Motion Test

Result

Motion Test - 36

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

An object is thrown vertically upward with a velocity of $$10\ ms^{-1}$$. It strikes the ground after _______ seconds. (Take $$g = 10\ ms^{-2}$$)

A
10

B
2

C
5

D
1

Solution

Given:
Initial velocity $$u=10\ m/s$$
Gravitational acceleration $$g=10\ m/s^2$$

$$\text{Time to ascend maximum height}$$:
When the object is ascending, $$g$$ is causing deceleration. At the maximum height, the velocity of the object is zero.
Let $$t_1$$ is the time required to reach the maximum height.
Thus, using the equation $$v=u-gt$$ we get,
$$0=10-(10\times t_1)$$
$$\Rightarrow t_1=1\ sec$$

$$\text{Time required to reach the maximum height} = \text{Time required to fall down to the ground from the maximum height}$$

Thus, the total time required is $$(1+1)=2\ sec$$
Question 2
1 / -0

An object is released from a balloon rising up with a constant speed of $$2\ ms^{-1}$$. Its magnitude of velocity after $$1\ s$$ in $$\ ms^{-1}$$ is

A
5.3

B
7.8

C
2.8

D
6.4

Solution

The ball initially will have the same velocity as that of the balloon
Using the equations of motion , taking velocity to be positive upwards and negative downwards
$$ v = u + at$$
$$v = 2 - g(1) $$
$$ v = 7.8 m/s$$
Question 3
1 / -0

When a body is projected vertically up from the ground its velocity is reduced to $$\frac{1}{4}$$th of its velocity at ground at height $$h$$. Then the maximum height reached by the body is

A
$$\dfrac{15}{32}h$$

B
$$\dfrac{15}{16}h$$

C
$$\dfrac{15}{8}h$$

D
$$\dfrac{5}{4}h$$

Solution

Using the equation of motion ,
$$v^2 = u^2 - 2gh $$
$$\dfrac{u^2}{16} = u^2 - 2gh $$
$$h = \dfrac{15 u^2}{32 g} $$

At Max. height, $$v=0$$ and equation of motion is
$$v^2 = u^2 - 2gh $$
$$H = \dfrac{u^2}{2g} =\dfrac{15h}{16}$$
Question 4
1 / -0

A body projected vertically up with a velocity of 10 m s$$^{-1}$$ reaches a height of 20 m. If it is projected with a velocity of 20 m s$$^{-1}$$, then the maximum height reached by the body is :

A
20 m

B
10 m

C
80 m

D
40 m

Solution

$$u = 10 m s^{-1} ; h = 20 m$$
$$u_1 = 20 m s^{-1}; h_1 = ?$$

Formula for maximum height
$$h=\frac{u^2}{2g}$$
hence

$$\displaystyle \dfrac{h}{h_1}= \dfrac{u^2}{u_1^2} ; \dfrac{20}{h_1} = \dfrac{10^2}{20^2}$$

$$h_1 \displaystyle = \dfrac{20 \times 400}{100} = 80 m$$
Question 5
1 / -0

To reach the same height on the moon as on the earth, a body must be projected up with:

A
Higher velocity on the moon

B
Lower velocity on the moon

C
Same velocity on the moon and earth

D
It depends on the mass of the body

Solution

$$\textbf{Hint:}$$ Maximum height attained should be same on earth and moon

$$\textbf{Explanation:}$$

The maximum height attained is given as,

$${h_{ma \times }} = \dfrac{{{u^2}}}{{2g}}$$

here $$u$$ is the initial velocity, and $$g$$ is the acceleration due to gravity

So, initial velocity is directly proportional to gravity

$$u\alpha \sqrt g $$

And Acceleration due to gravity of earth is more than moon,

$${g_{earth}} = 6{g_{moon}}$$

$$\therefore {u_{moon}} < {u_{earth}}$$

Hence, a body must be projected up with low velocity for moon.
Question 6
1 / -0

If an object is thrown vertically up with a velocity of $$19.6\ ms^{-1}$$, it strikes the ground after $$x$$ second, find $$x$$.

A
1

B
2

C
4

D
8

Solution

$$u=19.6m/s$$
Total time taken $$T=\dfrac{2u}{g}$$
$$x=T=2\times \dfrac{19.6}{9.8}$$
$$x=4sec$$
Question 7
1 / -0

The velocity of a body which starts from rest with an acceleration $$2\ ms^{-2}$$ and covering a distance of $$10\ m$$ in $$ms^{-1}$$ is:

A
$$\sqrt{98}$$

B
$$\sqrt{40}$$

C
$$\sqrt{20}$$

D
$$\sqrt{12}$$

Solution

using the equations of motion ,
$$ v^2 = u^2 + 2as$$
Intial velocity is zero.
$$ v^2 = 2 \times 2 \times 10 $$
$$ v = \sqrt{40} m/s$$
Question 8
1 / -0

A stone is dropped from a rising balloon at a height of 300 m above the ground and it reaches the ground in 10 s. The velocity of the balloon when it was dropped is :

A
$$19 m s^{-1}$$

B
$$19.6 m s^{-1}$$

C
$$29 m s^{-1}$$

D
$$0 m s^{-1}$$

Solution

$$h = -ut + \displaystyle \dfrac{1}{2} gt^2$$
$$300 = - 10 u \displaystyle + \dfrac{1}{2} \times 9.8 \times 100$$
$$-10 u = 300 - 490 = -190$$
$$u \displaystyle = \dfrac{190}{10}= 19 m s^{-1}$$
Question 9
1 / -0

A particle is thrown vertically upwards. Its velocity at one fourth of the maximum height is $$20 \ m s^{-1}$$. Then, the maximum height attained by it is

A
16 m

B
10 m

C
8 m

D
18 m

Solution

Let us assume the initial velocity to be $$u$$.
$$H$$ be the maximum height reached by the ball.

Given: $$v=0$$

By third equation of motion,
$$v^2-u^2=2as$$
$$0-u^2=2(-g)H$$
$$\Rightarrow H=\dfrac{u^2}{2g}$$ . . . . (i)

Given, for $$h=\dfrac{1}{4}H$$, $$v=20 \ ms^{-1}$$

$$h=\dfrac{1}{4} \times \dfrac{u^2}{2g}=\dfrac{u^2}{8g}$$

Putting $$h$$ and $$v$$ and $$a=-g$$ in third equation of motion,
$$(20)^2-u^2= 2(-g)\left( \dfrac{u^2}{8g}\right)$$
$$\Rightarrow u^2- \dfrac{u^2}{4}=400$$
$$\Rightarrow 3u^2 =1600$$
$$\Rightarrow u^2 =\dfrac{1600}{3}$$

Putting the value of $$u^2$$ is eq. (i),
Maximum height
$$H=\dfrac{\left( \dfrac{1600}{3}\right)}{2g}$$

$$H=\dfrac{1600}{3 \times 2 \times 10}=26.66 \ m$$

$$\displaystyle \dfrac{u^2}{8g} \times 2(+g) = (20)^2 - u^2+ \dfrac{u^2}{4} + u^2 = 400$$
$$5 u^2 = 1600$$
$$ u \displaystyle = \sqrt{\dfrac{1600}{5}} m s^{-1}$$
$$\displaystyle h_{max} = \dfrac{u^2}{2g} = \dfrac{1600}{5 \times 2 \times 10} = 16 m$$
Question 10
1 / -0

A ball is thrown vertically upwards. It has a speed of $$10 m s^{-1}$$, when it has reached on half of its maximum height. How high does the ball rise? (Take g $$=$$ 10 m s$$^{-2}$$)

A
10 m

B
5 m

C
15 m

D
20 m

Solution

Let initial velocity be u
$$\displaystyle h_{max} = \dfrac{u^2}{2g} ; \dfrac{h_{max}}{2} = \dfrac{u^2-10^2}{+2g}$$
$$\displaystyle \dfrac{u^2}{4g} = \dfrac{u^2-10^2}{2g}$$
$$u^2 = 2u^2-200$$
$$u^2 =200$$
$$u = 10 \sqrt{2} m s^{-1}$$
$$\therefore h_{max} = \displaystyle \dfrac{(10 \sqrt{2})^2}{2 \times 10} = 10 m$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Motion Test - 36

Result

Motion Test - 36

Score

-

Rank

-

SHARING IS CARING

Question 1

10

2

5

1

Solution

Question 2

5.3

7.8

2.8

6.4

Solution

Question 3

$$\dfrac{15}{32}h$$

$$\dfrac{15}{16}h$$

$$\dfrac{15}{8}h$$

$$\dfrac{5}{4}h$$

Solution

Question 4

20 m

10 m

80 m

40 m

Solution

Question 5

Higher velocity on the moon

Lower velocity on the moon

Same velocity on the moon and earth

It depends on the mass of the body

Solution

Question 6

1

2

4

8

Solution

Question 7

$$\sqrt{98}$$

$$\sqrt{40}$$

$$\sqrt{20}$$

$$\sqrt{12}$$

Solution

Question 8

$$19 m s^{-1}$$

$$19.6 m s^{-1}$$

$$29 m s^{-1}$$

$$0 m s^{-1}$$

Solution

Question 9

16 m

10 m

8 m

18 m

Solution

Question 10

10 m

5 m

15 m

20 m

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.