In a uniform circular motion, speed is constants but velocity vector is changing continuously and also acceleration vector is also changing continuously. So all of them are wrong

**Question 5**
Which of the following remains constant during the motion of the projectile fired from a planet,

a. KE

b. Momentum

c. Vertical component of the velocity

d. Horizontal component of the velocity

Solution
Since velocity is changing, KE and momentum is not constant.

Now since acceleration is vertically downwards, vertical component of velocity is changing.

Since there is no acceleration is horizontal direction, horizontal components is constant

Hence the correct option is (d)

**Question 6**
Which one is wrong for a body having uniform circular motion?

a. Speed of the body is constant

b. Acceleration is directed towards the centre

c. Velocity and Acceleration vector are having an angle 45

d. none of the above

Solution
Speed of particle is constant in Uniform circular motion

Velocity vector is tangent to the path at any point on the path.

Acceleration vector is directly inwards towards center

So velocity and acceleration vector are perpendicular to each other in Uniform circular motion

Hence the answer is (a) and (b)

**Question 7**
A body moves along a semicircular track of Radius R. Which of the following statement is true

a. Displacement of the body is 2R

b. Distance travelled by the body is πR

c. Displacement of the body is πR

d. none of the above

Solution
Displacement is measured by the length of line joining the initial and final point.

While distance is the length measured along the path

So in this case

Displacement=2R and distance=πR

Hence the answer is (a) and (b)

**Question 8**
Which of the following is false

a. The speed of the particle at any instant is given by the slope of the displacement-time graph

b. The distance moved by the particle in a time interval from t

_{1} to t

_{2} is given by the area under the velocity -time graph during that time interval

c. Magnitude of the acceleration of the particle at any instant is given by the slope of the velocity time graph

d. none of the above

Solution
Speed =$\frac {dx}{dt}$

Acceleration=$\frac {dv}{dt}$

Distance =$\int vdt$

Hence the answer is (d)

**Question 9**
A particle is going moving along x-axis. Which of the following statement is false

a. At time t

_{1} (dx/dt)

_{t=t1}=0,then (d

^{2}x/dt

^{2})

_{t=t1}=0

b. At time t

_{1} (dx/dt)

_{t=t1} < 0 then the particle is directed towards origin

c. If the velocity is zero for a time interval, the acceleration is zero at any instant within the time interval.

d. At time t

_{1} (d

^{2}x/dt

^{2})

_{t=t1} < 0 then the particle is directed towards origin

Solution
If the velocity is zero at any point that does not mean acceleration will become zero at that point

Example when a body is thrown vertically upwards, its velocity becomes zero at highest point but acceleration does not become zero

If the velocity is negative then body is moving towards origin.

If the acceleration is negative that does not mean it is moving towards origin for example when a body is thrown vertically upwards its acceleration is negative but it is moving away from origin

Hence the answer is (b) and (c)

**Question 10**
A particle starts at time t=0 from x=0 along the positive x-axis with constant speed v .After time t,it return back towards the origin with the speed 2v and reaches the origin in t/2 sec .Which of the following is true for the whole process

a. Average velocity is zero for the whole process

b. Average speed is 4/3v for the whole process

c. Displacement at time t is equal to vt
d. Displacement at time 3t/2 is 2vt

Solution
Since net displacement is zero, average velocity is zero for the whole process

Total distance travelled in forwards direction is vt....total time taken is t

Total distance travelled in backwards direction is vt....total time taken is t/2

So total distance in the journey=2vt

Total time taken=3t/2

Average speed=total distance/total time

=4vt/3t

=4v/3

hence answer is (a),(b) and (c)

**Question 11**
The range of the projectile depends upon

a. Angle of the projection

b. Acceleration due to gravity

c. Mass of the projectile

d. magnitude of the velocity of projection

Solution
$R=\frac {u^2sin 2 \theta}{2g}$

So it depends on velocity, angle and acceleration due to gravity

Hence the answer is (a),(b) and (d)

**Question 12**
Two bullets A and B are fired horizontally with speed v and 2v respectively. Which of the following is true
a. Both will reach the ground in same time

b. Bullet with speed 2v will cover more horizontal distance on the ground

c. B will reach the ground in less time than A

d. A will reach the ground in less time than B

Solution
Time taken to reach the ground depends on the velocity in vertical direction...Since both the bodies are projected horizontally. There are no vertical components of velocity involved. So they will reach the ground in same time. Now time taken is same so body with more horizontal velocity will travel more on the ground.

Hence the answer is (a) and (b)

**Question 13**
A body is projected horizontally from a point above the ground.The motion of body is defined as

x=2t

y=2t

^{2}
where x and y are horizontal and vertical displacement respectivley at time t.Which one of the following is true

a. The trajectory of the body is a parabola

b. The trajectory of the body is a straight line

c. the velocity vector at point t is 2

**i**+4t

**j**
d. the acceleration vector at time t is 4

**j**
Solution
Given

x=2t

y=2t^{2}

Eliminating t we get

y=x^{2}/2

So it is parabola

V_{x}=dx/dt=2

v_{y}=dy/dt=4t

So velocity at any time t is given by

v=2**i**+4t**j**

Now similarly

a_{x}=dV_{x}/dt=0

a_{y}=dV_{y}/dt=4

So acceleration vector is a=4**j**

Hence Answer is (a),(c) and (d)

**Question 14**
A car, starting from rest is accelerated at constant rate a until it attains speed v. It is then retarded at a constant rate b until it comes to rest. which of the following is true

a. the average speed for the whole motion is av/2b

b. the average speed for the whole motion is v/2

c. Total time taken for the journey is v(1/a+1/b)

d. none of the above

Solution
the distance s_{1} covered by the car during the time it is accelerated is given by

2as_{1}=v^{2}

Which gives

s_{1}=v^{2}/2a

Similarly in the decelerated motion, distance covered is

2bs_{2}=v^{2}

Which gives

s_{2}=v^{2}/2b

So total distance travelled during the whole journey

s=s_{1}+s_{2}=v^{2}/2(1/a+1/b)

Let t_{1} be the time taken during accelerated motion then

v=at_{1} or t_{1} =v/a

and Let t_{2} be the time taken during decelerated motion then

v=bt_{2} or t_{2}=v/b

Total time taken

t=t_{1}+t_{2} =v(1/a+1/b)

So average speed =total distance/total time taken =s/t=v/2

Hence Answer is (b) and (c)

**Question 15**
The initial velocity of the particle is u=4

**i**+3

**j** m/s. It is moving with uniform acceleration a=.4

**i**+.3

**j **m/s

^{2}.which of the following is true

a. the magnitude of the velocity after 10 sec is 10m/s

b. The velocity vector at time t is given by (4+.4t)

**i** +(3+.3t)

**j**
c. the displacement at time t is (4t+.2

^{2})

**i** +(3t+.15

^{2})

**j**
d . None of the above

Solution
Given

u=4**i**+3**j** m/s

a=.4**i**+.3**j** m/s^{2}

So velocity vector at any time t

v=u+at = 4**i**+3**j**+(.4**i**+.3**j**)t =(4+.4t)**i**+(3+.3t)**j**

So velocity at 10 sec

v=8**i**+6**j**

|v|=10

displacement vector at any time t

s=ut+(1/2)at^{2} =(4**i**+3**j**)t+(1/2)(.4**i**+.3j)t^{2} =**i**(4t+.2t^{2})+**j**(3t+.15t^{2})

Hence Answer is (a) and (b)

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