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

Result

Motion Test - 31

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Rank

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SHARING IS CARING

Question 1

$$2u$$

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

$$\sqrt{2}u$$

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

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

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

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

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

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

Solution

Question 6

$$1.25$$ m

$$21.25$$ m

$$30$$ m

$$32.5$$ m

Solution

Question 7

4.9 m

9.8 m

19.6 m

24.5 m

Solution

Question 8

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

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

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

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

Solution

Question 9

91 m

182 m

293 m

111 m

Solution

Question 10

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

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

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

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

Solution

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