Motion Test

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Motion Test - 52

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Weekly Quiz Competition

Question 1
1 / -0

The ratio of magnitude of velocity of projection in the upward motion to the magnitude of velocity of the body on reaching the point of projection in its downward motion is_____ .

A
9 : 1

B
1 : 1

C
2 : 1

D
4 :1

Solution

At the same horizontal level, the velocities of an object thrown up have equal magnitude but are opposite in direction. Thus the ratio is $$1:1$$

Let us assume an object was thrown up with a velocity $$u_1$$ and it reaches a height $$h.$$ Its final velocity $$v_1=0.$$
Using third equation of motion,
$$v_1^2-u_1^2=2(-g)h$$
$$u_1^2=2gh \Rightarrow u_1=\sqrt{2gh}$$ upwards

Now it comes down again through the same height $$h$$. Let us assume its final velocity is $$v_2.$$ We know its initial velocity for the descend will be $$u_2=0$$
$$v_2^2-u_2^2=2(-g)(-h)$$
$$v_2^2=2gh \Rightarrow v_2= \sqrt{2gh}$$ downwards

So we see that the magnitude of $$u_1$$ and $$v_2$$ is same.
Question 2
1 / -0

An automobile travelling with the speed of $$\displaystyle 72\ km {h}^{-1} $$ can be stopped within a distance of $$20\ m$$, by applying brakes. Determine the retardation.

A
$$\displaystyle -100 {ms} ^{-2} $$

B
$$\displaystyle -10 {ms} ^{-2} $$

C
$$\displaystyle 5 {ms} ^{-2} $$

D
$$\displaystyle -5 {ms} ^{-2} $$

Solution

Initial velocity of automobile $$u = 72 \ km/h = 72\times \dfrac{5}{18} = 20 \ m/s$$
Final velocity $$v = 0 \ m/s$$

Distance covered to stop $$S = 20 \ m$$

Using $$v^2 - u^2 = 2aS$$

$$\therefore$$ $$0 - 20^2 = 2\times a\times 20$$

$$\implies \ a = -10 \ m/s^2$$
Question 3
1 / -0

In the above shown figure, the velocity

A
increases between point O and A

B
increases between point A and B

C
decreases between points A and B

D
is zero throughout
Question 4
1 / -0

Tripling the speed of a motor car multiplies the distance needed for stopping it by

A
3

B
6

C
9

D
some other number

Solution

Hint: Use kinematics equation of motion

Step1:Relation between velocity and distance is

$$\mathrm{v}^{2}=\mathrm{u}^{2}+2 \mathrm{aS}$$

If the final velocity is Zero then the relation will become

$$\mathrm{u}=\sqrt{2 \mathrm{a} \mathrm{S}}$$

Now if we want to multiply the velocity $$\mathbf{3}$$ time then we have to multiply the distance $$\boldsymbol{9}$$ time to make the equation balanced.

$$\textbf{Hence option C correct}$$
Question 5
1 / -0

A displacement-time graph of a body moving with uniform velocity is shown in the figure. Find out its velocity and its displacement at the end of 5 seconds.

A
10 m/s, 125 m

B
5 m/s, 225 m

C
3m/s, 15 m

D
15 m/s, 75 m

Solution

From observing the graph, we say that displacement of the body at the end of 5 seconds is $$15 \ m$$.
Slope of the displacement-time graph gives the velocity of the body.
So, velocity of body $$v = slope = \dfrac{15-0}{5-0} = 3 \ m/s$$
Question 6
1 / -0

A ball is thrown vertically upwards from the top of a tower with a velocity of $$\displaystyle { 10\quad ms }^{ -1 }$$. The ball reaches the ground with the velocity $$\displaystyle { 30\quad ms }^{ -1 }$$. What is the height of the tower? $$\displaystyle (Take \quad g ={ 10\quad ms }^{ -2 })$$

A
40 m

B
20 m

C
60 m

D
80 m

Solution

Taking downward direction to be positive.
Initial velocity $$u = -10 \ m/s$$
Final velocity   $$v = 30\ m/s$$
Let the height of the tower be $$H$$.
Acceleration   $$a= g = 10 \ m/s^2$$
Using   $$v^2 - u^2 = 2aH$$
$$\therefore$$   $$30^2- (-10)^2 = 2\times 10 \ \times H$$
$$\implies\ H = 40 \ m$$
Question 7
1 / -0

A body whose speed in a particular direction is constant

A
must be accelerating

B
must be retarding

C
has a constant velocity

D
all the above

Solution

Velocity is a vector, which is speed plus direction. The body is moving in a particular direction with constant speed, means direction is constant and speed is also constant which means velocity is constant.
Question 8
1 / -0

Figure shows the displacement-time graphs (a) and (b) for a body moving in a straight path drawn on the same scales. Then the velocity in case of (b) is

A
more than the velocity in case of (a)

B
less than the velocity in case of (a)

C
equal to the velocity in case of (a)

D
square of the velocity in case of (a)

Solution

The slope of displacement-time curve gives us velocity. Larger the slope, higher will be velocity and vice versa.
As slope of (b) is less than slope of (a), velocity of (b) is less than that of (a).
Question 9
1 / -0

Two bodies of different masses $$ {m} _ {1}$$ and $$ {m} _ {2}$$ are dropped from two different heights $${h} _ {1}$$ and $$ {h} _ {2}$$. What is the ratio of time taken by the two to reach the ground?

A
$$\displaystyle \sqrt { { h }_{ 1 } } :\sqrt { { h }_{ 2 } } $$

B
$$\displaystyle \sqrt { { h }_{ 2 } } :\sqrt { { h }_{ 1 } } $$

C
$$\displaystyle {m} _ {1} {h} _ {2} : {m} _ {2} {h} _ {1} $$

D
$$\displaystyle {m} _ {1} {h} _ {1} : {m} _ {2} {h} _ {2} $$

Solution

Initial velocity of the stone
$$u = o$$
Acceleration
$$a = g$$
Let the height of the tower be $$h$$.
Using second equation of motion-
$$h = ut+\dfrac{1}{2}gt^2$$

$$\therefore$$ $$h = 0+\dfrac{1}{2}gt^2$$

$$\Rightarrow t = \sqrt{\dfrac{2h}{g}}$$
We get,
$$t\propto \sqrt{h}$$

$$\Rightarrow t_1:t_2 = \sqrt{h_1} : \sqrt{h_2}$$
Question 10
1 / -0

When brakes are applied the velocity of a car changes from $$\displaystyle { 30\ ms }^{ -1 } to\ { 10\ ms }^{ -1 }$$ in 5 s. The acceleration produced in it is ____ $$\displaystyle { ms }^{ -2 }$$.

A
$$4$$

B
$$-4$$

C
$$20$$

D
$$-20$$

Solution

Initial speed    $$u = 30 \ m/s$$
Final speed $$v = 10 \ m/s$$
Time    $$t = 5 \ s$$
Acceleration   $$a = \dfrac{v-u}{t}$$
$$\implies \ a = \dfrac{10-30}{5} = -4 \ m/s^2$$

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Motion Test - 52

Result

Motion Test - 52

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SHARING IS CARING

Question 1

9 : 1

1 : 1

2 : 1

4 :1

Solution

Question 2

$$\displaystyle -100 {ms} ^{-2} $$

$$\displaystyle -10 {ms} ^{-2} $$

$$\displaystyle 5 {ms} ^{-2} $$

$$\displaystyle -5 {ms} ^{-2} $$

Solution

Question 3

increases between point O and A

increases between point A and B

decreases between points A and B

is zero throughout

Question 4

3

6

9

some other number

Solution

Question 5

10 m/s, 125 m

5 m/s, 225 m

3m/s, 15 m

15 m/s, 75 m

Solution

Question 6

40 m

20 m

60 m

80 m

Solution

Question 7

must be accelerating

must be retarding

has a constant velocity

all the above

Solution

Question 8

more than the velocity in case of (a)

less than the velocity in case of (a)

equal to the velocity in case of (a)

square of the velocity in case of (a)

Solution

Question 9

$$\displaystyle \sqrt { { h }_{ 1 } } :\sqrt { { h }_{ 2 } } $$

$$\displaystyle \sqrt { { h }_{ 2 } } :\sqrt { { h }_{ 1 } } $$

$$\displaystyle {m} _ {1} {h} _ {2} : {m} _ {2} {h} _ {1} $$

$$\displaystyle {m} _ {1} {h} _ {1} : {m} _ {2} {h} _ {2} $$

Solution

Question 10

$$4$$

$$-4$$

$$20$$

$$-20$$

Solution

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