(A) A particle moves such a way that
where
i and
j are unit vectors along x and y direction. A and ω are constant
Question 1.
Find the trajectory of the particle
a. Circle
b. Parabola
c. Eclipse
d. None of the above
Solution
Given $\mathbf{R}= (Asin \omega t)\mathbf{i} + (Acos \omega t)\mathbf{j}$
So $x= Acos \omega t$ => $ cos \omega t =\frac {x}{A}$
$y=Asin \omega t$ => $sin \omega t =\frac {y}{A}$
Now we know that
$Sin^ 2 \omega t+ cos^2 \omega t =1$
So, $(\frac {x}{A})^2+ (\frac {y}{A})^2=1$
Or, $x^2+y^2=A^2$
Which is a equation of a circle
Question 2.
Find the
Average velocity of the particle from t=0 to
Solution
Given $\mathbf{R}= (Asin \omega t)\mathbf{i} + (Acos \omega t)\mathbf{j}$
At t=0
$\mathbf{R}= A \mathbf{i}$
At $t= \frac {2 \pi}{\omega }$
$\mathbf{R}= A\mathbf{j}$
Now average velocity=$ \frac {Displacement}{t} = \frac {A\mathbf{j}-A \mathbf{i}}{\frac {2 \pi}{\omega }} =\frac {2A \omega}{\pi} (\mathbf{j}-\mathbf{i})$
Question 3.
Find the scalar Product of Acceleration and velocity at any point t
a. 1
b. -1
c. 0
d. None of the above
Solution
Given $\mathbf{R}= (Asin \omega t)\mathbf{i} + (Acos \omega t)\mathbf{j}$
$\mathbf{v}= \frac {d\mathbf{R}}{dt}$
$= (-A \omega sin \omega t)\mathbf{i} + (A \omega cos \omega t)\mathbf{j}$
$\mathbf{a}= \frac {d\mathbf{v}}{dt}$
$=(-A \omega ^2 cos \omega t)\mathbf{i} + (-A \omega ^2 sin \omega t)\mathbf{j}$
$\mathbf{a}.\mathbf{v} = [(-A \omega sin \omega t)\mathbf{i} + (A \omega cos \omega t)\mathbf{j}]. [(-A \omega ^2 cos \omega t)\mathbf{i} + (-A \omega ^2 sin \omega t)\mathbf{j}]$
=0
Question 4.
Find the tangential acceleration at any point of time t
Solution
We know from above
$\mathbf{a}= \frac {d\mathbf{v}}{dt}$
$=(-A \omega ^2 cos \omega t)\mathbf{i}+ (-A \omega ^2 sin \omega t)\mathbf{j}$
$=-A \omega ^2 \mathbf{R}$
So it means acceleration directed towards the radial direction
So, tangential acceleration is zero.
Question 5.
Find the radial acceleration at any point of time t
Solution
Same as above
Radial acceleration= $A \omega ^2$
Question 6.
Solution
Given $\mathbf{R}= (Asin \omega t)\mathbf{i} + (Acos \omega t)\mathbf{j}$
At t=0
$\mathbf{R}= A\mathbf{i}$
At $t = \frac {\pi}{ \omega}$
$\mathbf{R}= -A\mathbf{j}$
So displacement= $-2A\mathbf{i}$
Distance =Distance traveled on the half circle = $\pi A$
(B) Two particles start from the origin of the horizontal x-y plane. Particle A moves along +x axis with uniform velocity u. Particle B moves along -y axis with velocity u. i and j are the unit vectors along x and y direction
Question 7.
Find the relative velocity of particle B w.r.t to A at time t
a. $-u(\mathbf{i}+\mathbf{j})$
b. $u(\mathbf{i}+\mathbf{j})$
c. $0$
d. none of the above
Solution
Velocity of A=$u\mathbf{i}$
Velocity of B =$-u\mathbf{j}$
Velocity of B w.r.t to A=Velocity of B-Velocity of A =$-u(\mathbf{i}+\mathbf{j})$
Question 8.
Find the relative velocity of particle A w.r.t to B at time t
a. $-u(\mathbf{i}+\mathbf{j})$
b. $u(\mathbf{i}+\mathbf{j})$
c. $0$
d. none of the above
Solution
Velocity of A=$u\mathbf{i}$
Velocity of B ==$-u\mathbf{j}$
Velocity of A w.r.t to B=Velocity of A-Velocity of B
=$u(\mathbf{i}+\mathbf{j})$
Question 9.
find the relative position vector of particle A w.r.t to B at time t
a. $-ut(\mathbf{i}+\mathbf{j})$
b. $ut(\mathbf{i}+\mathbf{j})$
c. $0$
d. none of the above
Solution
Position vector of A =$ut\mathbf{i}$
Position vector of B=$-ut\mathbf{j}$
Relative position vector of particle A wrt to B at time t=Position vector of A - Position vector of B =$ut(\mathbf{i}+\mathbf{j})$
Subjective questions
Question 10.
A particle moves 4 km North East and reach position A and then moves 3 Km South East to reach the position B. Let assume the initial point has co-ordinates (0, 0).And Co-ordinates of the position A and B are
A -> (x
_{a},y
_{a)}
B -> (x
_{b},y
_{b)}
Also
i and
j are the unit vector across the x and y direction
Find out the following
a.Find out the co-ordinates of Position A and B
b.Find out the distance of point B from the Origin
c.Find out the position vector of point A and B
Solution
(a) Let us draw the displacement diagram
From the diagram, it is clear for position A
$x_a =4 \times (cos 45)=2.83$ Km
$y_a =4 \times (sin 45) =2.83$ Km
Now taking components of AB vector across horizontal and vertical direction
$x=3cos45=2.12$ km
$y=3sin45=2.12$ km
From the figure
$x_b = x_a+x=2.12+2.83=4.95$ km
$y_b=y_a-y=2.83-2.12=.71$ km
(b) Distance of Point B from the origin is given by
$= \sqrt {x_b^2 + y_b^2}$
=5 km
(c) Position vector of point A = $x_a \mathbf{i} + y_a\mathbf{j} = 2.83 \mathbf{i}+2.83 \mathbf{j}$
Position vector of point B =$x_b \mathbf{i} + y_b \mathbf{j} = 4.95 \mathbf{i} +.71 \mathbf{j}$
Question 11.
An object is fired upward at an angle 40° to the horizontal .The object is fired with an initial speed of 20m/s.
Find out the following
a. How high up will it strike a wall which is 8 m away
b. How much time it will take to strike the wall
c. What are the horizontal and vertical components of velocity when it strike the wall
Solution
(a) and (b)
Horizontal component of velocity
$v_{0x}=v_0 cos 40=15.3$ m/sec
Vertical component of velocity
$v_{0y}=v_0 sin40=15.3$ m/sec
Now let take t be the time in which it strike the wall.
Then
$ x= v_{0x}t$
$y= v_{0y}t-\frac {gt^2}{2}$
Now x=8 m
So $8=15.3 \times t$
Or t=.52 sec
Now
$y= v_{0y}t- \frac {gt^2}{2}$
Substituting all the values
Y=5.33 m
(c) Horizontal velocity will remain same during all the motion while vertical velocity will have downward acceleration of g.
So Horizontal component of velocity at the point of strike
$v_x =15.3$ m/sec
Vertical component of velocity at the given point
$v_y =v_{0y}-gt=12.8 - (9.8)(.52)=7.704$ m/sec
Question 12.
The position of a particle is given by the below equation
$\mathbf{R}= (2sin2 \pi t)\mathbf{i} + (3cos2 \pi t)\mathbf{j} $
a. Find out the trajectory of the particle
b. Find out the velocity and acceleration vector and the relation between the acceleration and position vector.
c. Find out the times when velocity becomes maximum and minimum.
d. Find out the time dependence of the angle α between velocity and acceleration vector
e. Find the angle α at t=0 and t=1/4
Solution
a. Given
$\mathbf{R}= (2sin2 \pi t)\mathbf{i} + (3cos2 \pi t)\mathbf{j} $
So co-ordinates of the particles are
$x=2sin2 \pi t$ or $sin2 \pi t =\frac {x}{2}$
$y=3cos2 \pi t$ or $cos2 \pi t =\frac {y}{3}$
Now we know that
$(sin2 \pi t)^2+ (cos2 \pi t)^2=1$
So
$(\frac {x}{2})^2 + (\frac {y}{3})^2 =1$
Or
$ \frac {x^2}{4} + \frac {y^2}{9} =1 $
This is an equation of an ellipse
So the trajectory of the particle is an ellipse
(b) Given
$\mathbf{R}= (2sin 2 \pi t)\mathbf{i} + (3 cos 2 \pi t)\mathbf{j} $
Velocity vector is given by
$\mathbf{v} = \frac {d\mathbf{R}}{dt}$
or
$\mathbf{v} = (4 \pi cos 2 \pi t) \mathbf{i}+ (-6 \pi sin 2 \pi t)\mathbf{j}$
Acceleration vector is given by
$\mathbf{a} = \frac {d\mathbf{v}}{dt}$
Or
$\mathbf{a}= (-8 \pi ^2 sin 2 \pi t ) \mathbf{i}+(-12 \pi ^2 cos 2 \pi t)\mathbf{j}$
$=-4 \pi ^2 \mathbf{R}$
Also it indicates that acceleration is towards the center at every point of the motion.
(c)Magnitude of velocity is given by
$v = \sqrt {v_x^2 + v_y^2}$
or
$v= \sqrt {16 \pi ^2 cos^2 2 \pi t + 36 \pi ^2 sin^2 2\pi t}$
$= 2 \pi \sqrt {4 cos^2 2 \pi t + 9 sin^2 2\pi t}$
$=2 \pi \sqrt { 9 - 5 cos^2 2 \pi t}$
As $( sin2 \pi t )^2+( cos2 \pi t )^2=1$
So $v_{max} will occur at times when $| cos2 \pi t |=0$
Or $2 \pi t=(2n+1) (\pi /2)$ where n=0,1,2.....
Or t=1/4,3/4,5/4 ....
Also value of $v_{max} =6 \pi $
Now $v_{min} will occur at times when $| cos2 \pi t |=1$
Or $2 \pi t=n \pi $ where n=0, 1,2 ......
Or t=0,1/2,1.....
Also value of $v_{min} =4 \pi$
(d)
(e)Now when t=0 , $cos \alpha =0$
or $ \alpha =90$ °
When t=1/4 , $cos \alpha =0$ or 90°
Question 13.
A ball is thrown upward from a point O on the side of a hill which slopes upward uniformly at an angle 30°.Intial Velocity of the ball is $v_0$ and it is thrown at an angle 60° with respect to horizontal.
a. Find out the range along the slope of the hill
b. Find out the Time period of the Projectile
c. What height above the point O ,ball strike the incline plane
d. What velocity does the ball strike the plane
Solution
(a) The situation is depicted the figure below.
Now we know that, the trajectory of the ball is given by
$y= (tan \theta) x - \frac {gx^2}{2v_0^2 cos^2 \theta}$
Now here $ \theta = 60$ so $tan \theta= \sqrt {3}$ and $cos \theta = \frac {1}{2}$
Or $y=(\sqrt {3} )x - \frac {2gx^2}{v_0^2}$ ---(1)
Now the equation of the slope is given by
$y=x tan \theta$
Now here $ \theta = 30$ so $ tan \theta = \frac {1}{\sqrt {3}}$
So, $y= \frac {x}{\sqrt {3}}$ ---- (2)
Now the point of strike is the point of intersection of the equation (1) and (2)
Or
$\frac {x}{\sqrt {3}} = (\sqrt {3} )x - \frac {2gx^2}{v_0^2}$
or
$x= \frac {v_0^2}{g \sqrt {3}}$
Now distance along the slope is given by
$x=S cos30$
or $S= \frac {x}{cos30}$
$=\frac {2 v_0^2}{3g}$
(b) Time to reach the point is given by
$x=v_{0x} t$
=> $t= \frac {v_0 cos60}{x}$
=> $ t= \frac {2v_0}{g \sqrt {3}}$
(c) Height above the point O can be found by the equation (2)
$y= \frac {x}{\sqrt {3}}$ ---- (2)
We know that
$x= \frac {v_0^2}{g \sqrt {3}}$
So $y= \frac {v_0^2}{3g}$
(d)Horizontal velocity will remain same throughout the motion as there is no acceleration
$v_x= v_0 cos 60=v_0 / 2$
$v_y =v_{0y}-gt = v_0 sin60-g \times ( \frac {2v_0}{g \sqrt {3}} ) = \frac {-v_0}{2\sqrt {3}}$
So velocity of strike
$ v = \sqrt {v_x^2 + v_y^2} = \frac {v_0}{ \sqrt {3}}$
Question 14.
A particle moves in the plane
xy with velocity given by $\mathbf{v}=a \mathbf{i}+bx \mathbf{j}$ where
i and
j are the unit vectors of the
x and
y axis and a and b are constants. At the initial moment of time the particle was located at point x=y=0. Find
a. The equation of particle’s trajectory y(x)
b. The curvature radius of trajectory as a function of x.
Solution
(a) As given in the question velocity vector of particle is
$\mathbf{v}=a \mathbf{i}+bx \mathbf{j}$
From this we have x and y components of velocity i.e.
and $v_x = \frac {dx}{dt}=a $ and $v_y = \frac {dy}{dt} = bx$ ------- (1)
From equation 1 we can calculate equation describing motion of particle along x and y axis. Thus integrating for x
$\int_{0}^{x} dx = a \int_{0}^{t} dt$
or $x=at$ ------ (2)
Now from equation 2 we have
$dy = bx dt = bat dt$
Integrating it we get
$\int_{0}^{y} dy = ab \int_{0}^{t} t dt$
Or
$y = \frac {1}{2} abt^2$ --------- (3)
From equations 2 and 3 we get
$y = \frac {b}{2a} x^2 $ -------- (4)
This is the required equation of particle's trajectory.
(b) Radius of curvature of trajectory y(x) is given as
$ R= \frac {[1 + (\frac {dy}{dx})^2]^{3/2}}{\frac {d^y}{dx^2}}$ ------ (5)
Differentiating path given in equation 4 for its first and second derivatives we find
and $ \frac {dy}{dx} = \frac {b}{a} x$ and $ \frac {d^2y}{dx^2} =\frac {b}{a}$ -------- (6)
Using equation 5 and 6 we find radius of curvature of trajectory as following
$R = \frac {a}{b} [ 1 + (\frac {b}{a} x)^2 ] ^{3/2}$
Question 15.
A boy is standing at a distance $a_1$ from the foot of a tower. The boy throws an stone at a angle 45° which just touches the top of the tower and strikes the ground at a distance $a_2$ from the point the boy is standing. Find the height of the tower
Solution
Let H be the height of the tower
Now equation of trajectory is
$y=x tan \theta (1- \frac {x}{R})$
Now Range is given as $a_2$
when $y=H ,x=a_1$
so, $H=a_1 tan45(1- \frac {a_1}{a_2}) =\frac {a_1(a_2-a_1)}{a_2}$
Question 16.
A ball is thrown upward from a point on the side of a hill which slopes upward uniformly at angle 28°.Intial velocity of the ball is v
_{0}= 33 m/s and at an angle 65°(with respect to the horizontal. At what distance up the slope the ball strike and in what time?
Solution
The situation is depicted in fig
$v_0=33$ m/s and $ \theta _0=65$
The equation of trajectory of the
$y_b=(tan \theta _0)x-\frac {gx^2}{2v_0^2 cos \theta _0}$
or
$y_b=2.14x-.025x^2$
The equation of the incline
$y_i=(tan28)x=.53x$
When the ball strike the incline
$y_b=y_i$
or
$.53x=2.14x-.025 x^2$
or x=64.4 m
The distance along the incline obeys
$x=Scos28$ or $S=72.9$ m
The time to reach that point is given by
$x=v cos65 t$
or
t=4.63 sec
Question 17.
A Cannon on a level plain is aimed at an angle θ above the horizontal. A shell is fired with a muzzle velocity v
_{0} towards a pole which is distance R away. It hits the pole at height H.
a find the time taken to reach the pole
b. find the value of H in terms of θ,R and v
_{0}
Solution
The time taken to reach the pole is given
$R=v_0 cos? t$
or $t= \frac {R}{v_0 cos \theta}$
Now equation of motion of vertical motion
$h=v_0 sin \theta t- \frac {1}{2}gt^2$
at $t= \frac {R}{v_0 cos \theta}$, $h=H$
so
$H=(v_0 sin \theta ) \times ( \frac {R}{v_0 cos \theta}) - \frac {1}{2} g( \frac {R}{v_0 cos \theta}^2$
or $H=R tan? - \frac {gR^2}{2v_0^2 cos^2 \theta}$
Question 18.
The current velocity of a river grows in proportion to the distance from its bank and reaches its maximum v
_{0} in the middle. The width of the river is b. The velocity near the banks is zero. A boat is so moving on the river that its velocity u relative to the water is constant and perpendicular to the current
a. Find the velocity of the current at a distance y from the bank ( y<b/2)
b. Find the velocity of the boat relative to ground at any point on it path ( y<b/2)
c. Find the velocity of the boat relative to ground at any point on it path ( b/2< y < b)
d. Find the acceleration also on both the above distances
e. Find the distance through which the boat crossing the river will be carried away by the current
f. Find the trajectory of the boat
Solution
Let us take point A from which the boat departs as the origin of the coordinates system. The direction of the axis are shown below in Figure
The boat moves in a direction perpendicular to the current at constant velocity u.
Let us take a point B on the boat path which is at a distance y from bank. y< b/2
The current velocity will be
$ v= \frac {2v_0 y}{b}$ ---(1)
Now time taken to reach the boat at this point B
$t= \frac {y}{u}$
or $y=ut$
Substituting this value in equation (1)
$ v= \frac {2v_0 ut}{b}$
The velocity of the boat with respect to ground
=Velocity of the boat with respect to river + velocity of river with respect to ground
$\mathbf{v}= \frac {2v_0 ut}{b} \mathbf{i} + u \mathbf{j}$
Differentiating it
$ \mathbf{a} = \frac {2v_0 u}{b} \mathbf{i}$
This will be the acceleration
Now the boat will reach middle point in time
$T= \frac {b}{2u}$
The horizontal distance travelled during this time
$s = \frac {aT^2}{2}$
Or,
$s= \frac {v_0 b}{4u}$
Now let take the velocity for y > b/2
The current velocity will be
$v= \frac {2v_0(b- y)}{b} $ --- (1)
Then
$\mathbf{v}= \frac {2v_0(b- y)}{b} \mathbf{i} + u \mathbf{j}$
Differentiating it
$ \mathbf{a} = -\frac {2v_0 u}{b} \mathbf{i}$
Now the boat will reach bank
$T= \frac {b}{2u}$
The horizontal distance travelled during this time
$s = ut + \frac {aT^2}{2}$
$s = u \frac { b}{u} - \frac {2v_0 u}{2b} (\frac { b}{u})^2$
$s = \frac {v_0 b}{4u}$
So total distance
$s = \frac {v_0 b}{2u}$
Now we know from above calculation for y< b/2
$y=ut$
$x = \frac {at^2}{2}$
Now $a= \frac {2v_0u}{b}$
Substituting this in above
$x = \frac {v_0 u_0}{b} t^2 $
Now $t= \frac {y}{u}$
So
$y^2 = \frac {bu}{v_0} x$
This is a parabola
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