Question 1.
The radius vector of point x relative to origin varies with time as
$\mathbf{r}=a cos(kt)\mathbf{i}+b sin(kt)\mathbf{j}$
Where a and b are constants and i and j are vectors along x and y axis. Which one of the following is the mean velocity vector?
a. $ \frac {a cos(kt)\mathbf{i}- b sin(kt)\mathbf{j}}{t}$
b. $a cos(kt)\mathbf{i}+ b sin(kt)\mathbf{j}$
c. $a cos(kt)\mathbf{i}- b sin(kt)\mathbf{j}$
d. $ \frac {a cos(kt)\mathbf{i}+ b sin(kt)\mathbf{j}}{t}$
Solution
Average velocity vector is given as
$<\mathbf{v}>=\frac{\int \mathbf{v}dt}{t}$
Now given
$\mathbf{r}=a cos(kt)\mathbf{i}+b sin(kt)\mathbf{j}$
or
$\mathbf{v}= \frac {d\mathbf{r}}{dt}= - a sin(kt)\mathbf{i} + b cos(kt)\mathbf{j}$
Now we calculate Average velocity vector as
$<\mathbf{v}>=\frac{\int (- a sin(kt)\mathbf{i} + b cos(kt)\mathbf{j})dt}{t}$
$<\mathbf{v}>= \frac {a cos(kt)\mathbf{i}+ b sin(kt)\mathbf{j}}{t}$
Question 2.
A wind is blowing in the North direction at the speed of 5 km/hr. An airplane moves to a point in the East which is 2000 km away in 40 hr. Find the velocity of the airplane with respect to the wind.
Solution
The situation is depicted in the figure below
Let take v as the velocity with respect to wind. Lets it makes an angle $\theta$ with the east direction
Velocity of the airplane with respect to ground=2000/40=50 km/hr
From the figure
$vsin \theta =5$
and $vcos \theta =50$
So $v= \sqrt {2525}$ and $\theta = tan^{-1} \frac {1}{10}$
Question 3.
Two cars A and B run at constant speed $u_1$ and $u_2$ along the highways intersecting at an angle $\theta$. They start at t=0 at the intersection point. Find the time required to have distance s between the two cars
Solution
Let us assume
$u_{12}$= Speed of second car with respect to first car
$u_{21}$= Speed of first car with respect to second car
also $u_{12}= u_{21}$
From the figure given below and from mathematics theorem
$u_{12}= \sqrt {u_1^2 + u_2^2 + 2u_1 u_2 cos \theta}$
Now time will given by
$t = \frac {s}{\sqrt {u_1^2 + u_2^2 + 2u_1 u_2 cos \theta}}$
Question 4.
A bar XY of length l which always remains in the same vertical plane has its ends X and Y constrained to remain in contact with the horizontal floor and in vertical wall as shown below in the figure.
The bar starts from a vertical position and end X is moved along the floor with a constant velocity v
_{0} so that its displacement OX=v
_{0}t. The displacement time graph for Y end is
(a) Ellipse
(b) Circle
(c) Parabola
(d) Straight line
Solution
Choosing for the X-axis the vertical line of motion OY with the origin at O , we see from the figure that the displacement x of the end Y is given by the equation
$x = \sqrt {l^2 - (v_0 t)^2 }$
$x^2 = l^2 - v_0^2 t^2 $
$x^2 + v_0^2 t^2 = l^2$
or
$\frac {x^2}{l^2} + \frac {t^2}{(l\v_0)^2} = 1 $
This is the equation of an ellipse.
Question 5.
Three particles of mass m
_{1},m
_{2},m
_{3} moves in plane with initial velocity
v_{1 }, v_{2} , v_{3} . A constant acceleration
a acts on the three particles. At t=t
_{0}, there velocities becomes
v_{1}^{'}, v_{2}^{'} , v_{3}^{'} Which of the following expression are correct?
Solution
Now
$\mathbf{v_1^{'}} = \mathbf{v_1} + \mathbf{a}t_0$
Multiplying by $m_1$ on both sides
$m_1 \mathbf{v_1^{'}} = m_1 \mathbf{v_1} + m_1 \mathbf{a}t_0 $ ----(1)
Similarly
$m_2 \mathbf{v_2^{'}} = m_2 \mathbf{v_2} + m_2 \mathbf{a}t_0 $ ----(2)
$m_3 \mathbf{v_3^{'}} = m_3 \mathbf{v_3} + m_3 \mathbf{a}t_0 $ ----(3)
Adding (1) ,(2) and (3)
$ \frac {(m_1 \mathbf{v_1} + m_2 \mathbf{v_2} + m_3 \mathbf{v_3}) - (m_1 \mathbf{v_1^{'}} + m_2 \mathbf{v_2^{'}} + m_3 \mathbf{v_3^{'}})}{m_1 + m_2 + m_3} = - \mathbf{a}t_0$
Adding (1) and (2) only
$ \frac {(m_1 \mathbf{v_1} + m_2 \mathbf{v_2} ) - (m_1 \mathbf{v_1^{'}} + m_2 \mathbf{v_2^{'}} )}{m_1 + m_2 } = - \mathbf{a}t_0$
Now we have
$\mathbf{v_1^{'}} = \mathbf{v_1} + \mathbf{a}t_0$ -- (4)
$\mathbf{v_2^{'}} = \mathbf{v_2} + \mathbf{a}t_0$ -- (5)
$\mathbf{v_3^{'}} = \mathbf{v_3} + \mathbf{a}t_0$ --(6)
Adding (4),(5) and (6)
$ \frac {(\mathbf{v_1} + \mathbf{v_2} + \mathbf{v_3}) - (\mathbf{v_1^{'}} + \mathbf{v_2^{'}} + \mathbf{v_3^{'}})}{3} = - \mathbf{a}t_0$
Subtracting (4) and (6)
$ \mathbf{v_1^{'}} - \mathbf{v_2^{'}} = \mathbf{v_1} - \mathbf{v_2}$
Question 6.
Two particles X and Y travel along the x and y axis with respective velocities
$\mathbf{v_1} = 2\mathbf{i}$ m/sec
$\mathbf{v_2} = 3\mathbf{j}$ m/sec
At t=0 they are at
$x_1 = -3m$ and $y_1=0$
$x_2= 0$ and $y_2=-3 m$
Find the vector which represents the position of Y relative to X as a function of t
i and j are respective unit vectors along x and y direction
a. $(3t -3) \mathbf{j} + (3-2t) \mathbf{i}$
b. $(3t +3) \mathbf{j} + (3+2t) \mathbf{i}$
c. $(3t -2) \mathbf{j} + (2-2t) \mathbf{i}$
d. None of the above
Solution
Position of X after time t
$\mathbf{r_1}=(-3 + 2t)\mathbf{i}$
Position of Y after time t
$\mathbf{r_2}=(-3 + 3t)\mathbf{j}$
Relative vector of Y w.r.t to X
$= \mathbf{r_2} - \mathbf{r_1}$
Or =$(3t -3) \mathbf{j} + (3-2t) \mathbf{i}$
(B) Two ring X and Y are put on two vertical stationary rods AB and A
^{’}B
^{’}. An inextensible thread is fixed at point A
^{’} and on ring X and is passed through ring Y.
The system is shown below in Figure
Question 7
The ring X moves with constant velocity v downwards. The angle AXY is θ. Find the velocity of the ring Y
Solution
Let x be the distance of ring X from Fixed point A
Let y be the distance of ring Y from fixed point A'
Let L is distance between the two rods
Now length of the string remains constant
$ x + \sqrt {L^2 + (y-x)^2 } = C$
So, it is vertically upwards
Question 8
Find the velocity of ring Y with respect to ring X
Solution
Now velocity of Ring Y with respect to Ring X
$w_x = w -v = v - \frac {v}{cos \theta} -v = - \frac {v}{cos \theta}$
It is also vertically upwards
(C) A man X can swim at a speed of 10 m/s with respect to river . He wants to cross the river which is 50 m wide and has a current of 5 m/s
If he wishes to land on the other bank at a point directly across the river from his starting point
Question 9
Velocity of man with respect to the person standing on the bank
a. 5 m/s
b. 10 m/s
c. 6 m/s
d. $5 \sqrt {3}$ m/s
Solution
Let $v_c$ -> Velocity of the current
$v_w$ -> Velocity w.r.t current
$v_s$ -> Velocity w.r.t shore
Now
$\mathbf{v_w}=\mathbf{v_s} - \mathbf{v_c}$
$\mathbf{v_s}= \mathbf{v_w} + \mathbf{v_c}$
To directly cross the river
$ \mathbf{v_s} \perp \mathbf{v_c}$
$ sin \theta = \frac {v_c}{v_w} = \frac {5}{10} = \frac {1}{2}$
So $\theta = 30$°
So angle made with the current is 90+30=120 °
$v_s = v_w cos \theta = \frac {10 \sqrt {3}}{2}= 5 \sqrt {3} $ m/s
$t = \frac {d}{v_s} = \frac {50}{5 \sqrt {3}} = \frac {10}{ \sqrt {3}} $ seconds
Hence (d) is the correct answer
Question 10
Find the angle made by the man from the current
a. 120°
b. 60°
c. 30°
d. 90°
Solution
Here (a) is the answer as solved above
Question 11
Find the time taken by the man to cross the river
Solution
Here (a) is the answer as solved above
(D)Now if he instead decides to cross the river in shortest possible time
Question 12
what direction it should swim
a. 90° to current
b. 30° to current
c. 60 ° to current
d. none of these
Solution
To maximize the components of his velocity perpendicular to the river bank, the swimmer should head straight across the stream
Hence (a) is the correct option
Question 13.
How much time
a. 5
b. 10
c. 6
d. 4
Solution
T=50/10=5 sec
Question 14.
How much distance it will land from his starting point
a. 30
b. 20
c. 25
d. none of these
Solution
$D=v_c \times t=5 \times 5=25 $m/s
(E)The radius vector describing the position of the particle A relative to origin
Question 15.
Find the rectangular components of the average velocity in the time interval between $t$ and $t + \Delta t$
Solution
Question 16.
Which of the following statements are true about the motion
a. The particle is experiencing uniform acceleration motion
b. The particle starts at y axis and touched x axis in 1sec
c. The initial velocity is towards negative Y axis
d. The velocity at t=1 is towards positive x axis
Solution
For these, it is clear that all the four options are correct
(F) A particle is moving along a parabola given by
$y=x^2$
So that at any time $v_x=3 m/s$
Question 17.
Find the magnitude and direction of velocity at x=1/3 m
Solution
$y=x^2$ -(1)
$v_x=\frac {dx}{dt}=3 $
Differentiating equation (1) w.r.t time
$ \frac {dy}{dt} = 2x \frac {dx}{dt}$
$ \frac {dy}{dt} = 2x \times 3 =6x$
So, at x=1/3
$ \frac {dy}{dt} = 2x \times 3 =6x =2$
$\frac {dx}{dt}=3 $
So, Magnitude and direction will be
$ \sqrt {13} ,tan^{-1} \frac {2}{3}$
Question 18
Find the acceleration at x=1/3 (both magnitude and direction)
a. 18 m/sec
^{2} ,90°
b. 14 m/sec
^{2} ,60°
c. 18 m/sec
^{2} ,60°
d. 14 m/sec
^{2} ,90°
Solution
$ \frac {dy}{dt} =6x$
$\frac {dx}{dt}=3 $
Differentiating equations w.r.t time
$ \frac {d^y}{dt^2}= 6 \frac {dx}{dt} = 18$
$ \frac {d^x}{dt^2}=0$
So magnitude and direction will be
18 m/sec^{2} ,90^{0}
link to this page by copying the following text
Also Read
Go back to Class 11 Main Page using below links
Class 11 Maths
Class 11 Physics
Class 11 Chemistry
Class 11 Biology