Motion Test

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Motion Test - 60

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Weekly Quiz Competition

Question 1
1 / -0

A body loses half of its velocity on penetrating $$6\ cm$$ in a wooden block. How much will it penetrate more before coming to rest?

A
$$1\ cm$$

B
$$2\ cm$$

C
$$3\ cm$$

D
$$4\ cm$$

Solution

$$v^2-u^2=2as$$
$$(\dfrac{v}{2})^2-v^2=2a6=-\dfrac{3v^2}{4}$$
$$a=\dfrac{v^2}{16}$$

$$0-(\dfrac{v}{2})^2=2as$$
$$\dfrac{v^2}{4}=2\dfrac{v^2}{16}s$$
$$s=2cm$$
Question 2
1 / -0

A space shuttle is launched into space. During the first 8 minutes of its launch the average acceleration of the shuttle is $$17.5 m s^{-2}$$. What is its speed after 8 minutes?

A
$$8000 m s^{-1}$$

B
$$8400 m s^{-1}$$

C
$$1200 m s^{-1}$$

D
$$1500 m s^{-1}$$

Solution

Given,
$$u=0, a=17.5ms^{-2}, t=8\times 60=480s$$

$$v=u+at$$

$$v=0+17.5\times 480=8400ms^{-1}$$
Question 3
1 / -0

A bike accelerates uniformly from rest to a speed of 7.10 m $$s^{-1}$$ over a distance of 35.4 m. Determine the acceleration of the bike.

A
0.412 m $$s^{-2}$$

B
0.512 m $$s^{-2}$$

C
0.612 m $$s^{-2}$$

D
0.712 m $$s^{-2}$$

Solution

Given:
Initial velocity $$u = 0 \space m/s$$
Final velocity $$v = 7.10 \space m/s$$
Distance covered $$s = 35.4 \space m$$
Acceleration $$a = ?$$

Using the equation of motion:
$$v^2=u^2+2as$$
$$(7.10)^2 = 0 + 2a\times 35.4$$
$$a=\dfrac{(7.10)^2}{2\times 35.4}=0.712\ ms^{-2}$$
Question 4
1 / -0

The velocity graphs pictured below represent the motions of five different colored objects of identical mass, with the color of the graph corresponding to the color of the object.
What is the order of objects, according to the magnitudes of the net forces they are experiencing during the time represented by their graphs, greatest first?

A
Blue, purple, red, yellow, green

B
Purple, red, yellow, blue and green tie

C
Blue and green tie, yellow, purple, red

D
Purple, red, blue, yellow, green

E
Yellow, red, purple, blue and green tie

Solution

The magnitude of force acing on object is given by
$$\left| F\right|=\left| ma\right|$$
$$=m\left|\dfrac{dv}{dt}\right|$$
$$\implies \left|F\right|\propto \left|\dfrac{dv}{t}\right|$$
Hence magnitude of force on object is proportional to the magnitude of slope(m) of the given v-t curve.
$$m_{yellow}>m_{red}>m_{purple}>m_{green}=m_{blue}$$
Therefore this is the order of the magnitude of force acting on objects.
Question 5
1 / -0

The speed of a car reduces from 15 m $$s^{-1}$$ to 5 m $$s^{-1}$$ over a displacement of 10 m. What is the retardation of the car?

A
-10 m $$s^{-2}$$

B
+10 m $$s^{-2}$$

C
2 m $$s^{-2}$$

D
0.5 m $$s^{-2}$$

Solution

Here, initial velocity $$u=15 m/s$$ and final velocity $$v=5 m/s$$
Using the formula, $$v^2=u^2+2aS$$
$$\Rightarrow 5^2=15^2+2\ a\ (10)$$
$$\Rightarrow -20\ a=15^2- 5^2$$
$$a=-10 m/s^2$$ (negative sign indicates the car is deaccelerating)
Question 6
1 / -0

A race car accelerates uniformly from $$18.5 m/s$$ to $$46.1 m/s$$ in $$2.47$$ seconds. Determine the acceleration of the car. (in $$m/s^2$$)

A
$$11.2 m/s^2$$

B
$$11 m/s^2$$

C
$$1.2 m/s^2$$

D
$$51.2 m/s^2$$

Solution

Initial velocity of the car   $$u = 18.5 \ m/s$$
Final velocity of the car $$v = 46.1 \ m/s$$
Time taken   $$t = 2.47 \ s$$
Acceleration of the car   $$a = \dfrac{v-u}{t}$$
$$\implies \ a = \dfrac{46.1-18.5}{2.47} = 11.2 \ m/s^2$$
Question 7
1 / -0

Directions For Questions

A ball is dropped from rest from a height of $$100\ m$$ above ground level. Neglect the effects of air resistance.

...view full instructions

How long does it take for the ball to strike the ground?

A
$$4.52 s$$

B
$$5 s$$

C
$$6 s$$

D
$$7 s$$

Solution

Given,
Initial velocity, $$u=0$$ (at rest)
Height, $$h=100\ m$$
Acceleration due to gravity, $$a=g=9.8\ m/s^2$$
Time, $$t$$

Using Newton' second equation of motion,
$$s=ut+\dfrac12at^2$$

$$h=\dfrac12at^2$$

$$\implies t=\sqrt{\dfrac{2h}{a}}$$

$$t=\sqrt{\dfrac{2\times 100}{9.8}}$$

$$t=\sqrt{20.408}$$

$$t=4.52\ s$$
Question 8
1 / -0

Read the following statements and state whether they are True or False:
i) If $$u$$ and $$a$$ both are negative, motion is only retarded.
ii) If $$u$$ is negative but $$a$$ is positive, then displacement of the particle can never be positive.
iii) If u is positive but $$a$$ is negative particle comes to rest for a moment at some time but if $$u$$ is negative and $$a$$ positive it never comes to rest.

A
F, F, F

B
T, T, T

C
T, F, T

D
F, T, F

Solution

If both $$u$$ and $$a$$ are in the same direction then $$u$$ increases in direction decrease to zero then increase in direction of $$a$$.
So on the basis of the above statement, we come to a conclusion that all three statements are false.
For i) Velocity will increase in negative a direction always as initial velocity is also in -a direction.
For ii) First, $$u$$ decreases to zero then increases in the positive direction and keep on moving in the positive direction so ultimately will go the positive direction.
iii) Opposite sign of $$u$$ and a means at least one time $$u$$ will become zero.
Question 9
1 / -0

Rocket-powered sleds are used to test the human response to acceleration. If a rocket-powered sled is accelerated to a speed of $$444 m/s$$ in $$1.83$$ seconds, then what is the acceleration? (in $$m/s^2$$)

A
$$243 m/s/s$$

B
$$24 m/s/s$$

C
$$23 m/s/s$$

D
$$43 m/s/s$$

Solution

Given, initial velocity $$u=0 m/s, $$ final velocity $$v=444 m/s$$ and time $$t=1.83 s$$
If $$a $$ be the acceleration.
Using $$v=u+at$$,
we get $$444=0+a(1.83)$$
$$\implies$$ $$a=444/1.83=242.62 \sim 243 m/s^2$$
Question 10
1 / -0

What will be ratio of speed in first two seconds to the speed in next 4 s?

A
$$\sqrt 2$$ : 1

B
3 : 1

C
2 : 1

D
1 : 2

Solution

Magnitude of slope of distance-time graph gives the speed of the particle.
Slope of line AB, $$m_1 = \dfrac{BO}{AO} =\dfrac{x}{2}$$
Thus speed in first two seconds, $$v_1 = |m_1| = \dfrac{x}{2}$$
Slope of line BC, $$m_2 = \dfrac{-BO}{CO} =\dfrac{-x}{4}$$
Thus speed in first two seconds, $$v_2 = |m_2| = \dfrac{x}{4}$$
Thus ratio of speed $$\dfrac{v_1}{v_2} = \dfrac{x/2}{x/4} = \dfrac{2}{1}$$

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Re-Attempt Weekly Quiz Competition

Motion Test - 60

Result

Motion Test - 60

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SHARING IS CARING

Question 1

$$1\ cm$$

$$2\ cm$$

$$3\ cm$$

$$4\ cm$$

Solution

Question 2

$$8000 m s^{-1}$$

$$8400 m s^{-1}$$

$$1200 m s^{-1}$$

$$1500 m s^{-1}$$

Solution

Question 3

0.412 m $$s^{-2}$$

0.512 m $$s^{-2}$$

0.612 m $$s^{-2}$$

0.712 m $$s^{-2}$$

Solution

Question 4

Blue, purple, red, yellow, green

Purple, red, yellow, blue and green tie

Blue and green tie, yellow, purple, red

Purple, red, blue, yellow, green

Yellow, red, purple, blue and green tie

Solution

Question 5

-10 m $$s^{-2}$$

+10 m $$s^{-2}$$

2 m $$s^{-2}$$

0.5 m $$s^{-2}$$

Solution

Question 6

$$11.2 m/s^2$$

$$11 m/s^2$$

$$1.2 m/s^2$$

$$51.2 m/s^2$$

Solution

Question 7

$$4.52 s$$

$$5 s$$

$$6 s$$

$$7 s$$

Solution

Question 8

F, F, F

T, T, T

T, F, T

F, T, F

Solution

Question 9

$$243 m/s/s$$

$$24 m/s/s$$

$$23 m/s/s$$

$$43 m/s/s$$

Solution

Question 10

$$\sqrt 2$$ : 1

3 : 1

2 : 1

1 : 2

Solution

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