Motion in A Straight Line Test

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Motion in A Straight Line Test - 20

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Question 1
1 / -0

A ball dropped from a height of $$10$$ m, rebounds to a height of $$2.5$$ m. If the ball is in contact with the floor for $$0.01$$ second, its acceleration during contact is ($$g = 9.8$$ $$ms^{-2}$$)

A
$$20 $$ $$ms^{-2}$$

B
$$21 $$ $$ms^{-2}$$

C
$$210$$ $$ms^{-2}$$

D
$$2100 $$ $$ms^{-2}$$

Solution

When ball is hitting the ground,
It's velocity, $$v$$ is $$= \sqrt {2gh} = \sqrt {2 \times 9.8 \times 10}$$
After rebounding up its velocity be $$u$$.
$$u^2 = 2gh \Rightarrow \sqrt {2 \times g \times 2.5}$$
So, $$ v = u + at$$
$$\Rightarrow \sqrt {2 \times 9.8 \times 10} = - \sqrt {2 \times g \times 2.5} + a (0.01)$$
$$\Rightarrow \dfrac {\sqrt {2 \times 9.8}}{0.01} (\sqrt {10} + \sqrt {2.5}) = a$$
$$\Rightarrow a = 2100\ m/s^2$$
Question 2
1 / -0

A car moving with a constant acceleration covers the distance between two points $$180\;m $$ apart in $$6\;s$$. If its speed as it passes the second point is $$45\;ms^{-1}$$, its speed at the first point is

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

B
15 $$ms^{-1}$$

C
30 $$ms^{-1}$$

D
45 $$ms^{-1}$$

Solution

$$180 = u (6) + \frac {1}{2} a (6)^2$$
$$180 = 6u + 18a$$
$$30 = u +3a .................. (1)$$
$$45 = u + at$$
$$45 = u + a (6) ................. (2)$$
$$eq^n (2) - eq^n (1)$$
$$\Rightarrow 15 = 3a \Rightarrow a = 5$$
$$u = 15 m/s$$
Question 3
1 / -0

A sharp stone of mass $$2$$ kg falls from a height of $$10$$ m on sand and buries into the sand. It comes to rest in a time of $$0.029$$ second. The depth through which it buries into sand is

A
0.2 m

B
0.15 m

C
0.25 m

D
0.30 m

Solution

Velocity attained just before contact:
$$v = \sqrt {2gh} = \sqrt {2 \times 9.8 \times 10} = \sqrt {196} = 14\ m/s$$
$$u_1 = v$$
$$v_1 = u_1 + at$$
$$0 = 14 + a (0.029)$$
$$a = -(\dfrac {14}{0.029})$$

$$s = ut + \dfrac {1}{2} at^2 = 14 (0.029) + \dfrac {1}{2} (\dfrac {-14}{0.029})(0.029)^2$$
$$s = 0.029 (14 + \dfrac {1}{2} (-14))$$
$$ = 0.029 \times 7$$
$$ = 0.203\ m$$
$$= 0.2\ m$$
Question 4
1 / -0

A bus is moving along a straight road with a uniform acceleration. It passes through two points A and B separated by a certain distance with velocity of $$30$$ kmph and $$40$$ kmph respectively.Velocity of the car exactly midway between A and B is

A
38.3 kmph

B
35.35 kmph

C
33.3 kmph

D
None

Solution

$$v = u + at$$
$$40 = 30 + at$$
$$at = 10$$
$$v^2 = u^2 + 2as$$
$$40^2 = 30^2 + 2as$$
$$2as = 700$$

Now at midway $$s^1 = \dfrac {s}{2}$$
$$v^2 = u^2 + 2a \dfrac {s}{2}$$
$$v^2 = 30^2 + \dfrac {700}{2} = 900 + 350$$
$$v^2 = 1250$$
$$v = \sqrt {1250}=35.35\ kmph$$
Question 5
1 / -0

A body travels $$200\;cm$$ in the first two seconds with cosntant speed and $$220\;cm$$ in the next $$4$$ seconds with constant deceleration. The speed of the body at the end of the $$7^{th}$$ second is

A
$$22.5 cm/s$$

B
$$12.5 cm/s$$

C
$$57.5 cm/s$$

D
$$0 cm/s$$

Solution

Let the initial velocity of the body be $$U$$.

In the first $$2$$ seconds, $$200 = 2U \implies U = 100\ cm/s$$
In the next $$4$$ seconds, $$220 = 4U - \dfrac{1}{2} 16a = 4U - 8a = 4(100) - 8a \rightarrow a= 22.5\ cm/s^2 $$

Velocity of the body at the end of $$7^{th}$$ second is $$V= U - at = 100 - ( 22.5) ( 7) = - 57.5\ cm/s $$
Question 6
1 / -0

While moving with uniform acceleration, a body has covered $$550 \;m$$ in $$10$$ second and attained a velocity of $$105 \;ms^{-1}$$. Its initial velocity $$u$$ and acceleration $$a$$ respectively are

A
$$10 ms^{-1}, 5 ms^{-2}$$

B
$$10 ms^{-1}, -5 ms^{-2}$$

C
$$5 ms^{-1}, 10 ms^{-2}$$

D
$$10 ms^{-1}, 0 ms^{-2}$$

Solution

$$v = u +at$$
$$105 = u + 10a \ \ \ ..... (1)$$
$$550 = u (10) + \dfrac {1}{2} \times a \times 10^2$$
$$\Rightarrow 550 = 10u + 50 a \Rightarrow 55 = u + 5a \ \ \ ..... (2)$$
$$eq^n (1) - eq^n (2)$$
$$\Rightarrow 50 = 5a$$
$$a = 10\ m/s^2$$
$$u = 5\ m/s$$
Question 7
1 / -0

The average velocity of a body moving with uniform acceleration after travelling a distance of $$3.06\ m$$ is $$0.34\ {m/s}$$. The change in velocity of the body is $$0.18\ {m/s}$$. During this time, its acceleration is

A
$$0.01\ {m/s}^{2}$$

B
$$0.02\ {m/s}^{2}$$

C
$$0.03\ {m/s}^{2}$$

D
$$0.04\ {m/s}^{2}$$

Solution

$$V_{avg} = \dfrac {D}{t} \Rightarrow 0.34 = \dfrac {3.06}{t}$$

$$\Rightarrow t = \dfrac {3.06}{0.34}$$ = $$9\ sec$$

We have, $$v = u + at$$
$$v - u = at$$
$$0.18 = a \times 9$$

$$\Rightarrow a = \dfrac {0.18}{9} = 0.02 \ m/s^2$$
Question 8
1 / -0

A body travels $$200\;m$$ in the first two second and $$220\;m$$ in the next four second. The velocity at the end of the seventh second from the start will be

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

B
15 $$ms^{-1}$$

C
220 $$ms^{-1}$$

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

Solution

$$200 = u(2) + \dfrac {1}{2} \times a (2)^2 = 2u + 2a = 200$$
$$\Rightarrow u + a =100 - (1)$$

Distance in next $$4\ sec$$ $$ \Rightarrow s_6 - s_2$$
$$\Rightarrow 220 = (u (6)] + \dfrac {1}{2} a (6)^2) - (u (2) + \dfrac {1}{2} a (2)^2)$$
$$420 = 6u + 18a - (2)$$
$$420 = 6 (100 - a) + 18a (from 1)$$
$$\Rightarrow -180 = 12a \Rightarrow a = - 15 m/s^2 , u = 115 m/s$$

$$v\ (at \;end \;of \;7 \;sec) = u + at$$$$= 115 + (-15 \times 7) = 10 m/s$$
Question 9
1 / -0

A ball after having fallen from rest under the influence of gravity for $$6s$$, crashes through a horizontal glass plate, thereby losing two-third of its velocity. Then it reaches the ground in $$2s$$, height of the plate above the ground is

A
$$19.6m$$

B
$$39.2m$$

C
$$58.8m$$

D
$$78.4m$$

Solution

$$v_B = v_A + gt = 0 + g(6) = 6g$$
Now, $$\dfrac {2}{3}^{rd} $$ of the velocity is lost.
Hence, $$ \dfrac {1}{3} \times 6g = 2g $$ velocity is left.
$$h = ut + \dfrac {1}{2} gt = (2g) (2) + \dfrac {1}{2} g (2)^2 = 6g$$$$= 58.8\ m$$
Question 10
1 / -0

A ball is thrown straight upward with a speed $$v$$ from a point $$h$$ meters above the ground. The time taken for the ball to strike the ground is

A
$$\dfrac{v}{g}\left [1+\sqrt{1+\dfrac{2hg}{v^{2}}} \right ]$$

B
$$\dfrac{v}{g}\left [1-\sqrt{1-\dfrac{2hg}{v^{2}}} \right ]$$

C
$$\dfrac{v}{g}\left [1-\sqrt{1+\dfrac{2hg}{v^{2}}} \right ]$$

D
$$\dfrac{v}{g}\left [{2+\dfrac{2hg}{v^{2}}} \right ]$$

Solution

$$s = vt - \dfrac {1}{2} gt^2$$
Now, at $$C$$ (maximum height point),
$$h^` = \dfrac {v^2}{2g} $$
So, for journey from $$C$$ to $$B$$,
time $$= \sqrt {\dfrac {2(h+h^`)}{g}} = \sqrt {\dfrac {2(h + \dfrac {v^2}{2g})}{g}} = t_2$$
$$\Rightarrow $$ total time $$ = t_1 + t_2 = \dfrac {v}{g} (1 + \sqrt {1 + \dfrac {2hg}{v^2}})$$

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Motion in A Straight Line Test - 20

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Motion in A Straight Line Test - 20

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SHARING IS CARING

Question 1

$$20 $$ $$ms^{-2}$$

$$21 $$ $$ms^{-2}$$

$$210$$ $$ms^{-2}$$

$$2100 $$ $$ms^{-2}$$

Solution

Question 2

10 $$ms^{-1}$$

15 $$ms^{-1}$$

30 $$ms^{-1}$$

45 $$ms^{-1}$$

Solution

Question 3

0.2 m

0.15 m

0.25 m

0.30 m

Solution

Question 4

38.3 kmph

35.35 kmph

33.3 kmph

None

Solution

Question 5

$$22.5 cm/s$$

$$12.5 cm/s$$

$$57.5 cm/s$$

$$0 cm/s$$

Solution

Question 6

$$10 ms^{-1}, 5 ms^{-2}$$

$$10 ms^{-1}, -5 ms^{-2}$$

$$5 ms^{-1}, 10 ms^{-2}$$

$$10 ms^{-1}, 0 ms^{-2}$$

Solution

Question 7

$$0.01\ {m/s}^{2}$$

$$0.02\ {m/s}^{2}$$

$$0.03\ {m/s}^{2}$$

$$0.04\ {m/s}^{2}$$

Solution

Question 8

10 $$ms^{-1}$$

15 $$ms^{-1}$$

220 $$ms^{-1}$$

5 $$ms^{-1}$$

Solution

Question 9

$$19.6m$$

$$39.2m$$

$$58.8m$$

$$78.4m$$

Solution

Question 10

$$\dfrac{v}{g}\left [1+\sqrt{1+\dfrac{2hg}{v^{2}}} \right ]$$

$$\dfrac{v}{g}\left [1-\sqrt{1-\dfrac{2hg}{v^{2}}} \right ]$$

$$\dfrac{v}{g}\left [1-\sqrt{1+\dfrac{2hg}{v^{2}}} \right ]$$

$$\dfrac{v}{g}\left [{2+\dfrac{2hg}{v^{2}}} \right ]$$

Solution

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