Motion in a Straight Line — Page 8 | NEET Physics

Q71

From a building two balls A and B are thrown such that A is thrown upwards and B downwards ( both vertically with the same speed ). If vA and vB are their respective velocities on reaching the ground, then

A

{v_B} > {v_A}

B

{v_A} = {v_B}

C

{v_A} > {v_B}

D their velocities depend on their masses.

Correct Answer

Option B

Solution

Assume the initial velocity of each particle is = u And height of building = h If final velocity of A is vA then vA2 = u2 + 2(-g)(-h) = u2 + 2gh If final velocity of B is vB then vB2 = u2 + 2gh $\therefore$ vA = vB Sign Rule : Take the direction of initial velocity positive opposite direction as negative.

Here for ball A initial velocity u is upward so upward is positive and downward is negative.

That is why gravity is = - g and height = - h And for ball B initial velocity u is downward so downward is positive and upward is negative.

That is why gravity is = + g and height = + h

Q72

A particle starts with an initial velocity of

10.0 \mathrm{~ms}^{-1}

along

x

-direction and accelerates uniformly at the rate of

2.0 \mathrm{~ms}^{-2}

. The time taken by the particle to reach the velocity of

60.0 \mathrm{~ms}^{-1}

is __________.

A 30s

B 6s

C 3s

D 25s

Correct Answer

Option D

Solution

To find the time taken by the particle to reach the velocity of

60.0 \mathrm{~ms}^{-1}

, we can use the formula:

v = u + at

Where:

v

is the final velocity,

u

is the initial velocity,

a

is the acceleration, and

t

is the time taken. Plugging in the given values:

60.0 \mathrm{~ms}^{-1} = 10.0 \mathrm{~ms}^{-1} + 2.0 \mathrm{~ms}^{-2} \cdot t

Solve for

t

:

50.0 \mathrm{~ms}^{-1} = 2.0 \mathrm{~ms}^{-2} \cdot t

t = \frac{50.0 \mathrm{~ms}^{-1}}{2.0 \mathrm{~ms}^{-2}} = 25 \mathrm{s}

So, the time taken by the particle to reach the velocity of

60.0 \mathrm{~ms}^{-1}

is 25 seconds.

Q73

Two balls A and B are placed at the top of 180 m tall tower. Ball A is released from the top at t = 0 s. Ball B is thrown vertically down with an initial velocity 'u' at t = 2 s. After a certain time, both balls meet 100 m above the ground. Find the value of 'u' in ms

-

1. [use g = 10 ms

-

2] :

A 10

B 15

C 20

D 30

Correct Answer

Option D

Solution

As the meeting point lies $100 \mathrm{~m}$ above ground, displacement of ball will be $80 \mathrm{~m}$ . For ball $A$

\begin{aligned} &u=0, \mathrm{~S}=80 \mathrm{~m}, a =+\mathrm{g}=+10 \mathrm{~m} / \mathrm{s}^2 \text{, time }=t_1 \\\\ &\Rightarrow S =u t+\frac{1}{2} a t^2 \\\\ &\Rightarrow 80 =0+\frac{1}{2} \times 10 \times t_1{ }^2 \\\\ &\Rightarrow \frac{160}{10} =t_1{ }^2 \\\\ &\Rightarrow \quad t_1 =4 \mathrm{~s} \end{aligned}

As ball $B$ is thrown after 2 seconds after release of $A$ .

Thus, time available for ball $B$ is 2 seconds to cover a distance of $80 \mathrm{~m}$ .

Let speed be ' $u$ ' $\mathrm{m} / \mathrm{s}, t_2=4-2=2 \mathrm{~s}, \mathrm{~S}=80 \mathrm{~m}$ , $a=+g=+10 \mathrm{~m} / \mathrm{s}^2$

\therefore 80 =u \times 2+\frac{1}{2} \times 10 \times(2)^2

\Rightarrow 80 =u+20

\Rightarrow 2 u =60

\Rightarrow u =30 \mathrm{~m} / \mathrm{s}

Q74

A body projected vertically upwards with a certain speed from the top of a tower reaches the ground in

t_1

. If it is projected vertically downwards from the same point with the same speed, it reaches the ground in

t_2

. Time required to reach the ground, if it is dropped from the top of the tower, is :

A

\sqrt{\mathrm{t}_1+\mathrm{t}_2}

B

\sqrt{\mathrm{t}_1 \mathrm{t}_2}

C

\sqrt{\dfrac{\mathrm{t}_1}{\mathrm{t}_2}}

D

\sqrt{\mathrm{t}_1-\mathrm{t}_2}

Correct Answer

Option B

Solution

To solve this problem, we'll use the equations of motion under constant acceleration due to gravity.

Let's define: $h$ : Height of the tower $u$ : Initial speed of projection (same magnitude in both cases) $g$ : Acceleration due to gravity (positive downward) $t_1$ : Time taken to reach the ground when projected upwards $t_2$ : Time taken to reach the ground when projected downwards $t$ : Time taken to reach the ground when simply dropped Case 1: Projectile Thrown Upwards When the body is projected upwards from the top of the tower, its initial velocity is $-u$ (since upward direction is negative), and it reaches the ground in time $t_1$ .

The equation of motion is: $h = -u t_1 + \dfrac{1}{2} g t_1^2$ Case 2: Projectile Thrown Downwards When the body is projected downwards from the top of the tower, its initial velocity is $u$ , and it reaches the ground in time $t_2$ .

The equation is: $h = u t_2 + \dfrac{1}{2} g t_2^2$ Case 3: Body Dropped When the body is simply dropped, its initial velocity is $0$ , and it reaches the ground in time $t$ : $h = \dfrac{1}{2} g t^2$ Step 1: Equate the Heights From cases 1 and 2, equate the expressions for $h$ : $-u t_1 + \dfrac{1}{2} g t_1^2 = u t_2 + \dfrac{1}{2} g t_2^2$ Step 2: Solve for $u$ Simplify the equation: $-u t_1 - u t_2 = \dfrac{1}{2} g t_2^2 - \dfrac{1}{2} g t_1^2$ $-u (t_1 + t_2) = \dfrac{1}{2} g (t_2^2 - t_1^2)$ $-u (t_1 + t_2) = \dfrac{1}{2} g (t_2 - t_1)(t_2 + t_1)$ Divide both sides by $t_1 + t_2$ : $-u = \dfrac{1}{2} g (t_2 - t_1)$ $u = \dfrac{1}{2} g (t_1 - t_2)$ Step 3: Express $h$ in Terms of $t_1$ and $t_2$ Substitute $u$ back into one of the equations for $h$ : $h = -\left( \dfrac{1}{2} g (t_1 - t_2) \right) t_1 + \dfrac{1}{2} g t_1^2$ Simplify: $h = \dfrac{1}{2} g t_1 t_2$ Step 4: Equate the Height for the Dropped Case From the dropped case: $h = \dfrac{1}{2} g t^2$ Set the two expressions for $h$ equal to each other: $\dfrac{1}{2} g t^2 = \dfrac{1}{2} g t_1 t_2$ $t^2 = t_1 t_2$ Step 5: Solve for $t$ $t = \sqrt{t_1 t_2}$ Conclusion: The time required for the body to reach the ground when dropped is the geometric mean of $t_1$ and $t_2$ : Answer: Option B $t = \sqrt{t_1 \, t_2}$

Q75

An object, moving with a speed of 6.25 m/s, is decelerated at a rate given by :

{{dv} \over {dt}} = - 2.5\sqrt v

where v is the instantaneous speed. The time taken by the object, to come to rest, would be :

A 2 s

B 4 s

C 8 s

D 1 s

Correct Answer

Option A

Solution

Given

{{dv} \over {dt}} = - 2.5\sqrt v

\Rightarrow {{dv} \over {\sqrt v }} = - 2.5dt

On integrating,

\int_{6.25}^0 {{v^{ - {{1}/{2}}}}} \,dv = - 2.5\int_0^t {dt}

\Rightarrow \left[ {{{{v^{ + {{1}/{2}}}}} \over {\left( {{{1}/{2}}} \right)}}} \right]_{6.25}^0 = - 2.5\left[ t \right]_0^t

\Rightarrow - 2{\left( {6.25} \right)^{{{1}/{2}}}} = - 2.5t

\Rightarrow t = 2\,sec

Q76

A particle moving in a straight line covers half the distance with speed

6 \mathrm{~m} / \mathrm{s}

. The other half is covered in two equal time intervals with speeds

9 \mathrm{~m} / \mathrm{s}

and

15 \mathrm{~m} / \mathrm{s}

respectively. The average speed of the particle during the motion is :

A 9.2 m/s

B 8.8 m/s

C 10 m/s

D 8 m/s

Correct Answer

Option D

Solution

Let's denote the total distance covered by the particle as

2d

, where

d

is the distance for each half.

To calculate the average speed, we need to find the total distance traveled and divide it by the total time taken.

For the first half of the journey, the particle covers the distance

d

at a speed of

6 \, \text{m/s}

. The time taken for this part of the journey can be calculated using the formula

\text{time} = \frac{\text{distance}}{\text{speed}}

. So,

\text{time}_1 = \frac{d}{6}

For the second half of the journey, the distance

d

is further divided into two parts, each covered in equal time intervals. Given the speeds are

9 \, \text{m/s}

and

15 \, \text{m/s}

respectively, let's call the equal time intervals

t

. The distances covered in these intervals can be found by

\text{distance} = \text{speed} \times \text{time}

. For the part covered at

9 \, \text{m/s}

:

d_1 = 9t

For the part covered at

15 \, \text{m/s}

:

d_2 = 15t

Since these two parts together make up the second half of the journey,

d_1 + d_2 = d

9t + 15t = d

This gives us

24t = d

, and from this, we can find

t = \frac{d}{24}

. The total time for the second half of the journey is the sum of the times for the two parts, which are equal (

t

each), so the total time for the second half is

2t

. Since

t = \frac{d}{24}

,

\text{time}_2 = 2 \times \frac{d}{24} = \frac{d}{12}

The total time taken for the entire journey is the sum of the times for the first and second halves:

\text{total time} = \text{time}_1 + \text{time}_2 = \frac{d}{6} + \frac{d}{12}

\text{total time} = \frac{2d}{12} + \frac{d}{12} = \frac{3d}{12} = \frac{d}{4}

The total distance is

2d

, and the total time is

\frac{d}{4}

. Therefore, the average speed is calculated as:

\text{average speed} = \frac{\text{total distance}}{\text{total time}} = \frac{2d}{\frac{d}{4}} = \frac{2d}{1} \times \frac{4}{d} = 8 \, \text{m/s}

Thus, the correct answer is Option D: 8 m/s.

Q77

If

\mathrm{t}=\sqrt{x}+4

, then

\left(\dfrac{\mathrm{d} x}{\mathrm{~d} t}\right)_{\mathrm{t}=4}

is :

A 4

B zero

C 8

D 16

Correct Answer

Option B

Solution

Given, $t=\sqrt{x}+4$ , Squaring on both

\begin{aligned} & x=(t-4)^2=t^2-8 t+16 \\\\ & \frac{d x}{d t}=2 t-8\\\\ & \text{ at } t=4 \\\\ & \frac{d x}{d t}=8-8=0 \end{aligned}

Q78

Train A is moving along two parallel rail tracks towards north with speed

72 \mathrm{~km} / \mathrm{h}

and train B is moving towards south with speed

108 \mathrm{~km} / \mathrm{h}

. Velocity of train B with respect to A and velocity of ground with respect to B are (in

\mathrm{ms}^{-1}

):

A -50 and -30

B -50 and 30

C -30 and 50

D 50 and -30

Correct Answer

Option B

Solution

To find the velocity of Train B with respect to Train A, we have to subtract the velocity of Train A from the velocity of Train B, keeping in mind that they are moving in opposite directions.

Since they are moving in opposite directions, the relative velocity is calculated by adding their magnitudes when converting into the same unit, which in this case is meters per second.

First, let's convert the speeds from km/h to m/s by multiplying by the conversion factor

\frac{1000 \text{ m/km}}{3600 \text{ s/h}} = \frac{5}{18} \text{ m/s}

. For Train A:

v_A = 72 \frac{\text{km}}{\text{h}} \times \frac{5}{18} \frac{\text{m/s}}{(\text{km/h})} = 20 \frac{\text{m}}{\text{s}}

For Train B:

v_B = 108 \frac{\text{km}}{\text{h}} \times \frac{5}{18} \frac{\text{m/s}}{(\text{km/h})} = 30 \frac{\text{m}}{\text{s}}

To find the velocity of B relative to A ( $v_{B/A}$ ), we consider the direction: Train B is moving towards the south and Train A is moving towards the north.

Therefore, relative to Train A, Train B is moving even faster towards the south, we calculate:

v_{B/A} = v_B + v_A = 30 \frac{\text{m}}{\text{s}} + 20 \frac{\text{m}}{\text{s}} = 50 \frac{\text{m}}{\text{s}}

Since Train B is moving towards the south and Train A towards the north, we take the southward direction as negative in our coordinate system for this calculation.

That means the velocity of B with respect to A is:

v_{B/A} = -50 \frac{\text{m}}{\text{s}}

Next, we calculate the velocity of the ground with respect to Train B ( $v_{ground/B}$ ).

The ground is stationary, thus it has a velocity of 0 m/s in any direction.

The velocity of an object with respect to another object moving is just the opposite of the second object's velocity.

Thus:

v_{ground/B} = -v_B = -30 \frac{\text{m}}{\text{s}}

However, because we are considering the southward direction as negative, the negative of a southward velocity is a northward velocity.

Hence we get:

v_{ground/B} = 30 \frac{\text{m}}{\text{s}}

So the velocity of Train B with respect to Train A is -50 m/s, and the velocity of the ground with respect to Train B is 30 m/s.

Therefore, the correct answer is: Option A: -50 m/s and -30 m/s.

(Incorrect, because the velocity of ground with respect to B is positive in our chosen coordinate system) Option B: -50 m/s and 30 m/s.

(Correct) Option C: -30 m/s and 50 m/s.

(Incorrect) Option D: 50 m/s and -30 m/s.

(Incorrect) Thus, the correct answer is Option B: -50 m/s and 30 m/s.

Q79

A NCC parade is going at a uniform speed of

9 \mathrm{~km} / \mathrm{h}

under a mango tree on which a monkey is sitting at a height of

19.6 \mathrm{~m}

. At any particular instant, the monkey drops a mango. A cadet will receive the mango whose distance from the tree at time of drop is: (Given

g=9.8 \mathrm{~m} / \mathrm{s}^{2}

)

A 5 m

B 10 m

C 19.8 m

D 24.5 m

Correct Answer

Option A

Solution

A NCC parade is moving at a steady speed of 9 km/h under a mango tree where a monkey is perched at a height of 19.6 meters.

Suddenly, the monkey drops a mango.

To determine which cadet will catch the mango, we need to calculate the distance of the cadet from the tree at the moment the mango is dropped, considering the acceleration due to gravity $g = 9.8 \, \text{m/s}^2$ .

First, we use the formula to find the time $t$ it takes for the mango to fall: $H = \dfrac{1}{2}\,g\,t^2$ Substituting the given values: $19.6 = 4.9\,t^2$ Solving for $t$ : $t^2 = \dfrac{19.6}{4.9}$ $t^2 = 4$ $t = 2 \, \text{sec}$ Next, we need to find the distance $D$ the cadet travels in these 2 seconds: $D = \text{speed} \times \text{time}$ Since the speed is given in km/h, we first convert it to m/s: $9 \, \text{km/h} = 9 \times \dfrac{1000}{3600} \, \text{m/s} = 2.5 \, \text{m/s}$ So, the distance covered by the cadet is: $D = 2.5 \, \text{m/s} \times 2 \, \text{sec} = 5 \, \text{m}$ Thus, the cadet who is 5 meters away from the tree at the moment the mango is dropped will catch the mango.

Q80

A passenger sitting in a train A moving at

90 \mathrm{~km} / \mathrm{h}

observes another train

\mathrm{B}

moving in the opposite direction for

8 \mathrm{~s}

. If the velocity of the train B is

54 \mathrm{~km} / \mathrm{h}

, then length of train B is:

A 80 m

B 200 m

C 120 m

D 320 m

Correct Answer

Option D

Solution

To find the length of train B, we first need to determine the relative velocity between train A and train B.

Since they are moving in opposite directions, their velocities add up:

v_{AB} = v_A + v_B = 90 \mathrm{~km/h} + 54 \mathrm{~km/h} = 144 \mathrm{~km/h}

Now, we need to convert this relative velocity to meters per second:

v_{AB} = \frac{144 \mathrm{~km/h} × 1000 \mathrm{~m/km}}{3600 \mathrm{~s/h}} = 40 \mathrm{~m/s}

The passenger in train A observes train B for 8 seconds. To find the length of train B, we can use the formula:

\text{length} = \text{relative velocity} × \text{time}

\text{length} = 40 \mathrm{~m/s} ~×~ 8 \mathrm{~s} = 320 \mathrm{~m}

So, the length of train B is 320 meters.