Motion in A Straight Line Test

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Motion in A Straight Line Test - 29

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Weekly Quiz Competition

Question 1
1 / -0

(a) How long will a stone take to fall to the ground from the top of a building $$80 m$$ high and (b) what will be the velocity of the stone on reaching the ground? (Take $$g=10 m {s}^{-2}$$)

A
(a) $$14$$ $$s$$, (b) $$30$$ $$m$$ $${s}^{-1}$$

B
(a) $$4$$ $$s$$, (b) $$30$$ $$m$$ $${s}^{-1}$$

C
(a) $$10$$ $$s$$, (b) $$40$$ $$m$$ $${s}^{-1}$$

D
(a) $$4$$ $$s$$, (b) $$40$$ $$m$$ $${s}^{-1}$$

Solution

a.) $$s=ut+0.5at^2$$
$$80=0+0.5(10)t^2$$
$$t=4\ s$$
b.) $$v=u+at$$
$$=10\times 4=40\ m/s$$
Question 2
1 / -0

A stone is just released from the window of a moving train moving along a horizontal straight track .When observed by a person on the ground,the stone will hit the ground following a

A
straight line path

B
circular path

C
parabolic path

D
hyperbolic path

Solution

The stone will follow the motion of a projectile, because:
It has a initial horizontal velocity, which is same as that of the train .it acquires a vertical component under the force of gravity .
Since it has a constant speed on horizontal direction and a constant acceleration in vertical direction, the trajectory is parabolic.
Question 3
1 / -0

A ball is thrown vertically upwards. It returns $$6 s$$ later. Calculate:$$(i)$$ the greatest height reached by the ball, and $$(ii)$$ the initial velocity of the ball. (Take $$g=10 m {s}^{-2}$$)

A
(i) $$40$$ $$m$$, (ii) $$30$$ $$m$$ $${s}^{-1}$$

B
(i) $$45$$ $$m$$, (ii) $$30$$ $$m$$ $${s}^{-1}$$

C
(i) $$45$$ $$m$$, (ii) $$60$$ $$m$$ $${s}^{-1}$$

D
(i) $$45$$ $$m$$, (ii) $$20$$ $$m$$ $${s}^{-1}$$

Solution

The ball is thrown up and it returns in $$6 sec$$,

Time of accent = Time of decent

So, time to reach the highest point $$=6/2 = 3 s$$

By 2nd equation of motion

$$s = ut + \dfrac{1}{2}gt^{2}$$

To calculate height, consider motion of the ball from highest point to the ground
$$H = \dfrac{1}{2}gt^{2} = \dfrac{1}{2} \times 10 \times 3^{2} =45\ m$$

To calculate projection velocity, consider motion of the ball from projection to return to the point of projection.

$$0=ut-\dfrac{1}{2}gt^2$$

$$u=10 \times 6/2=30\ m/s$$
Question 4
1 / -0

A ball is thrown from rear end of the compartment of train to the front end which is moving at a constant horizontal velocity. An observer $$A$$ sitting in compartment and another observer $$B$$ standing on the ground draw the trajectory. They will have

A
equal horizontal and equal vertical ranges.

B
equal vertical ranges but different horizontal ranges.

C
different vertical ranges but equal horizontal ranges.

D
different vertical ranges and different horizontal ranges.

Solution

In vertical direction, ball has zero initial velocity but same value of $$g$$ w.r.t observer A & B,so they will draw trajectory of equal vertical range
But along horizontal direction, for observer $$A$$, ball is moving with velocity at which it is thrown w.r.t. train while for $$B $$, it is moving with a velocity equal to train $$+$$ velocity at which it is thrown w.r.t groundtrain
Question 5
1 / -0

A rocket is fired upward from the earth's surface such that it creates an acceleration of $$19.6 { m }{ { s }^{- 2 } }$$. If after $$5 s$$, its engine is switched off, the maximum height of the rocket from earth's surface would be

A
$$980\ m$$

B
$$735\ m$$

C
$$490\ m$$

D
$$245\ m$$

Solution

concept are attached, rest calculation is:
Velocity when the engine is switched off
$$v=19.6\quad *\quad 5\quad =\quad 9.8\quad { m }/{ s }$$
$${ h }_{ max }={ h }_{ 1 }+{ h }_{ 2 }$$
where, $${ h }_{ 1 }=\dfrac { 1 }{ 2 } a{ t }^{ 2\quad }\& \quad { h }_{ 2 }=\dfrac { { v }^{ 2 } }{ 2a } $$

$${ h }_{ max }=\dfrac { 1 }{ 2 } *19.6*5*5+\dfrac { 98*98 }{ 2*9.8 } $$

$$ = 245+490=735 m$$

$$B$$ is correct answer
Question 6
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From a $$200 m$$ high tower, one ball is thrown upwards with speed of $$10 { m }/{ s }$$ and another is thrown vertically downwards at the same speed simultaneously. The time difference of their reaching the ground will be nearest to

A
$$12 s$$

B
$$6 s$$

C
$$2 s$$

D
$$1 s$$

Solution

The ball thrown upward will have zero velocity in$$1\: s$$. It returns back to thrown point in another $$ 1\: s $$ with the same velocity as second. Thus the difference will be $$ 2\: s $$.
Question 7
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A car starting from rest accelerates uniformly to acquire a speed $$20\ kmh^{-1}$$ in 30 min. The distance travelled by car in this time interval will be

A
600 km

B
5 km

C
6 km

D
10 km

Solution

Here, $$u = 0, v = 20 km/h, t = 30 min = 0.5 h$$

Putting these values in $$v=u+at$$
$$20 = 0+a*0.5$$
We get, $$a = 40 km/h^2$$

Putting the values in $$S = ut+\dfrac{1}{2}at^2$$

$$S = 0*0.5+\dfrac{1}{2}*40*0.5^2$$

$$S = 5 km$$
Question 8
1 / -0

A body moves from rest with uniform acceleration and travels 270 m in 3 s. Find the velocity of the body at 10 s after the start.

A
$$600\, m\, s^{-1}$$

B
$$6\, m\, s^{-1}$$

C
$$60\, m\, s^{-1}$$

D
$$100\, m\, s^{-1}$$

Solution

$$u=0, s=270 m, t=3 s$$
$$s = ut+\dfrac{1}{2}at^2$$
$$270 = 0\times 3+\dfrac{1}{2}\times a\times 3^2$$
$$a= 60\ m/s^2$$

Now putting the value of $$a$$ in $$v=u+at$$,
Velocity after $$t=10 s$$ is given by
$$v=0+60\times 10 = 600\ m/s$$
Question 9
1 / -0

A particle starts to move in a straight line from a point with velocity $$10 m s^{-1}$$ and acceleration $$-2.0 m s^{-2}$$. Find position(S) and velocity(v) of the particle at $$t = 5$$ s

A
$$S = 25 m, v = 0 m/s$$

B
$$S = -25 m, v = 0 m/s$$

C
$$S = 5 m, v = 10 m/s$$

D
$$S = 0 m, v = 0 m/s$$

Solution

Displacement at $$t =5 s$$ is

$$ S= u \, t +\dfrac{1}{2} at ^2$$

$$=10 \times 5 + \dfrac{1}{2} \times (-2.0) \times (5)^2$$

$$=50-25=25 \, m$$

i.e., after $$5 s$$, the particle will be at distance $$25 m$$ from the starting point.

Velocity at $$t = 5 s$$ is

$$v = u + at$$

or $$v = 10+(-2)\times 5=0$$
Question 10
1 / -0

A ball is thrown vertically upwards from the top of a tower with an initial velocity of $$19.6 m {s}^{-1}$$. The ball reaches the ground after $$5 s$$. Calculate :$$(i)$$ the height of the tower, $$(ii)$$ the velocity of ball on reaching the ground. Take $$g=9.8 m {s}^{-2}$$.

A
$$(i) 24.5 m$$, $$(ii) 29.4$$ $$m$$ $${s}^{-1}$$

B
$$(i) 24.5 m$$, $$(ii) 19.4$$ $$m$$ $${s}^{-1}$$

C
$$(i) 24.5 m$$, $$(ii) 28$$ $$m$$ $${s}^{-1}$$

D
$$(i) 25 m$$, $$(ii) 29.4$$ $$m$$ $${s}^{-1}$$

Solution

$$s=ut+\dfrac{1}{2}at^2$$
$$=19.6 \times 5 - \dfrac{1}{2} \times 9.8 \times 5^2$$
$$=-24.5\ m$$

Hence, height of tower $$= 24.5 m$$
$$v=u+at$$
$$=19.6-9.8\times 5=-29.4\ m/s$$

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Motion in A Straight Line Test - 29

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Motion in A Straight Line Test - 29

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Question 1

(a) $$14$$ $$s$$, (b) $$30$$ $$m$$ $${s}^{-1}$$

(a) $$4$$ $$s$$, (b) $$30$$ $$m$$ $${s}^{-1}$$

(a) $$10$$ $$s$$, (b) $$40$$ $$m$$ $${s}^{-1}$$

(a) $$4$$ $$s$$, (b) $$40$$ $$m$$ $${s}^{-1}$$

Solution

Question 2

straight line path

circular path

parabolic path

hyperbolic path

Solution

Question 3

(i) $$40$$ $$m$$, (ii) $$30$$ $$m$$ $${s}^{-1}$$

(i) $$45$$ $$m$$, (ii) $$30$$ $$m$$ $${s}^{-1}$$

(i) $$45$$ $$m$$, (ii) $$60$$ $$m$$ $${s}^{-1}$$

(i) $$45$$ $$m$$, (ii) $$20$$ $$m$$ $${s}^{-1}$$

Solution

Question 4

equal horizontal and equal vertical ranges.

equal vertical ranges but different horizontal ranges.

different vertical ranges but equal horizontal ranges.

different vertical ranges and different horizontal ranges.

Solution

Question 5

$$980\ m$$

$$735\ m$$

$$490\ m$$

$$245\ m$$

Solution

Question 6

$$12 s$$

$$6 s$$

$$2 s$$

$$1 s$$

Solution

Question 7

600 km

5 km

6 km

10 km

Solution

Question 8

$$600\, m\, s^{-1}$$

$$6\, m\, s^{-1}$$

$$60\, m\, s^{-1}$$

$$100\, m\, s^{-1}$$

Solution

Question 9

$$S = 25 m, v = 0 m/s$$

$$S = -25 m, v = 0 m/s$$

$$S = 5 m, v = 10 m/s$$

$$S = 0 m, v = 0 m/s$$

Solution

Question 10

$$(i) 24.5 m$$, $$(ii) 29.4$$ $$m$$ $${s}^{-1}$$

$$(i) 24.5 m$$, $$(ii) 19.4$$ $$m$$ $${s}^{-1}$$

$$(i) 24.5 m$$, $$(ii) 28$$ $$m$$ $${s}^{-1}$$

$$(i) 25 m$$, $$(ii) 29.4$$ $$m$$ $${s}^{-1}$$

Solution

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