In this page we have Motion in a plane Solved examples for Class 11. This worksheet contains all the format of questions like Multiple choice question,Long answer type questions, Fill in the blanks ,True and False,Text based questions. Hope you like them and do not forget to like , social share
and comment at the end of the page.
Numerical Questions
Question 1
A stone is thrown from ground level over horizontal ground. It just clears three walls, the successive distances between them being r and 2r. The inner wall is 15/7 times as high as the outer walls which are equal in height. The total horizontal range is nr, where n is an integer. Find n. Solution
Let us just assume that both the outer walls are equal in height say h and they are at equal distance x from the end points of the parabolic trajectory as can be shown below in the figure.
Now equation of the parabola is
y=bx-cx2 ----(1)
y=0 t x=nr=R
where R is the range of the parabola.
Putting these values in equation (1) we get
b=cnr (2)
Now the range RR of the parabola is
R=a+r+2r+a=nr
This gives
a=(n-3)r/2 (3)
The trajectory of the stone passes through the top of the three walls whose coordinates are
(a,h),(a+r,15h/7),(a+3r,h)
Using these co-ordinates in equation 1 we get
h=ab-ca2 (4)
15h/7=b(a+r)-c(a+r)2 (5)
h=b(a+3r)-c(a+3r)2 (6)
After combining (2), (3), (4), (5) and (6) and solving them we get n = 4.
Question 2
A particle moves such that
x = (18.0)t
and y = 4t - 4.90t2
(a) Write a vector expression for the particle position as a function of time, using the unit vectorsi and j
(b) Obtain the expression for the velocity vector as a function of time
(c) Obtain the expression for the acceleration vector a as a function of time.
(d) Find the the position, the velocity, and the acceleration of the particle at t = 1.00 s. Solution
(a) Position vector is given by r= xi + yj Therefore r= (18.0)t i + [4t - 4.90t2 ] j
(b)v= dr/dt =d[(18.0)t i + [4t - 4.90t2 ] j ]/dt =18 i + (4-9.8t)j
(c) a= dv/dt =-9.8 j
(d) Att=1 sec r= (18.0)t i + [4t - 4.90t2 ] j r= (18.0)i - .9 j
(e) At t = 100 s, v=18 i + (4-9.8t)j =18i -5.8 j
(f)At1= 3.00 s, a = -9.8 j
Multiple Choice Questions
Question 3
Which two quantities are constant throughout projectile motion when air resistance is negligible?
(a) The speed and acceleration.
(b) The vertical component of velocity and acceleration.
(c) The magnitude of the acceleration and the speed.
(d) The horizontal component of velocity and acceleration.
(e) The direction of the velocity and acceleration. Solution
(a) The speed and acceleration.
- speed is not constant. a = constant = g.
(b) The vertical component of velocity and acceleration.
- ay ≠ 0 ⇒ vy not constant.
(c) The magnitude of the acceleration and the speed.
- speed is not constant.
(d) The horizontal component of velocity and acceleration.
- ax = 0 ⇒ vx = constant.
(e) The direction of the velocity and acceleration.
- Direction of v changes.
Therefore answer is D.
Question 4
A truck goes around a circular track of radius R at speed v so that it makes one circuit every T seconds As it does so it experiences a centripetal acceleration of magnitude p.If the truck now goes around a different circular track of radius 4R so that it now takes a time ½T to go around once, what is the magnitude of its centripetal acceleration?
(a) ½ p
(b) 16p.
(c) 4p.
(d) 8p. Solution
Initial situation
a1 = v2 /R
Now v=2πR/T
or a1= 4π2 R/T2
On the second track:
a2= 4π2 (4R)/ (.5T)2
=64π2 R/T2
Now
a2 = 16 a1 =16p
Hence (b) is correct
Question 5
Which one of the following statements is true?
(a) A scalar quantity is the one that is conserved in a process.
(b) A scalar quantity is the one that can never take negative values.
(c) A scalar quantity is the one that does not vary from one point to another in space.
(d) A scalar quantity has the same value for observers with different orientations of the axes Solution
(d)
Question 6
The component of a vector r along X-axis will have maximum value if
(a) r is along positive Y-axis
(b) r is along positive X-axis
(c) r makes an angle of 45° with the X-axis
(d) r is along negative Y-axis Solution
(b) as component is $r \cos \theta$ as this is maximum when $\theta$ is zero
Question 7
Consider the quantities, pressure, power, energy, impulse, gravitational potential, electrical charge, temperature, area. Out of these, the only vector quantities are
(a) Impulse, pressure and area
(b) Impulse and area
(c) Area and gravitational potential
(d) Impulse and pressure Solution
(b) As potential and Pressure are scalar quantities
Question 8
It is found that |A+B|=|A|.This necessarily implies,
(a) B = 0
(b) A,B are anti-parallel
(c) A,B are perpendicular
(d) $\mathbf{A}.\mathbf{B} \leq 0$ Solution
|A+B|=|A|
Squaring both the sides
|A+B|2=|A|2
$(\mathbf{A} + \mathbf{B}).(\mathbf{A} + \mathbf{B})=\mathbf{A}.\mathbf{A}$
$\mathbf{A}.\mathbf{A} + \mathbf{B}.\mathbf{B} + 2 \mathbf{A}.\mathbf{B}=\mathbf{A}.\mathbf{A}$
$\mathbf{B}.\mathbf{B} + 2 \mathbf{A}.\mathbf{B}=0$
$|\mathbf{B}|^2 + 2 \mathbf{A}.\mathbf{B}=0$
Now $|\mathbf{B}|^2$ cannot be negative, hence $\mathbf{A}.\mathbf{B} \leq 0$
Therefore (d) is the correct option
Question 9
if A= 3i + 4j and B= 7i + 24j. Which is of the below vector having the same magnitude as B and parallel to A
(a) 15i + 20j
(b) 15i - 20j
(c) 24i - 7j
(d) 7i - 24j Solution
$|\mathbf{A}|= \sqrt {3^2 + 4^2} =5$
Unit vector in the direction of A will be given by
$\widehat{\mathbf{A}} = \frac {\mathbf{A}}{|\mathbf{A}|} = \frac {1}{5} (3 \mathbf{i} + 4 \mathbf{j})$
Also
$|\mathbf{B}|=\sqrt {7^2 +24^2} = 25$
The vector having the same magnitude as B and parallel to A
$=|\mathbf{B}| \widehat{\mathbf{A}}= 25 \times \frac {1}{5} (3 \mathbf{i} + 4 \mathbf{j})=15 \mathbf{i} + 20 \mathbf{j}$
Hence (a) is the correct option
Question 10
For a particle performing uniform circular motion, choose the correct statement(s) from the following:
(a) Magnitude of particle velocity (speed) remains constant.
(b) Particle velocity remains directed perpendicular to radius vector.
(c) Direction of acceleration keeps changing as particle moves.
(d) Angular momentum is constant in magnitude but direction keeps changing Solution
(a),(b) and (c)
Conceptual Questions
Question 11
A ball is thrown from a roof top at an angle of 30° above the horizontal. It hits the ground a few seconds later. At what point during its motion, does the ball have
(a) greatest speed.
(b) smallest speed.
(c) greatest acceleration?
Explain Solution
Horizontal velocity remains constant through out the motion. Vertical velocity first decreases and reaches zero at highest point and then start increasing and reaches maximum Just before it hits the ground.
(a) So greatest speed happens Just before it hits the ground.
(b) So smallest speed happens when vertical velocity becomes zero.So At the highest point ,smallest speed is achieved
(c) Acceleration is due to acceleration due to gravity and it remains constants through out the motion.a = g = constant.
Question 12
A football is kicked into the air vertically upwards. What is its
(a) acceleration at the highest point
(b) velocity at the highest point? Solution
At highest point, velocity becomes zero and acceleration remains same as acceleration due to gravity
Question 13
What will the effect on horizontal range of a projectile when its initial velocity is doubled keeping the angle of projection same? Solution
$R= \frac {u^2 \sin 2\theta}{g}$
So when the initial velocity is doubled, range will becomes four times
Question 14
Two spheres of 1 kg and 10 kg are dropped simultaneously from the same height . which sphere will reach the ground first? Solution
Both the spheres will reach simultaneously, if air friction is negligible
Question 15
What is the angle between the direction of velocity and acceleration at the highest point of a projectile? Solution
90°
Question 16
a. Can a scalar quantity be added to vector quantity
b. Can the product of scalar quantity and vector quantity possible
c. Is the magnitude and direction same for A + B and B + A
d. Is the magnitude and direction same for A - B and B - A Solution
a. No
b. Yes
c. Yes
d. False. Magnitude is same but direction is opposite
Numerical Questions
Question 17
A bomb is fired from a cannon with a velocity of 1000 m/s making an angle 30° with the horizontal
a. What is the time taken for the bomb to reach highest point
b. What is the time of flight
c. What is the maximum height reached
d. what is the range
Take g=9.8 m/s2 Solution
Here u=1000 m/s, $\theta =30^0$
a. Time taken by the bomb to reach maximum height is given by
$t= \frac {u \sin \theta}{g} =\frac {1000 \times .5}{9.8} = 51 s$
b. Time of flight = 2t= 102 sec
c. The maximum height reached is given by
$h=\frac {u^2 \sin^2 \theta}{2g} = 1.27 \times 10^4$ m
d. Horizontal Range is given by
$R= \frac {u^2 \sin 2\theta}{g}=8.83 \times 10^4$ m
Question 18
The sum and difference of the two vectors are equal in magnitude
|A + B| = |A - B|
Prove that the vectors A and B are mutually perpendicular Solution
|A + B| = |A - B|
|A + B|2 = |A - B|2
or
$(\mathbf{A} + \mathbf{B}).(\mathbf{A} + \mathbf{B})=(\mathbf{A} - \mathbf{B}).(\mathbf{A} - \mathbf{B})$
$\mathbf{A}.\mathbf{A} + \mathbf{B}.\mathbf{B} + 2 \mathbf{A}.\mathbf{B}=\mathbf{A}.\mathbf{A} + \mathbf{B}.\mathbf{B} - 2 \mathbf{A}.\mathbf{B}$
$4 \mathbf{A}.\mathbf{B}=0$
$\mathbf{A}.\mathbf{B}=0$
As the scalar product of A and B is zero,vectors A and B are mutually perpendicular
Question 19
A bullet is fired from a gun at a speed of 5000 m/s. At what height should the gun be aimed above a goal if it has to strike the goal at a distance of 500 m?
Take g=10m/s2 Solution
Let t be the time taken to cover the horizontal distance of 500 m from the gun to the target ,then
$t= \frac {500}{5000} = .1s$
During this time,bullet will fall down vertically due to acceleration due to gravity.
Vertical distance covered
$h= \frac {1}{2} gt^2 = .005$ m = 5cm.
Hence the gun should be aimed at 5cm above the target
Question 20
A person sees a bird on a tree 39.6 m high and at a distance of 59.2 m. With what velocity the person should throw an arrow at an angle of 45° so that it may hit the bird? Solution
Let u be the initial velocity abd t be the time to hit the bird
Horizontal distance covered =$u \cos \theta \times t = \frac {ut}{\sqrt {2}}$
Now
$\frac {ut}{\sqrt {2}}= 59.2$ -(1)
Vertical distance covered in t sec
$h= u \sin \theta t - \frac {1}{2}gt^2 = \frac {ut}{\sqrt {2}}- \frac {1}{2} \times 9.8 \times t^2$
Now
$\frac {ut}{\sqrt {2}}- \frac {1}{2} \times 9.8 \times t^2=39.6$ -(2)
From (1) and (2)
$39.6 = 59.2 - \frac {1}{2} \times 9.8 \times t^2$
t= 4 sec
Substituting this value in (1),we get
$u=29.6 \sqrt {2}$ m/s
Text based Questions
Question 21 Define Centripetal acceleration and Derive an expression for centripetal acceleration for an body moving with uniform speed v along a circular path of radius r Question 22 What is projectile motion. Derive the expression for Time of flight, Maximum height and Horizontal range Question 23 Establish the relationship between linear velocity and angular velocity in a uniform circular motion Question 24 Suppose you have two force F and F. How would you combine them in order to have a resultant force of magnitude
a. zero
b. 2 |F|
c. |F| Solution
a. Anti-parallel
b. Parallel
c. both at an angle of 120°
