Motion in A Straight Line Test

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Motion in A Straight Line Test - 75

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Question 1
1 / -0

A stone $$A$$ is thrown vertically upward with speed $$u$$ and another stone $$B$$ is thrown vertically downward with same speed $$u$$ from a height $$h$$. The time taken by stone $$A$$ and $$B$$ to reach the ground are $$16\ s$$ and $$4\ s$$ respectively. The time taken by another stone to reach the ground after it has been dropped from height $$H$$ is [Take $$g=10\ m/s^{2}$$]

A
$$2\sqrt{10}\ s$$

B
$$10\ s$$

C
$$8\ s$$

D
$$40\ s$$
Question 2
1 / -0

Two persons start running towards each other from two points that are 120 m apart. First person runs with a speed of $$5\;m{s^{ - 1}}$$ and the other with a speed of $$7\;m{s^{ - 1}}$$.Both the persons meet after

A
10 s

B
24 s

C
1 min

D
48 s

Solution
Question 3
1 / -0

A particle slides down a smooth inclined plane of elevation $$\theta$$ fixed in the elevator going up with an acceleration $$a_0$$ as shown. The base of the incline has length $$L$$. Then the time taken by the particle to reach the bottom of the plane is given by:

A
$$\sqrt{\dfrac{2L}{(g+a_0)\sin\theta}}$$

B
$$\sqrt{\dfrac{2L}{(g+a_0)\sin\theta \cos\theta}}$$

C
$$\sqrt{\dfrac{2L}{g\sin\theta \cos\theta}}$$

D
None of the above

Solution

Let a be the acceleration of the particle with respect to the incline.

Taking components of the forces parallel to the incline and applying Newton’s law, we get,

$$mg sinθ + ma_0 sinθ = ma$$

or $$a $$= $$(g + a_0) sinθ$$

This is the acceleration with respect to the elevator.

Now, $$cosθ = L/x$$, where $$ x$$ is the distance moved by the particle.

$$x = L/cosθ$$
From the equation of motion
$$S=ut+\frac { 1 }{ 2 } { at }^{ 2 }$$
u is 0
$$S=\frac { 1 }{ 2 } { at }^{ 2 }$$

$$\dfrac { L }{ cos\theta } =\dfrac { 1 }{ 2 } { \left( g+{ a }_{ 0 } \right) sin\theta \quad t }^{ 2 }$$

$$t=\left( \dfrac { 2L }{ { \left( g+{ a }_{ 0 } \right) sin\theta \quad }cos\theta ^{ } } \right) { }^{ 1/2 }$$
Question 4
1 / -0

Balls are dropped from the roof of a tower at a fixed interval of time. At the moment when $$9 ^ { \text { th } }$$ ball reaches the ground $$\mathrm { n } ^ { \text { th } }$$ ball is at $$3 / 4 ^ { \text { th } }$$ height of tower. The value of $$n$$ is:

A
$$13$$

B
$$7$$

C
$$6$$

D
$$5$$

Solution

The depth of the ground from the roof of the tower be D.

3/4 th Height of the tower is 1/4 th D or D/4 from the roof.

We know the depth of fall is proportional to the square of the time of fall.

Therefore, if the ball takes T second to reach the ground, i.e, to fall a depth of D,

in T/2 second , the depth of fall will be D/4.

If balls are dropped in equal interval of time, when the first ball reaches the ground , ( it takes T second to reach the ground) , nine th ball will be just dropped and 5 th ball would have traveled D/4 depth.

When the 9 th ball reaches the ground, 17 th ball will be just dropped and 13 th ball would have travelled D/4 depth.

Answer n = 13
Hence option (A) is correct
Question 5
1 / -0

A stone thrown upwards with a velocity u reaches upto a height h. If the initial velocity is 2u the height attained would be

A
$$2h$$

B
$$4h$$

C
$$8h$$

D
$$16h$$

Solution
Question 6
1 / -0

A clean body of mass $$100\ g$$ starts moving with a velocity of $$2\ m/s$$ on a smooth horizontal plane and it accumulates dust at the rate of $$5\ g/s$$ . The velocity of body after $$20\ s$$ will be

A
$$0.5\ m/s$$

B
$$1\ m/s$$

C
$$2\ m/s$$

D
$$4\ m/s$$

Solution
Question 7
1 / -0

A body moves with a velocity of 3 m/s due east and then turns due north to travel with the same velocity. If the total time of travel is 6s, the acceleration of the body

A
$$\sqrt 3 m/s^2$$ towards north west

B
$$\frac{1}{\sqrt 2} m/s^2$$ towards north west

C
$$\sqrt 2 m/s^2$$ towards north east

D
all the above

Solution

$$\textbf{Hint}$$: Rate of change of velocity is called as Acceleration.
$$\textbf{Step 1}:\textbf{Change in velocity}$$:
Given The body is travelling with a velocity of 3m/s due East and after turning due north it is travelling with the same velocity.
Total time taken by the body is 6 seconds.
$$ Initial \quad velocity=3 \hat{i}$$ (East or towards x-direction)
$$Final \quad velocity=3\hat{j}$$ (North or towards y-direction)
Change in velocity= Final Velocity-Initial Velocity
$$=3\hat{j}-3\hat{i}=3(\hat{j}-\hat{i})$$
Total time=6 seconds
$$\textbf{Step 2}:\textbf{Acceleration}$$:
$$Acceleration=\dfrac{|3(\hat{j}-\hat{i})}{6}|$$
$$=\dfrac{|(\hat{j}-\hat{i})|}{2}=\dfrac{\sqrt{(1)^2+(-1)^2}}{2}=\dfrac{\sqrt(2)}{2}$$
$$=\dfrac{1}{\sqrt{2}}m/s^{2}$$ in the North-west direction.

$${\textbf{Correct option: B}}$$
Question 8
1 / -0

Burglars, after looting a bank, jump into a waiting van at rest and speed up along a straight road at a constant rate of $$2.0{\text{ m}}{{\text{s}}^{ - 2}}$$ . At that instant, a police jeep moving at a constant speed of 30 m/s is 300m behind the burglars moving in the same direction. Choose the correct statement from the following;

A
the police with not be able to catch the burglars

B
the shortest distance between the police and the burglars is 75m

C
The shortest distance between the police and the burglars is 125m

D
The police can catch the burglars if they accelerate at a constant rate of $$0.5{\text{ m}}{{\text{s}}^{ - 2}}$$
Question 9
1 / -0

A ball is dropped freely while another is thrown vertically downward with an initial velocity of $$\dfrac{v}{2}$$ from the same point simultaneously. After 't' second they are separated by a distance of?

A
$$\dfrac{vt}{2}$$

B
$$\dfrac{1}{2}gt^2$$

C
$$vt$$

D
$$\dfrac{vt+gt^2}{2}$$

Solution

ball 1:
$$u=0$$
$$v=u+at$$
$$v=gt$$
$$S1=\dfrac { ut+{ at }^{ 2 } }{ 2 } $$
$$S1=\dfrac { { gt }^{ 2 } }{ 2 } $$
ball 2:
$$u=v$$
$$vf=u+at$$
$$vf=v+gt$$
$$S2=\dfrac { vt+{ gt }^{ 2 } }{ 2 } $$
distance of separation $$=S2-S1=vt.$$
Question 10
1 / -0

A rocket is projected at the surface of earth like that in which acceleration 19.6 $$m/s^{2}$$ produces. After 5 seconds if stops its engine then calculate the following before to come at the surface of the earth.

A
Maximum height obtained by rocket

B
The velocity of the rocket at the time of reach it in the surface.

C
Total time of journey

D
The graph between velocity-time and acceleration - time.

Solution