Motion Test

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Motion Test - 31

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Weekly Quiz Competition

Question 1
1 / -0

A body projected up with a velocity $u$ reaches maximum height $h$ . To reach double the height, it must be projected up with a velocity of:

A
$2u$

B
$\dfrac{u}{2}$

C
$\sqrt{2}u$

D
$\dfrac{u}{\sqrt{2}}$

Solution

$\textbf{Step 1 - Calculating height Refer Figure}$
Given,
initial height, $h_{1} = h$
final height, $h_{2} = 2h$
As we know, at maximum height final velocity $(v)$ become zero.
$v^{2} = u^{2} - 2gh$
$0 = u^{2} - 2gh$
$\Rightarrow h = \dfrac {u^{2}}{2g}$ $....(1)$
Similarly,
$2h = \dfrac {(u_{2})^{2}}{2g}$ $....(2)$

$\textbf{Step 2 - Dividing equation (1) and (2)}$

$\dfrac {h}{2h} = \dfrac {u^{2}}{2g}\times \dfrac {2g}{(u_{2})^{2}}$

$\Rightarrow \dfrac {1}{2} = \dfrac {u^{2}}{(u_{2})^{2}}$

$\Rightarrow u_{2} = \sqrt {2}u$

Correct option : C.
Question 2
1 / -0

Two bodies are projected simultaneously with the same speed of $19.6 \;ms^{-1}$ from the top of a tower, one vertically upwards and the other vertically downwards. As they reach the ground, the time gap is:

A
0 s

B
2 s

C
4 s

D
6 s
Question 3
1 / -0

Two balls are dropped to the ground from different heights. One ball is dropped $2\ s$ after the other but both of them strike the ground simultaneously $5\ s$ after the first is dropped. The difference in the heights, when they are dropped is ( $g=10\ m/s^2$ )

A
80 m

B
45 m

C
95 m

D
100 m

Solution

Ball B is dropped first. Then, ball A is dropped after 2 s.
$h_B = \dfrac {1}{2}gt^2 = \dfrac {1}{2}g(5)^2 = 12.5g = \dfrac {25g}{2}$ ....... (1)
$h_A = \dfrac {1}{2}gt^2 = \dfrac {1}{2}g(3)^2 = \dfrac {9g}{2}$ ........ (2)
$eq^n (1) - eq^n (2) \Rightarrow h_B - h_A = \dfrac {25g}{2} - \dfrac {9g}{2} = \dfrac {16g}{2} = \dfrac {16 \times 10}{2}=80\ m$
Question 4
1 / -0

A body is thrown vertically up with certain velocity. If $h$ is the maximum height reached by it, its position when its velocity reduces to $(1/3)^{rd}$ of its velocity of projection is at

A
8h/9 from the ground

B
8h/9 below the top-most point

C
4h/9 from the ground

D
h/3 below the top-most point

Solution

Let initial velocity is $u$ ,
$h = \dfrac {u_1^2}{2g}$

$v^2 = u^2 - 2gx$

$(\dfrac {u_1}{3})^2 = u_1^2 - 2gx$

$2gx = \dfrac {8}{9} u_1^2$

$x = \dfrac {8}{9} (\dfrac {u_1^2}{2g}) = \dfrac {8}{9}h$
Question 5
1 / -0

A ball is dropped from the top of a building. The ball takes $0.5$ s to fall past the window $3$ m in length at certain distance from the top of the building. ( $g=10\;ms^{-2}$ )
Speed of the ball as it crosses the top edge of the window is

A
$3.5 \;ms^{-1}$

B
$8.5 \;ms^{-1}$

C
$5 \;ms^{-1}$

D
$12\; ms^{-1}$

Solution
Question 6
1 / -0

A person standing on the edge of a well throws a stone vertically upwards with an initial velocity $5ms^{-1}$ . The stone goes up, comes down and falls in the well making a sound. If the person hears the sound 3 second after throwing, then the depth of water is (neglect time travel for the sound and take $g = 10ms^{-2}$ ):

A
$1.25$ m

B
$21.25$ m

C
$30$ m

D
$32.5$ m

Solution

$h = ?$ Taking vertically downward direction as negative,
$h = ut - \dfrac {1}{2}gt^2$
$t = 3 s$
$h = 5(3) - \dfrac {1}{2}g(3)^2 = 15-\dfrac {9g}{2}$ $= 15 - 45 = 30\ m$
Hence, depth $= 30\ m$
Question 7
1 / -0

A body released from a height falls freely towards the earth. Another body is released from the same height exactly a second later. Then the separation between the two bodies two seconds after the release of the second body is

A
4.9 m

B
9.8 m

C
19.6 m

D
24.5 m

Solution

Both the bodies are falling freely with acceleration $g = 9.8\ m/s^2$ .
Equation to be used: $s=\dfrac{1}{2}\times g\times t^2$
The first body descends for 3 sec.
The height covered by this body is $s_1 = \dfrac {1}{2} \times (9.8) (3^2)=44.1\ m$
The second body descends for 2 sec.
The height covered by this body is $s_2 = \dfrac {1}{2} \times (9.8) (2^2)=19.6\ m$
Separation between the bodies is $s_1 - s_2 = (44.1-19.6) = 24.5\ m$
Question 8
1 / -0

A stone is dropped from a balloon at an altitude of $280\ m$ . If the balloon ascends with a velocity of $5\ ms^{-1}$ and descends with a velocity of $5\ ms^{-1}$ , times taken by the stone to reach the ground in the two cases respectively are ( $g=10\ ms^{-2}$ ):

A
$8\ s$ and $9\ s$

B
$9\ s$ and $8\ s$

C
$3\ s$ and $4\ s$

D
$8\ s$ and $7\ s$

Solution

Initial speed of the particle becomes same as the ballon have.
Assume upward direction as positive and downward direction as negative.
$1^{st}$ Case:
Ascend with $5\ m/s$
$u = 5\ m/s$ : Initial velocity of particle
Apply second equation of motion.
$s = ut - \dfrac {1}{2}gt^2$
$\Rightarrow s = 5t - 5t^2$
Displacement is $-280\ m$ , in downward direction
$\Rightarrow 5t-5t^2= -280$
$\Rightarrow t = 8\ sec$

$2^{nd}$ Case:
Descend with $5\ m/s$
$u = -5\ m/s$ : Initial velocity of particle
Apply second equation of motion.
$s = ut - \dfrac {1}{2}gt^2$
Displacement is $-280\ m$ , in downward direction
$\Rightarrow -280 = -5t - 5t^2$
$\Rightarrow t = 7\ sec$
Question 9
1 / -0

A parachutist after bailing out falls 50 m without friction. When parachute opens, it decelerates at $2\ m/ s^{2}$ . He reaches the ground with a speed of $3\ m/ s$ . At what height, did he bail out ? (Given $g=9.8\ m/s^2$ approximately)

A
91 m

B
182 m

C
293 m

D
111 m

Solution

The speed of the person just before the parachute opens is:
$u=\sqrt {2gh}=\sqrt {2\times 9.8\times 50}=\sqrt {980}$

When the parachute opens and descends,
$v^2=u^2+2(-a)h$
$v^2-u^2=-2(2)(h)$
$9-980=-4h$
$\Rightarrow h=971/4=242.75\ m$

$\therefore$ The total height of fall is $242.75+50=292.75\ m\approx293\ m$
Question 10
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, $v = u + at$
$v - u = at$
$0.18 = a \times 9$

$\Rightarrow a = \dfrac {0.18}{9} = 0.02 \ m/s^2$

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Re-Attempt Weekly Quiz Competition

Motion Test - 31

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Motion Test - 31

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SHARING IS CARING

Question 1

2u2u2u

u2\dfrac{u}{2}2u​

2u\sqrt{2}u2​u

u2\dfrac{u}{\sqrt{2}}2​u​

Solution

Question 2

0 s

2 s

4 s

6 s

Question 3

80 m

45 m

95 m

100 m

Solution

Question 4

8h/9 from the ground

8h/9 below the top-most point

4h/9 from the ground

h/3 below the top-most point

Solution

Question 5

3.5 ms−13.5 \;ms^{-1}3.5ms−1

8.5 ms−18.5 \;ms^{-1}8.5ms−1

5 ms−15 \;ms^{-1}5ms−1

12 ms−112\; ms^{-1}12ms−1

Solution

Question 6

1.251.251.25 m

21.2521.2521.25 m

303030 m

32.532.532.5 m

Solution

Question 7

4.9 m

9.8 m

19.6 m

24.5 m

Solution

Question 8

8 s8\ s8 s and 9 s9\ s9 s

9 s9\ s9 s and 8 s8\ s8 s

3 s3\ s3 s and 4 s4\ s4 s

8 s8\ s8 s and 7 s7\ s7 s

Solution

Question 9

91 m

182 m

293 m

111 m

Solution

Question 10

0.01 m/s20.01\ {m/s}^{2}0.01 m/s2

0.02 m/s20.02\ {m/s}^{2}0.02 m/s2

0.03 m/s20.03\ {m/s}^{2}0.03 m/s2

0.04 m/s20.04\ {m/s}^{2}0.04 m/s2

Solution

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$2u$

$\dfrac{u}{2}$

$\sqrt{2}u$

$\dfrac{u}{\sqrt{2}}$

$3.5 \;ms^{-1}$

$8.5 \;ms^{-1}$

$5 \;ms^{-1}$

$12\; ms^{-1}$

$1.25$ m

$21.25$ m

$30$ m

$32.5$ m

$8\ s$ and $9\ s$

$9\ s$ and $8\ s$

$3\ s$ and $4\ s$

$8\ s$ and $7\ s$

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

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

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

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