Motion Test

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Motion Test - 38

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Question 1
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The ratio of the heights from which two bodies are dropped is 3 : 5 respectively. The ratio of their final velocities is?

A
$$\sqrt{5} : \sqrt{3}$$

B
$$\sqrt{5} : \sqrt{2}$$

C
$$\sqrt{3} : \sqrt{5}$$

D
none of the above

Solution

Initial speed, $$u = 0$$ (at rest)
Final velocity, $$v$$
Acceleration due to gravity, $$a = g$$
Let the height from which it is dropped be $$s=h$$.
Using Newton's third equation of motion,
$$2as=v^2 - u^2$$
$$2gh=v^2$$
$$\implies v=\sqrt{2gh}$$

Let the heights from which two bodies are dropped are $$h_1$$ and $$h_2$$
Given,
$$h_1:h_2=3:5$$
Final velocity of $$1^{st}$$ body, $$v_1=\sqrt{2gh_1}$$ ...............(1)
Final velocity of $$2^{nd}$$ body, $$v_2=\sqrt{2gh_2}$$ ..............(2)
Diviing equation (1) by (2)
$$\dfrac{v_1}{v_2}=\dfrac{\sqrt{2gh_1}}{\sqrt{2gh_2}}$$

$$\dfrac{v_1}{v_2}=\sqrt {\dfrac{h_1}{h_2}}$$

$$\dfrac{v_1}{v_2}=\sqrt {\dfrac{3}{5}}$$

$$v_1:v_2=\sqrt3:\sqrt5$$
Question 2
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An object may have
(I) varying speed without having varying velocity.
(II) varying velocity without having varying speed.
(III) non-zero acceleration without having varying velocity.
(IV) non-zero acceleration without having varying speed.

A
I and II are correct.

B
II and III are correct.

C
II and IV are correct.

D
None of the above.

Solution

Speed is a scalar quantity while velocity is a vector quantity.
An object can have constant speed but velocity will change since velocity is a vector quantity.
Without having a varying speed non-zero acceleration. Uniform circular motion is an example of options B and D
Thus, B and D options are correct
Question 3
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A balloonist is ascending at a velocity of $$12\ m{ s }^{ -1 }$$. A packet is dropped from it when it is at height of $$65 \ m$$ from the ground. Time taken by the packet to reach the ground is

A
$$5 \ s$$

B
$$-5 \ s$$

C
$$7 \ s$$

D
$$\cfrac{13}{5} \ s$$

Solution

Using the second equation of motion,

$$s=ut+\frac { 1 }{ 2 } a{ t }^{ 2 }$$

Here, $$s = 65 \ m,\ \ u = -12 \ m/s$$ (since it's moving upwards), $$a=g=10 \ m/s^2$$

$$65=-12t+\dfrac { 1 }{ 2 } 10{ t }^{ 2 }$$

$$ \Rightarrow 5{ t }^{ 2 }-12t-65=0\\ \Rightarrow t=5\quad ; t=-13/5$$
Since, time cannot be negative.
Thus, $$t=5s$$
Question 4
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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 speed $$3\ m{ s }^{ -1 }$$. At what height did he bail out?
($$g=9.81m/{ s }^{ 2 }$$)

A
$$91\ m$$

B
$$182\ m$$

C
$$293\ m$$

D
$$111\ m$$

Solution

$$\text{Initial velocity,}\ u=\sqrt{2gh}$$

$$\text{or,}\ u=\sqrt{2\times 9.8 \times 50}=14\sqrt{5}$$

$$\text{The velocity at ground,}\ v=3\ \text{m/s}$$

$$v^2 - u^2 =2as$$

$$\text{or,}\ {3}^{2}-980=2\times (-2) \times s$$

$$\text{Thus, }\ s=\dfrac{971}{4},\ \text{which is nearly 243 m.}$$

$$\text{Initially he has fallen 50 m}$$

$$\text{Thus total height at which he bailed out} = 243+50=293\ \text{m}$$
Question 5
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When a ball is $$h$$ metre high from a point $$O$$, its velocity is $$v$$. When it is $$h\ m$$ below $$O$$, its velocity is $$2v$$. Find the maximum height from $$O$$ it will acquire.

A
$$\cfrac { 2h }{ 3 } $$

B
$$\cfrac { 5h }{ 3 } $$

C
$$\cfrac { 3h }{ 2 } $$

D
$$2h$$

Solution

Using third equation of motion $$v^2-u^2=2as$$
For motion from Highest to lowest point $$,\ { \left( 2v \right) }^{ 2 }- \left( v \right)^{ 2 }=2g\left( h \right) \quad or\quad \dfrac { { v }^{ 2 } }{ 2g } =\dfrac { 2h }{ 3 } $$
We know Maximum height is given by $$H=\frac{v^2}{2g}$$
At TOP POINT velocity of ball is $$ v$$, so maximum height it will attain $$ =\dfrac { { v }^{ 2 } }{ 2g } =\dfrac { 2h }{ 3 } $$
From point O,maximum height ball acquire $$ =h+\dfrac { 2h }{ 3 } =\dfrac { 5h }{ 3 } $$
Question 6
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The displacement - time graphs of two bodies $$A$$ and $$B$$ are $$OP$$ and $$OQ$$ respectively. Velocities of A and B are related as

A
$$V_A > V_B$$

B
$$ V_A < V_B $$

C
$$V_A = V_B $$

D
Cannot Predict

Solution

Velocity is given by the slope of displacement time graph, therefore, graph making more angle with the x-axis has higher velocity, therefore we have
$$V_A > V_B$$
Question 7
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A point moves rectilinearly in one direction. Above figure shows the distance $$s$$ traversed by the point as a function of the time $$t$$. Using the plot, find the maximum velocity.

A
$$1.5\:m/s$$

B
$$2.5\:m/s$$

C
$$2 \:m/s$$

D
$$5\:m/s$$

Solution

Since velocity is the slope of distance versus time graph, thus maximum velocity means maximum slope or the point of inflation.

From the graph, it can be clearly seen that maximum slope occurs in the straight line region between $$10$$ seconds and $$14$$ seconds, which is given by

$$\quad v = \dfrac { 14-4 }{ 14-10} $$ or

$$\quad v = \dfrac { 10 }{ 4} = 2.5$$ m/s
Question 8
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The first stage of the rocket launches a satellite to a height of $$50\ km$$ and velocity attained is $$6000\ km{ h }^{ -1 }$$ at which point its fuel exhausted. How high the rocket will reach (Take $$g=10\ m/s^2$$ and assume $$g$$ is constant up to for rocket's entire journey?

A
$$138.9\ km$$

B
$$188.9\ km$$

C
$$88.9\ km$$

D
$$168.9\ km$$

Solution
Question 9
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Directions For Questions

The graph given shows the position of two cars A and B, as a function of time. The cars move along the x-axis on parallel but separate tracks, so that they can pass each others position without colliding.

...view full instructions

At which instant do the two cars have the same velocity?

A
$$ t_1 $$

B
$$ t_2 $$

C
$$ t_3 $$

D
$$ t_4 $$

Solution

Velocity is given by slope of distance-time graph. Thus same velocity means same slope. It can be clearly seen, the two curves have same slope at time $${ t }_{ 2 }$$
Question 10
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When a ball is thrown up vertically with a velocity $${ v }_{ 0 }$$ it reaches a height $$h$$. If one wishes to triple the maximum height then the ball be thrown with a velocity

A
$$\sqrt { 3 } { v }_{ 0 }$$

B
$$3{ v }_{ 0 }$$

C
$$9{ v }_{ 0 }$$

D
$$\dfrac{3}{2}{ v }_{ 0 }$$

Solution

Using the third equation of motion,

$$v^2 = u^2 - 2as$$

$$Case I$$: Given , $$u=v_o$$, $$v=0$$, $$s=h$$
Substituting in the equation, we get,
$$\ 0={ v }_{ o }^{ 2 }-2gh\quad or\quad h=\dfrac { { v }_{ o }^{ 2 } }{ 2g } $$

$$Case II$$: Given, $$v=0$$, $$h'=3h$$, $$u=v_1$$
Substituting in the equation, we get,

$$\ 0={ v }_{ 1 }^{ 2 }-2g\left( 3h \right) \quad or\quad h=\dfrac { { v }_{ 1 }^{ 2 } }{ 6g } $$

On equating both equations, we get,
$$ \dfrac { { v }_{ 1 }^{ 2 } }{ 6g } =\dfrac { { v }_{ o }^{ 2 } }{ 2g } \quad or\quad { v }_{ 1 }=\sqrt { 3 } { v }_{ o }$$