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Projectile Motion Worksheet



(A) A particle moves such a way that
Projectile Motion Worksheet for JEE Main and JEE Advanced
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

Answer

Given R=(Asinωt)i+(Acosωt)j
So x=Acosωt => cosωt=xA
y=Asinωt => sinωt=yA
Now we know that
Sin2ωt+cos2ωt=1
So, (xA)2+(yA)2=1
Or, x2+y2=A2
Which is a equation of a circle


Question 2.
Find the Average velocity of the particle from t=0 to
Projectile Motion Worksheet for JEE Main and JEE Advanced

Answer

Given R=(Asinωt)i+(Acosωt)j
At t=0
R=Ai
At t=2πω
R=Aj
Now average velocity=Displacementt=AjAi2πω=2Aωπ(ji)


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

Answer

Given R=(Asinωt)i+(Acosωt)j
v=dRdt
=(Aωsinωt)i+(Aωcosωt)j
a=dvdt
=(Aω2cosωt)i+(Aω2sinωt)j
a.v=[(Aωsinωt)i+(Aωcosωt)j].[(Aω2cosωt)i+(Aω2sinωt)j]
=0


Question 4.
Find the tangential acceleration at any point of time t
Projectile Motion Worksheet for JEE Main and JEE Advanced

Answer

We know from above
a=dvdt
=(Aω2cosωt)i+(Aω2sinωt)j
=Aω2R
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
Motion in a plane Worksheet for JEE Main and JEE Advanced

Answer

Same as above
Radial acceleration= Aω2


Question 6.
Motion in a plane Worksheet for JEE Main and JEE Advanced

Answer

Given R=(Asinωt)i+(Acosωt)j
At t=0
R=Ai
At t=πω
R=Aj
So displacement= 2Ai
Distance =Distance traveled on the half circle = πA



(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(i+j)
b. u(i+j)
c. 0
d. none of the above

Answer

Velocity of A=ui
Velocity of B =uj
Velocity of B w.r.t to A=Velocity of B-Velocity of A =u(i+j)


Question 8.
Find the relative velocity of particle A w.r.t to B at time t
a. u(i+j)
b. u(i+j)
c. 0
d. none of the above

Answer

Velocity of A=ui
Velocity of B ==uj
Velocity of A w.r.t to B=Velocity of A-Velocity of B
=u(i+j)


Question 9.
find the relative position vector of particle A w.r.t to B at time t
a. ut(i+j)
b. ut(i+j)
c. 0
d. none of the above

Answer

Position vector of A =uti
Position vector of B=utj
Relative position vector of particle A wrt to B at time t=Position vector of A - Position vector of B =ut(i+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 -> (xa,ya)
B -> (xb,yb)
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

Answer

(a) Let us draw the displacement diagram
Motion in a plane Worksheet for JEE Main and JEE Advanced

From the diagram, it is clear for position A
xa=4×(cos45)=2.83 Km
ya=4×(sin45)=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
xb=xa+x=2.12+2.83=4.95 km
yb=yay=2.832.12=.71 km

(b) Distance of Point B from the origin is given by
=x2b+y2b
=5 km

(c) Position vector of point A = xai+yaj=2.83i+2.83j
Position vector of point B =xbi+ybj=4.95i+.71j


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

Answer

(a) and (b)
Horizontal component of velocity
v0x=v0cos40=15.3 m/sec
Vertical component of velocity
v0y=v0sin40=15.3 m/sec
Now let take t be the time in which it strike the wall.
Then
x=v0xt
y=v0ytgt22
Now x=8 m
So 8=15.3×t
Or t=.52 sec
Now
y=v0ytgt22
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
vx=15.3 m/sec
Vertical component of velocity at the given point
vy=v0ygt=12.8(9.8)(.52)=7.704 m/sec


Question 12.
The position of a particle is given by the below equation
R=(2sin2πt)i+(3cos2πt)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

Answer

a. Given
R=(2sin2πt)i+(3cos2πt)j
So co-ordinates of the particles are
x=2sin2πt or sin2πt=x2
y=3cos2πt or cos2πt=y3
Now we know that
(sin2πt)2+(cos2πt)2=1
So
(x2)2+(y3)2=1
Or
x24+y29=1

This is an equation of an ellipse
So the trajectory of the particle is an ellipse

(b) Given
R=(2sin2πt)i+(3cos2πt)j
Velocity vector is given by
v=dRdt
or
v=(4πcos2πt)i+(6πsin2πt)j
Acceleration vector is given by
a=dvdt
Or
a=(8π2sin2πt)i+(12π2cos2πt)j
=4π2R
Also it indicates that acceleration is towards the center at every point of the motion.

(c)Magnitude of velocity is given by
v=v2x+v2y

or
v=16π2cos22πt+36π2sin22πt
=2π4cos22πt+9sin22πt
=2π95cos22πt
As (sin2πt)2+(cos2πt)2=1

So vmaxwilloccurattimeswhen| cos2 \pi t |=0Or2 \pi t=(2n+1) (\pi /2)wheren=0,1,2.....Ort=1/4,3/4,5/4....Alsovalueofv_{max} =6 \pi Nowv_{min} will occur at times when |cos2πt|=1
Or 2πt=nπ where n=0, 1,2 ......
Or t=0,1/2,1.....
Also value of vmin=4π
(d)
Projectile Motion Worksheet for JEE Main and JEE Advanced

(e)Now when t=0 , cosα=0
or α=90 °
When t=1/4 , cosα=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 v0 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

Answer

(a) The situation is depicted the figure below.
Projectile Motion Worksheet for JEE Main and JEE Advanced

Now we know that, the trajectory of the ball is given by
y=(tanθ)xgx22v20cos2θ
Now here θ=60 so tanθ=3 and cosθ=12
Or y=(3)x2gx2v20 ---(1)
Now the equation of the slope is given by
y=xtanθ
Now here θ=30 so tanθ=13
So, y=x3 ---- (2)
Now the point of strike is the point of intersection of the equation (1) and (2)
Or
x3=(3)x2gx2v20
or
x=v20g3
Now distance along the slope is given by
x=Scos30
or S=xcos30
=2v203g
(b) Time to reach the point is given by
x=v0xt
=> t=v0cos60x
=> t=2v0g3
(c) Height above the point O can be found by the equation (2)
y=x3 ---- (2)
We know that
x=v20g3
So y=v203g
(d)Horizontal velocity will remain same throughout the motion as there is no acceleration
vx=v0cos60=v0/2
vy=v0ygt=v0sin60g×(2v0g3)=v023
So velocity of strike
v=v2x+v2y=v03


Question 14.
A particle moves in the plane xy with velocity given by v=ai+bxj where i  andj 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.

Answer

(a) As given in the question velocity vector of particle is
v=ai+bxj
From this we have x and y components of velocity i.e.
and vx=dxdt=a and vy=dydt=bx ------- (1)
From equation 1 we can calculate equation describing motion of particle along x and y axis. Thus integrating for x
x0dx=at0dt
or x=at ------ (2)
Now from equation 2 we have
dy=bxdt=batdt
Integrating it we get
y0dy=abt0tdt
Or
y=12abt2 --------- (3)
From equations 2 and 3 we get
y=b2ax2 -------- (4)
This is the required equation of particle's trajectory.
(b) Radius of curvature of trajectory y(x) is given as
R=[1+(dydx)2]3/2dydx2 ------ (5)
Differentiating path given in equation 4 for its first and second derivatives we find
and dydx=bax and d2ydx2=ba -------- (6)
Using equation 5 and 6 we find radius of curvature of trajectory as following

R=ab[1+(bax)2]3/2


Question 15.
A boy is standing at a distance a1 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 a2 from the point the boy is standing. Find the height of the tower

Answer

Let H be the height of the tower
Now equation of trajectory is
y=xtanθ(1xR)
Now Range is given as a2
when y=H,x=a1
so, H=a1tan45(1a1a2)=a1(a2a1)a2


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 v0= 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?

Answer

The situation is depicted in fig
Projectile Motion Worksheet for JEE Main and JEE Advanced
v0=33 m/s and θ0=65
The equation of trajectory of the
yb=(tanθ0)xgx22v20cosθ0
or
yb=2.14x.025x2
The equation of the incline
yi=(tan28)x=.53x
When the ball strike the incline
yb=yi
or
.53x=2.14x.025x2
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=vcos65t
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 v0 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 v0

Answer

The time taken to reach the pole is given
R=v0cos?t
or t=Rv0cosθ
Now equation of motion of vertical motion
h=v0sinθt12gt2
at t=Rv0cosθ, h=H
so
H=(v0sinθ)×(Rv0cosθ)12g(Rv0cosθ2
or H=Rtan?gR22v20cos2θ


Question 18.
The current velocity of a river grows in proportion to the distance from its bank and reaches its maximum v0 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

Answer

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
Projectile Motion Worksheet for JEE Main and JEE Advanced

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=2v0yb ---(1)
Now time taken to reach the boat at this point B
t=yu
or y=ut
Substituting this value in equation (1)
v=2v0utb
The velocity of the boat with respect to ground
=Velocity of the boat with respect to river + velocity of river with respect to ground
v=2v0utbi+uj
Differentiating it
a=2v0ubi
This will be the acceleration
Now the boat will reach middle point in time
T=b2u
The horizontal distance travelled during this time
s=aT22
Or,
s=v0b4u
Now let take the velocity for y > b/2
The current velocity will be
v=2v0(by)b --- (1)
Then
v=2v0(by)bi+uj
Differentiating it
a=2v0ubi
Now the boat will reach bank
T=b2u
The horizontal distance travelled during this time
s=ut+aT22
s=ubu2v0u2b(bu)2
s=v0b4u
So total distance
s=v0b2u

Now we know from above calculation for y< b/2
y=ut
x=at22
Now a=2v0ub
Substituting this in above
x=v0u0bt2
Now t=yu
So
y2=buv0x

This is a parabola




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