Projectile motion problems for IIT JEE

Question 1
Consider the figure given below

1. Acceleration is positive during A to C and negative during C to B
2. Acceleration is maximum at C
3. Velocity is maximum at C
4. Average acceleration is zero during the journey

Question 2
Which one of the following statement(s) is true?

1. A body moving with uniform speed can have variable velocity
2. A body moving with uniform velocity can have variable speed
3. Average velocity is always equal to instantaneous velocity
4. x-t graph can be a straight line parallel to position axis

Question 3
A ball is projected upward at a certain angle with the horizontal .which of the following statement is/are correct. At highest point
a. velocity of the projectile is not zero
b. acceleration of the projectile is zero
c. velocity of the projectile is along the horizontal direction
d. Acceleration of the projectile is vertically upwards

Question 4
Which one of the following statement is correct?
a. A body has constant speed but varying velocity
b. A body has constant speed but varying acceleration
c. A body having constant speed cannot have acceleration
d. None of the above

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

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

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

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 t1 to t2 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

Question 9
 A particle is going moving along x-axis. Which of the following statement is false
a. At time t₁ (dx/dt)_t=t1=0,then (d²x/dt²)_t=t1=0
b. At time t₁ (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 t₁ (d²x/dt²)_t=t1 < 0 then the particle is directed towards origin

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

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
 

R=u2sin2θ2g
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

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=2t²
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 2i+4tj
d. the acceleration vector at time t is 4j

Given 
x=2t
y=2t²
Eliminating t we get
y=x²/2
So it is parabola
V_x=dx/dt=2
v_y=dy/dt=4t
So velocity at any time t is given by
v=2i+4tj
Now similarly
a_x=dV_x/dt=0
a_y=dV_y/dt=4
So acceleration vector is a=4j

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
 

the distance s₁ covered by the car during the time it is accelerated is given by
2as₁=v²
Which gives
s₁=v²/2a
Similarly in the decelerated motion, distance covered is
2bs₂=v²
Which gives
s₂=v²/2b

So total distance travelled during the whole journey
s=s₁+s₂=v²/2(1/a+1/b)

Let t₁ be the time taken during accelerated motion then
v=at₁ or t₁ =v/a
and Let t₂ be the time taken during decelerated motion then
v=bt₂ or t₂=v/b

Total time taken
t=t₁+t₂ =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=4i+3j m/s. It is moving with uniform acceleration a=.4i+.3j m/s².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+.2²)i +(3t+.15²)j
d . None of the above

Solution 1

Acceleration is defined as = dvdt which is a slope to the velocity –time diagram
From the graph, it is clear that slope is positive from A to C and negative from C to B
Also slope become maximum at point C
Now from the graph, it is clear that velocity is maximum at point C
Now Average acceleration is defined =changeinvelocityTime
As change in velocity is zero during the journey, so Average acceleration is zero for the journey.
So all options are correct

Solution 2

Only option (a) is correct since direction of body moving with constant speed can have directions changing with time and hence it can have variable velocity.
Option b is wrong because magnitude of body moving with uniform velocity remains same and so does its speed remain constant.
Option c is wrong as average velocity and average velocity of a particle along a path are not always same.
Option d is wrong because x-t graph parallel to position axis indicated that the position of the given object is changing at a given instant of time.

Solution 3

At highest point vertical component of velocity becomes zero but horizontal components remains.
Net velocity is along horizontal
Also acceleration is vertically downwards throughout the journey
Hence the correct option is (c)

Solution 4

 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 .

Hence the correct option is (a)

Solution 5

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)

Solution 6

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)

Solution 7

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)

Solution 8

Speed = dx/dt

Acceleration= dv/dt

Distance = ∫vdt

Hence the answer is (d)

Solution 9

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)

Solution 10

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)

Solution 11

 R=u2sin2θ2g

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

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

Solution 12

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)

Solution 13

Given
x=2t
y=2t²
Eliminating t we get
y=x²/2
So it is parabola
V_x=dx/dt=2
v_y=dy/dt=4t
So velocity at any time t is given by
v=2i+4tj
Now similarly
a_x=dV_x/dt=0
a_y=dV_y/dt=4
So acceleration vector is a=4j

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

Solution 14

the distance s₁ covered by the car during the time it is accelerated is given by
2as₁=v²
Which gives
s₁=v²/2a
Similarly in the decelerated motion, distance covered is
2bs₂=v²
Which gives
s₂=v²/2b

So total distance travelled during the whole journey
s=s₁+s₂=v²/2(1/a+1/b)

Let t₁ be the time taken during accelerated motion then
v=at₁ or t₁ =v/a
and Let t₂ be the time taken during decelerated motion then
v=bt₂ or t₂=v/b

Total time taken
t=t₁+t₂ =v(1/a+1/b)

So average speed =total distance/total time taken =s/t=v/2
Hence Answer is (b) and (c)

Solution 15

Given
u=4i+3j m/s
a=.4i+.3j m/s²
So velocity vector at any time t
v=u+at = 4i+3j+(.4i+.3j)t =(4+.4t)i+(3+.3t)j
So velocity at 10 sec
v=8i+6j
|v|=10
displacement vector at any time t
s=ut+(1/2)at² =(4i+3j)t+(1/2)(.4i+.3j)t² =i(4t+.2t²)+j(3t+.15t²)
Hence Answer is (a) and (b)