$$
$$0= 9.8 - 9.8 t$$
$$t=1$$
So, to reach the height point the ball takes $$1$$ sec.
So, time taken by the ball to come back
$$=2t = 2sec$$
in $$2$$ sec, the car advances
$$S=ut+ \frac{1}{2}+t^2$$
$$= 0 + \frac{1}{2} 1\times4 = 2m$$
So the ball will fall $$2$$ m behind the car.
Question 3
1 / -0

A parachutist after bailing out falls for $$10$$ s without friction. When the parachute opens he descends with an acceleration of $$2\ m/s^{2}$$ against his direction and reached the ground with $$4$$ $$m/s$$. From what height he has dropped himself? $$(g=10\ m/s^{2})$$

A
$$500$$ m

B
$$2496$$ m

C
$$2996$$ m

D
$$4296$$ m

Solution

$$u=gt\Rightarrow u=10g$$
$$h_{1}=\dfrac{1}{2}gt^{2}$$
$$h_{2}=\dfrac{v^{2}-u^{2}}{-2a}$$
total height $$=h_{1}+h_{2}$$

$$v = gt_{1}                                 (100)^{2} = u^{2} +2 \times 10 S_{1}$$
$$v = 10 \times 10 = 100 m/s. [S_{1} = 500 mt]$$
$$4 = 100-2 t_{2}$$
   $$t_{2} = 48 sec$$
$$\Rightarrow S_{2} = 100 t_{2} - 1/2 \times 2 t_{2}^{2}$$
   $$ = 4800-(48)^{2}$$
   $$ = 2496 $$ mt.
$$S_{1} + S_{2}= 2996$$ mt
Question 4
1 / -0

A $$5\ kg$$ stone falls from a height of $$1000\ m$$ and penetrates $$2\ m$$ in a layer of sand. The time of penetration is

A
14.285 s

B
0.0285 s

C
7.146 s

D
0.285 s

Solution

Speed of the stone just before touching the sand is
$$v=\sqrt {2gh}=\sqrt {2\times 9.81\times 1000}=140.07m/s$$
Deceleration of the stone due to sand is
$$v^2-u^2=-2as$$
$$0-140.07^2=-2(a)(2)$$
$$a=4905m/s^2$$
We have initial speed and the deceleration, from
$$v=u-at$$
$$u=at$$
$$\displaystyle t=\frac {u}{a}=\frac {140.07}{4905}=0.0285s$$
Option B.
Question 5
1 / -0

A ball of mass $$100$$ gm is projected vertically upward from the ground with a velocity of $$50 ms^{-1}$$ . At the same time another identical ball is dropped from a height of $$100$$ m to fall freely along the same path as that followed by the first ball. After some time the two balls collide, stick together and finally fall to the ground. The time taken by the combined mass to fall to the ground is approximately $$(g=10 ms^{-2})$$

A
4.5 s

B
6.5 s

C
9 s

D
13 s

Solution

$${V_{1}}^{2} = (50)^{2} - 2gx $$
$${V_{2}}^{2} = 2g(100 - x)$$
$$\Rightarrow {V_{2}}^{2} = 2g(100) + {V_{1}}^{2} - (50)^{2}$$
$$\Rightarrow {V_{2}}^{2} - {V_{1}}^{2} = 2000 - 2500$$
$$\Rightarrow (V_{1} + V_{2})(V_{1} - V_{2}) = \sqrt{-500}$$
$$V_{1} = 50 - gt$$
$$V_{2} = gt$$
$$\Rightarrow V_{1} + V_{2} = 50$$
$$\Rightarrow V_{1} - V_{2} = +10$$
$$V_{1} = 30 m/s (\uparrow)$$
$$V_{2} = 230 m/s (\downarrow)$$
$$\Rightarrow x =\dfrac{900-2500}{20} = 80$$ m
$$80 = -5t + \dfrac{1}{2}gt^{2}$$
$$t^{2} - t - 16 = 0$$
$$t = \dfrac{+1\pm \sqrt{1 + 64}}{2} = \dfrac{1\pm \sqrt{65}}{2}$$
$$=\dfrac{\sqrt{65}+1}{2}\approx 4.5$$ sec
Question 6
1 / -0

A ball is projected from the bottom of a tower and is found to go above the tower and is caught by the thrower at the bottom of the tower after a time interval $$t_{1}$$. An observer at the top of the tower finds the same ball go up above him and then come back to his level in a time interval $$t_{2}$$ .The height of the tower is

A
$$\frac{1}{2}gt_{1}t_{2}$$

B
$$\dfrac{gt_{1}t_{2}}{8}$$

C
$$\dfrac{g}{8}\left ( t^{2}_{1}-t^{2}_{2} \right )$$

D
$$\dfrac{g}{2}(t_{1}-t_{2})^{2}$$

Solution

$$v=u+at\\ \Rightarrow u=\dfrac { g{ t }_{ 1 } }{ 2 } \\ s=ut+\dfrac { 1 }{ 2 } a{ t }^{ 2 }\\ \Rightarrow s=\dfrac { g{ t }_{ 1 } }{ 2 } \dfrac { ({ t }_{ 1 }-{ t }_{ 2 }) }{ 2 } -\dfrac { g }{ 2 } \dfrac { { ({ t }_{ 1 }-{ t }_{ 2 }) }^{ 2 } }{ 4 } \\ \Rightarrow s=\dfrac { g({ t }_{ 1 }^{ 2 }-{ t }_{ 2 }^{ 2 }) }{ 8 } $$
Question 7
1 / -0

A particle starts from rest with uniform acceleration $$a$$. Its velocity after $$n$$ second is $$v$$. The displacement of the body in the last two second is

A
$$\dfrac{2v\left ( n-1 \right )}{n}$$

B
$$\dfrac{v\left ( n-1 \right )}{n}$$

C
$$\dfrac{v\left ( n+1 \right )}{n}$$

D
$$\dfrac{2v\left ( 2n+1 \right )}{n}$$

Solution

We know,
At $$t = n $$sec
$$s_n = \dfrac{1}{2}an^2$$
$$v^2 = 2as_n$$
$$= 2a\cdot \dfrac{1}{2}an^2$$
$$= a^2n^2$$
$$\therefore v = an $$
$$\therefore a = \dfrac{v}{n}$$ ...(i)
Now, $$s_n-s_{n-2} = \dfrac{1}{2}a\left[n^2-(n-2)^2\right]$$
$$= \dfrac{1}{2}a\left[4n-4\right]$$
$$= 2a(n-1)$$ from (i)
$$s_n - s_{(n-2)} = \displaystyle \dfrac{2v(n-1)}{n}$$
Question 8
1 / -0

A ball is dropped from a high rise platform at $$t =0$$ starting from rest. After $$6$$ seconds another ball is thrown downwards from the same platform with a speed $$v$$. The two balls meet at $$t = 18\ s$$. What is the value of $$v$$ ?
(take $$g = 10\ m/s^{2}$$)

A
$$75\ m/s$$

B
$$55\ m/s$$

C
$$40\ m/s$$

D
$$60\ m/s$$

Solution

Given , At t= 18 s ball meets
$$\Rightarrow $$ Distance covered by ball 1$$(S_1)$$ = Distance covered by ball 2$$(S_2)$$
Now, Using $$ S= ut +\dfrac{1}{2}at^2$$
$$ S_1 = \dfrac{1}{2}gt^2 = \dfrac{1}{2} 10(18)^2 = 1620m$$
And, $$S_2 = vt+\dfrac{1}{2}gt^2 = 12v + 720$$
But, $$ S_1 = S_2 $$
$$\Rightarrow 1620 = 720 +12v$$
$$\Rightarrow 900 = 12v$$
$$\Rightarrow v= 75 m/s$$
Therefore, A is correct option.
Question 9
1 / -0

Two balls $$A$$ and $$B$$ are thrown with same velocity $$u$$ from the top of a tower Ball $$A$$ is thrown vertically upward and the ball $$B$$ is thrown vertically downward. Choose the correct statement.

A
Ball $$B$$ reaches the ground with greater velocity

B
Ball$$A$$ reaches the ground with greater velocity

C
Both the balls reach the ground with same velocity

D
Cannot be interpreted

Solution

We know,
The direction of motion while touching the ground will be vertically downwards,
Now coming to the speed,
Speed is nothing but a reflection of Kinetic energy of a body,
More the kinetic energy more the speed.
Taking the ball-earth system,
Mechanical energy is conserved,

Gain in Kinetic energy=loss in potential energy

And as both the balls are at same height,
Irrespective of the initial direction of throw,
Both will lose same potential energy and gain the same Kinetic energy,
Hence having the same velocity when they are near the ground.

Option $$\textbf C$$ is the correct answer
Question 10
1 / -0

A stone is dropped into a well in which the water level is h below, the top of the well. If $$v$$ is velocity of sound, then time $$T$$ after which the splash is heard is equal to

A
$$\dfrac{2h}{v}$$

B
$$\sqrt{\dfrac{2h}{v}}+\dfrac{h}{g}$$

C
$$\sqrt{\dfrac{2h}{g}}+\dfrac{h}{v}$$

D
$$\sqrt{\dfrac{h}{2g}}+\dfrac{2h}{v}$$

Solution

Time taken for the sound to reach to the top = Time taken for the stone to reach the ground + Time taken for the sound to reach the top.

$$h = ut + \dfrac{1}{2}at^2 $$ (u is zero since the stone is freely falling)
$$h = \dfrac{1}{2}at^2 $$
$$ t_g = \sqrt{\dfrac{2h}{g}}$$

Time taken for the sound to reach the top is
$$t_c = \dfrac{h}{v} $$
Total time taken $$= \sqrt{\dfrac{2h}{g}} + \dfrac{h}{v} $$

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Motion in A Straight Line Test - 24

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Motion in A Straight Line Test - 24

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SHARING IS CARING

Question 1

$$1 s$$, $$1.9 m$$

$$1 s$$, $$2.9 m$$

$$1 s$$, $$4.9 m$$

$$1 s$$, $$3.9 m$$

Solution

Question 2

$$1\ m$$

$$2\ m$$

$$3\ m$$

$$4\ m$$

Solution

Question 3

$$500$$ m

$$2496$$ m

$$2996$$ m

$$4296$$ m

Solution

Question 4

14.285 s

0.0285 s

7.146 s

0.285 s

Solution

Question 5

4.5 s

6.5 s

9 s

13 s

Solution

Question 6

$$\frac{1}{2}gt_{1}t_{2}$$

$$\dfrac{gt_{1}t_{2}}{8}$$

$$\dfrac{g}{8}\left ( t^{2}_{1}-t^{2}_{2} \right )$$

$$\dfrac{g}{2}(t_{1}-t_{2})^{2}$$

Solution

Question 7

$$\dfrac{2v\left ( n-1 \right )}{n}$$

$$\dfrac{v\left ( n-1 \right )}{n}$$

$$\dfrac{v\left ( n+1 \right )}{n}$$

$$\dfrac{2v\left ( 2n+1 \right )}{n}$$

Solution

Question 8

$$75\ m/s$$

$$55\ m/s$$

$$40\ m/s$$

$$60\ m/s$$

Solution

Question 9

Ball $$B$$ reaches the ground with greater velocity

Ball$$A$$ reaches the ground with greater velocity

Both the balls reach the ground with same velocity

Cannot be interpreted

Solution

Question 10

$$\dfrac{2h}{v}$$

$$\sqrt{\dfrac{2h}{v}}+\dfrac{h}{g}$$

$$\sqrt{\dfrac{2h}{g}}+\dfrac{h}{v}$$

$$\sqrt{\dfrac{h}{2g}}+\dfrac{2h}{v}$$

Solution

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