Physics-11.3-Description of motion in two and three dimensions
DESCRIPTION OF MOTION IN TWO AND THREE DIMENSIONS
SYLLABUS
Scalars and vectors, vector addition, a real number, zero vector and it properties. Resolution of vectors. Scalar and vector products, uniform circular motion and its applications projectile motion.
1.
2.1. KINEMATICS AND DYNAMICS
The part of mechanics which deals with the description of the motion of an object without considering reason of the origin is called kinematics. Where as, the study of the motion of an object related to its cause is called dynamics.
(ii) Motion in two dimension:
Motion of an object in a plane is called two dimensional (2-D) motions. For 2-D motion velocity or acceleration can be described by two components in any two mutually perpendicular directions in cartesian coordinate system i.e. its position, velocity, displacement and acceleration can have two nonzero components.
3.2. Vector and Scalar
Those physical quantity in which we know magnitude and direction called vector.
Those physical quantity in which we know magnitude but unknown direction called Scalar.
Those vector whose magnitude is zero called zero vector.
Vector Scalar product of vectors
(i)
(ii)
(iii)
(iv)
Example 1. A particle of mass 10–2kg is moving along the positive X-axis under the influence of a force where k = 10–2 Nm2. At time t = 0, it is at
x = 1.0 m and its velocity is v = 0. find its velocity when it reaches
x = 0.5 m
Solution: Given
Here k and x2 are always positive. Hence F is always negative (whether x is positive or negative)
Now, we know that F = ma
In this case we have
[as F[x] = –k / 2x2]
or
or
or
or v = 1.0 m/s
So, here v = – 1.0 m/s (because velocity is along negative X-direction)
Example 2. A particle travels according to the equation where is the acceleration. A and B are constants, is the velocity of the particle. Find its velocity as a function of time. Also find terminal velocity.
Key concept: Keep remember that above question is not related with uniform acceleration.
At terminal velocity, acceleration = 0, that is,
Example 3. Velocity of body on incline plane in maximum at which Inclination angle?
(a) 0 (b) 90(c) 360 (d) 180
Solution:(B)
=90
Example 4. The radius vector of a point depends on time as
where and are constant vectors. Find the modulus of velocity and acceleration at any time
Solution: (a) Velocity
Modulus of velocity vector will be
Here c and b are modulus of and and is the angle between and which can be written
As
Hence,
(b) Acceleration
Hence,
5. PROJECTILE MOTION (MOTION IN TWO DIMENSIONS)
5.1 MOTION IN TWO DIMENSIONS:
In two-dimension motion a particle moves in a plane. e.g. a particle going in a circle, a cricket ball thrown in by a fielder (on a windless day). In the first case the particle can go round with a constant speed (uniform circular motion) or it could move with a non-constant speed (Non uniform circular motion). We will study both. In the second case the cricket ball in a projectile i.e. it has been projected (thrown) and it moves under the influence of gravity. This motion is projectile motion. Before we study this specialized case of two-dimension motion, let us first understand the general features of two-dimension motion.
(i) Position and Displacements:
As we had done in one-dimensional motion, the first task for us is to know where the particle is i.e. its position. The position here is a vector intends from a reference point (usually the origin of the coordinate system) to the object.
In the unit vector notation it can be written
Where are the vector components of and the coefficients x and y are its scalar components.
The coefficients x and y give the objects location along the axes and relative to the origin. Figure shows particle P whose position vector at an instant is
As the particle moves its position vector (P.V.) changes such that the vector always extends to the object from origin.
Assuming the particle has a P.V. at time t1 and P.V. at time t2, its displacement is
During time interval .
Example 5. The P.V. for a particle is initially and then later is is . What is the displacement?
Solution:
(ii) Velocity and Average velocity:
If a particle moves through a displacement in a time interval , then the average velocity is
….(1)
Here are x and y components of average velocity.
The instantaneous velocity
But
, then components of along x and y respectively.
Let us understand this using a particle, which is moving in 2D.
In following figure particle is at at t1 and at at t2 (= t1+ t2). The vector is the particle displacement in time. Then as is clear from (1) average velocity points in same direction as
Now if we reduce we see that
(1) vector moves towards
(2) becomes smaller and its direction approaches the tangent
(3) In the limit 0, , the instantaneous velocity at t1 i.e. average velocity becomes instantaneous velocity , which has the direction along tangent line.
CAUTION:
Recall that in one dimension motion the tangent of x – t graph point gave the direction (+ve or –ve slope) as well as magnitude ( = tan) of velocity at that instant. But in two dimension motion to get position vs. time graph we need three coordinates (x, y, & t).
Hence it is not possible, the x – y graph which shows the trajectory (path) only give the direction of velocity vector and it in tangent to the path of the particle at any instant.
5.2 DESCRIPTION OF PROJECTILE MOTION
If an object is given an initial velocity in any direction (except 90) and then allowed to travel freely under gravity, the motion is called projectile motion. In case of projectile motion it is assumed that acceleration due to gravity is constant and the effect of air resistance is negligible. Projectile motion is a two dimensional motion and can be regarded as simultaneous superposition of two motions one horizontal with velocity ucos and acceleration=0 and other vertical with initial velocity u sin and acceleration = -g. The acceleration due to gravity, g, is uniform and the surface of the earth is considered to be flat.
Following are for main features of a projectile:
(i) Such a particle will move horizontally and as well as vertically i.e. along a curve.
(ii) For convenience, we will take origin at the point from where the particle is thrown and horizontal and vertical components of velocity along x-axis and y-axis respectively.
(iii) The velocity of particle at any instant is directed along the tangent to the path and can have horizontal and vertical components.
(iv) The only force acting on the particle is its weight (mg) directed downwards. Hence acceleration is g directed vertically downwards. The air friction is neglected.
(v) As acceleration does not change with time, the projectile motion is a uniformly accelerated motion at all time instants, a1 (horizontal)=0 and a2(vertical) = -g.
5.3 TIME OF FLIGHT, HORIZONTAL RANGE AND TRAJECTORY OF THE PROJECTILE
Let P(x, y) represent the position of a projectile after a time t from the time of projection.
At the point O, horizontal component of velocity = ucos and
vertical component of velocity = u sin
Acceleration due to gravity acts vertically downwards. Hence vertical component of velocity changes. While horizontal component of velocity remains constant (ucos) during the complete motion.
Time of flight: It is the time taken by the projectile from the instant it is realised till it strikes a position the same horizontal plane as the point of projection.
To calculate time of flight, we can find the time when vertical displacement is zero i.e.
0 = usint – (usin )t = 0
t = 0 or t =
Thus
Time of flight T =
Range: During this time horizontal component of velocity has taken the particle through a distance x horizontally
Where, = ucos T = u cos =
This distance, we call as range on horizontal plane
R =
Rmax = for = 45
For a given velocity same range can be obtained for an angle and angle (90 – ) i.e.
R = =
R =
To calculate maximum height, the vertical component of velocity becomes zero, when the particle is at the highest point from the ground. At that time particle has only horizontal component of velocity i.e. = ucos
0 = (usin)2 – 2gHmax
Hmax =
The motion of projectile can be analysed through vector notation also for e.g. velocity of projectile (v) at any time t can be written as
Hence, the magnitude of velocity (speed) v at any time is given as
v =
= (The path of a projectile is called its trajectory)
Trajectory of projectile:
y = usint – . . . (1) (vertical displacement at any time t)
x = ucost . . . (2) (Horizontal displacement at any time t)
Using t =
We get,
y = x tan –
Which suggest that trajectory of projectile is parabola
Example 6. The figure shows two position A and B at the same height h above the ground. If the maximum height of the projectile is H, Then determine the time t elapsed between the positions A and B in terms of H.
Solution: Let T be the time of flight. We can now write
since
or
5.4 TIME OF FLIGHT & HORIZONTAL RANGE OF PROJECTILE MOTION ON INCLINED PLANE
Projectile Motion on Inclined Plane: Figure shows an inclined plane at an angle and a particle at an angle with the direction of plane with initial velocity u. In such cases we take our reference x- and y-axes in the direction along and perpendicular to the inclined as shown.
Unlike to the simple projectile motion, here the x-component of the velocity of the projectile will also be retarded by a gsin. Now y-component of the velocity is retarded by g cos instead of g. As shown here g is resolved in two directions.
As here y-direction component is retarded by gcos, to find the time of flight and maximum height, we can use equations T= and R = , replacing g by gcos,
Time of flight on inclined plane projectile is
Maximum height of the projectile with respect to inclined plane is
For evaluation of range on inclined plane we cannot use equation R= , just by replacing g by gcos, as here we also have
Acceleration in x-axis ax = -g sin
Now we again find the distance traveled by the particle along x-direction in the duration time of flight is
R =
On substituting the value of time of flight T, we get
Students are advised not to apply the above expression of range on inclined plane, as a standard result, it should be processed and evaluated according to the numerical problem. Above results we’ve derived for the projectile thrown up an inclined plane. If projectile is thrown down an inclined plane, the acceleration along the plane gsin will increase the velocity of the particle along the plane, thus in the expression for range we should use +ve sing as
To find the maximum range on incline plane, One can use maxima-minima as . The range on inclined plane has a maximum value given as
In above equation +ve sign is used for projectile up the plane and –ve sign is used for projectile down the plane.
Example 7. A ball is dropped from a height h above a point on an inclined plane, with angle of inclination . The ball makes an elastic collision with the surface and rebounds off the plane. Determine the distance from the point of first impact to the point where ball hit the plane second time.
Key concept: Take the point of first impact as the origin. Direction along the plane will be the x-axis and the direction perpendicular to the plane will be the y-axis. It is given that the ball rebounds elastically and implies that no change in kinetic energy of the ball before and after the collision. The ball rebounds with the same speed with which it will strike the plane after falling a distance h, which is . After rebound, the horizontal component of velocity u sin will be accelerated by g sin and the vertical component of the velocity ucos will be retarded by gcos.
Solution: Here time of flight from first impact to the second impact is given as
In this duration the distance travelled by the horizontal component is
R =
( since u= )
Example 8. A projectile is thrown with a speed u, at an angle to an inclined plane of inclination . Find the angle at which the projectile is thrown such that it strikes the inclined plane normally.
Key concept: At the time of striking, x-component of velocity is zero (vx=0) is zero.
Solution:
Here we have time of flight of particle is
Thus from speed equation in x- direction, we have
0 = ucos - g sin
or cot = tan
or = cot-1 (tan)
1. 6. UNIFORM AND NoN - UNIFORM CIRCULAR MOTION
Uniform circular motion: In uniform circular motion, a particle moves in a circular path of constant radius with constant speed. The velocity of a particle varies continuously as the direction changes. Thus, uniform circular motion is uniformly accelerated motion. Since the acceleration produces a change only in the direction of velocity vector, therefore, it must always be at right angles to the direction of motion. Otherwise, a component in the direction of motion would produce change in speed of the particle.
Centripetal Acceleration: The acceleration of a particle in uniform circular motion is directed towards the centre of the circular path. This radially inward acceleration is called the centripetal acceleration (means centre seeking). Its magnitude is given by
where v is the speed of the particle, and r is the radius of the circular path. In terms of angular speed , we may write ac = 2 r
If a body makes N revolution per minute, then its angular speed is given by
In uniform circular motion
(i) Velocity remains constant in magnitude but varies in direction
(ii) The acceleration is always normal to the velocity vector.
(iii) The acceleration is always directed towards the centre of the circular path.
Comparison of uniform circular motion with straight line motion and projectile motion:
Non-uniform circular motion:
(i) The velocity changes both in magnitude as well as in direction
(ii) The velocity vector is always tangential to the path
(iii) The acceleration vector is not perpendicular to the velocity vector
(iv) The acceleration vector has two components
(a) Tangential acceleration at changes the magnitude of velocity vector, i.e.
(b) Normal acceleration or centripetal acceleration ac changes the direction of the velocity vector, i.e.
(v) The total acceleration is the vector sum of the tangential and centripetal acceleration
Example 9. A particle is moving in a circular path of radius 10 cm. Its linear speed is given by cm/s. Find the angle between acceleration and radius at t = 2s.
Solution: Radial explanation
Tangential acceleration
Equations of motion:
1.
2.
3.
4. The distance travelled by the body in nth second
Projectile motion:
5. Time of flight T =
6. Range R =
7. Rmax = for = 45
8. Hight Hmax =
9. Trajectory of projectile
y = x tan – (i.e. parabolic)
For Inclined plane:
10. Time of flight
11. Hight
12. Range (for down inclined)
13. Range (for up inclined)
14. Maximum range
+ve sign is used for projectile up the plane and –ve sign is used for projectile down the plane.
Circular motion:
15. Centripetal acceleration (normal acceleration)
16. Tangential acceleration =
Relative motion:
17. Relative velocity of a particle while moving in the same direction.
Relative velocity
18. Elative velocity of a particle while moving in the opposite direction.
Relative velocity
Objective
1. A glass marble projected horizontally from the top of a table falls at a distance x from the edge of the table. If h is the highest of the table, then the velocity of projection is
(A) (B)
(C) gxh (D) gx + h.
Solution : (b)
2. The trajectory of a projectile in a vertical plane is y = ax – bx2, where a, b are constants, x and y are respectively the horizontal and vertical distances of the projectile from the point of projection. The maximum height attained is
(A) (B)
(C) (D)
Solution: y = ax- bx2 ;
For maximum y, < 0
= a – 2bx = 0 x =
for x = , y is maximum.
ymax = a .
3. Which of the following parameters of a particle executing uniform ircular motion remains constant?
(A) speed (B) radial acceleration (C) angular velocity (D) all of these
(A) (B)
(C) (D)
Solution: (C) The magnitude of the velocity of the particle, that is speed v remains constant. The velocity vector changes from time to time. Therefore, it accelerates radically inwards. This is known as Centripetal acceleration, ar. Its magnitude remains constant whereas its direction changes from time to time. Therefore is not constant. Since the sense of rotation of the particle remains constant its angular velocity remains constant in direction. Since = Constant & , remains constant.
A, & C are correct.
4. A particle is projected horizontally from the top of a cliff of height H with a speed . The radius of curvature of the trajectory at the instant of projection will be
(A) H/2 (B) H
(C) 2H (D)
Solution: (C) Since,
Radial acceleration ar = g
where r is the radius of curvature.
r = 2H
5. A body falling freely from a given height H hits an inclined plane in its path at a height ‘h’. As a result of this impact the direction of the velocity of the body becomes horizontal. Find the total time the body will take to reach the ground.
(A) (B)
(C) (D) none of the above
Solution : (A) Time taken by the body to strike the inclined plane
t1 =
Now as after impact the velocity of the body is horizontal, so time taken to reach the ground -
t2 =
So total time of motion -
t = t1 + t2 =
6. If R is the range for an angle of projection of 15o with the horizontal, then the other angle of projection for which the range is R, is
(A) 75o (B) 60o
(C) 45o (D) 30o.
Solution : (a) = (90o – 15o) = 75o
7. An object is thrown along a direction inclined at an angle of 45o with the horizontal direction. The horizontal range of the particle is
(A) four times the vertical height (B) thrice the vertical height
(C) twice the vertical height (D) equal to vertical height.
Solution : (a)
Clearly, R = 4 hmax.
8. A projectile thrown with a velocity at an angle has a range R on the surface of earth. For same and , its range on the surface of moon will be.
(A) 36 R (B) 60o
(C) (D) .
Solution : (d) is reduced by a factor of 6 on the moon.
9. If air resistance is ignored, then the horizontal motion of the oblique projectile takes place at
(A) uniform acceleration (B) variable acceleration
(C) uniform retardation (D) uniform velocity.
Solution : (d) The only force acting on a projectile is the force of gravity which acts vertically downwards. It has no horizontal component. So, horizontal motion of the oblique projectile takes place at uniform velocity.
10. A ball is projected with kinetic energy K at an angle of 45o to the horizontal. At the highest point during its flight, its kinetic energy will be
(A) K (B)
(C) (D) zero.
Solution : (c) Kinetic energy at highest point = K cos2 45o =
11. Two anti-aircraft guns fire two shells, one at an angle of 20o with the horizontal. If muzzle velocity is same in both the cases, then
(A) the first shell has a smaller horizontal range
(B) the second shell has a smaller horizontal range
(C) both the shells have to same vertical range
(D) both the shells have the same horizontal range.
Solution : (d) The horizontal range is the same whether the projectile is thrown at an angle with the horizontal or at an angle (90–) with the horizontal.
12. An aeroplane flying horizontally with a speed of 360 km h–1 releases a bomb at a height of 490 m from the ground. When will the bomb strike the bomb strike the ground ?
(A) 8s (B) 6s
(C) 7s (D) 10s.
Solution : (d)
13. It is possible to project a body with a given speed in two possible ways so that it has the same horizontal range R. The product of the times taken by it in the two possible ways is (g is the acceleration due to gravity)
(A) (B)
(C) (D)
Solution: If a body is projected with a given velocity u at angles and (90–) to the horizontal, it will have the same range R given by
The corresponding times of flight are
and
Hence the correct choice is (b)
14. The resultant of velocity of two cars moving with equal speeds, Ratio of magnitude , when the direction of either car is reversed. The angle between the direction of cars is
(A) 30° (B) 60°
(C) 120° (D) None of these
Solution: Let the vectors be
| | = | | (given)
=
Putting A = B we obtain = 120°.
15. The locus of a projectile relative to another projectile is a
(A) straight line (B) circle
(C) ellipse (D) parabola
Solution: Let these projectiles be projected with velocities
and
The velocities of the projectiles after a time t are given as
Relative velocity
Since the relative velocity is a constant and so does not vary with time, the locus of one w.r.t. the other is a straight line.
16. Given : is a unit vector. The value of n is
(A) (B)
(C) (D)
Solution: (C) or
or or .
17. Two particles A and B move with constant velocities v1 and v2 along two mutually perpendicular straight lines towards the intersection point O. At moment t = 0, the particles were located at distances l1 and l2 from O, respectively. The shortest distance between A & B is
(A) (B)
(C) (D)
Solution: After time t, the position of the point A and B are (l1 - v1t) and (l2 - v2t), respectively.
The distance L between the points A and B are,
L2 = (l1 - v1t)2 + (l2 - v2t)2
Differentiating with respect to time,
For minimum value of L,
or
Putting the value of t in equation
Lmin =
18. A man walks 4 m towards East and then 3 m towards North and there he fixes a pole 12 m high. The distance between the starting point and tip of the pole in space is
(A) 7 m (B) 11 m
(C) 13 m (D) 19 m
Solution: (C)
19. A particle is moving along a circular path of radius 5m and with uniform speed 5 m/s. What will be the average acceleration when the particle completes half revolution?
(A) zero (B) 10 m/s2
(C) 10 nm/s2 (D) 10/nm/s2
Solution: The change in velocity when the particle completes half revolution is given by v = 5 m/s - (-5 m/s) = 1 0 m/s Time taken to complete half revolution
Average acceleration
(D) is the correct choice.
20. A particle is projected horizontally from the top of a cliff of height H with a speed . The radius of curvature of the trajectory at the instant of projection will be
(A) H/2 (B) H
(C) 2H (D)
Solution: (C) Since,
Radial acceleration ar = g
where r is the radius of curvature.
r = 2H
21. Which of the following graph correctly represents velocity-time relationship for a particle released from rest to fall freely under gravity?
(A)
(B)
(C)
(D)
Solution: Releasing of the particle from rest means that v0 =0 at t = 0 and v =gt
at any time t.
the slope of v/t graph is a constant.
v/t graph is a straight line passing through the origin.
(A)
22. A body is moving in a circle at a uniform speed . What is the magnitude of the change in velocity when the radius vector describes an angle
(A) (B)
(C) (D)
Solution: (D)
23. What can be the possible velocity displacement (v – s) graph of a particle moving in a straight line under constant acceleration
(A) straight line (B) parabola
(C) ellipse (D) circle
Solution: (B) v2 4as parabola
24. A particle is moving eastward with a speed of 5 m/s. After 10 seconds, the direction changes towards north, but speed remains same. The average acceleration in this time is
(A) zero (B) m/s2 towards N-W
(C) m/s2 towards N-E (D) m/s2 towards S-W
Solution 6: (B)
(j - i) / 2 so direction is W N
25. If a particle takes t second less and acquires a velocity of v ms-2 more in falling through the same distance on two planets where the accelerations due to gravity are 2g and 8g respectively then
(A) v = 4gt (B) v = 5gt
(C) v = 2gt (D) v = l6gt
Solution: (A) V1
v 4gt m/s
26. An aeroplane flies along a horizontal circle of circumference 10 km, at a constant speed of 100 km/hr. The change in velocity in one fourth of a revolution is
(A) zero
(B) 141 km/hr at 90 from the original direction
(C) 141 km/hr at 135 from the original direction.
(D) 200 km/hr at 180 from the original direction.
Solution: (C)
135 to horizontal direction
27. Ratio or minimum kinetic energies of two projectiles of same mass is 4 : 1. The ratio of the maximum height attained by them is also 4 : 1. The ratio of their ranges would be
(A) 16 : 1 (B) 4 : 1
(C) 8 : 1 (D) 2 : 1
Solution: (A) minimum K.E. = 1/2m
28. A particle of mass m attached to a string of length I is describing circular motion on a sooth plane inclined at an angle with the horizontal. For the particle to reach the highest point its velocity at the lowest point should exceed.
(A)
(B)
(C)
(D)
Solution: (A) g effective g sin
V min
29. A particle is projected from the ground with an initial speed of v at an angle with horizontal. The average velocity of the particle between its point of projection and highest point of trajectory is
(A)
(B)
(C)
(D) v cos
Solution: (C)
displacement =
vavg =
30. A particle is projected with a certain velocity at an angle above the horizontal from the foot of an inclined plane of inclination 30. If the particle strikes the plane normally then is equal to
(A)
(B) 45
(C) 60
(D)
Solution: (D) v = 0 = ucos( - 30) – gsin(30)t (along x-axis)
…..(1)
along Y-axis
…(2)
by (1) and (2)
31. With what minimum speed must a particle be projected from origin so that it is able to pass through a given point (30m, 40m). Take g 10 m/s2
(A) 60 m/s (B) 30 m/s
(C) 50 m/s (D) None of these
Solution: (D)
(x, y) = (30, 40)
40 = 30 tan -
tan(2) = - 6/ 4 = -3/2
sin(2) = 3 /
cos(2) = -2 /
32. A ball is dropped vertically from a height h above the ground. It hits the ground and bounces up vertically to a height h/2. Neglecting subsequent motion and air resistance, its velocity varies with the height h as
(A)
(B)
(C)
(D)
Solution: (A) Velocity in downward motion is negative and upward motion will be positive. So either any will be A or C but v2 4 as the graph will be a parabola.
33. A ball is projected at an angle of 45, so as to cross a wall at a distance from the point of projection. It falls at a distance b on the other side of the wall. If h is the height of the wall then
(A) (B)
(C) (D)
Solution: (D)
tan 45 tan1 tan
1
34. At what angle the forces of 2 N and N act so that their combined effect is that of a single force of N ?
(A) 0 (B) 30
(C) 45 (D) 60
Solution: (C) or or = 45
35. Two bodies fall freely from the same height, but the second body starts falling T seconds after the first. The time (after which the first body begins to fall) when the distance between the bodies equals L is
(A) (B)
(C) (D) None of these
Solution: (D)
After time T velocity of I’st ball = gT
Relative velocity between ball 1 and ball2 = Gt
L = t =
Total time = T + t
36. Two projectiles, one fired from the surface of the earth with speed 5 m/s and the other fired from the surface of a planet with initial speed 3 m/s, trace identical trajectories. Neglecting air resistance, the value of acceleration due to gravity on the planet will be if g = 10 m/s2 on earth
(A) 5.9m/s2 (B) 3.6 ms/2
(C) 16.3 m/s2 (D) 8.5 m/s2
Solution: (B)
Trajectory is identical so
37. A particle is projected under gravity with velocity from a point at a height h above the level plane. The maximum range R on the ground is
(A)
(B)
(C)
(D)
Solution: (D) Co–ordinates of point P are (R, –h). These coordinates should satisfy the equation of projectile i.e.
or R2 tan2 – 4aRtan (R2 – 4ah) 0
For to be real (4aR) 2 4R2(R2 – 4ah)
or 4a2 4a (a h)
or R2 2
The maximum range is R max 2
38. A projectile has the same range R for two angles of projections. If T1 and T2 be the times of flight in the two cases, then
(A) T1T2 R (B) T1T2 R2
(C) (D)
Solution: (C)
39. If T is the total time of flight, H the maximum height and R is the horizontal range of a projectile. Then x and y co-ordinates at any time t are related as
(A) (B)
(C) (D)
Solution: (C)
By (C)
40. The horizontal range and maximum height attained by a projectile are R and H respectively. If a constant horizontal acceleration a g/4 is imparted to the projectile due to wind, then its horizontal range and maximum height will be
(A) (B)
(C) (R + 2H), H (D) (R + H), H
Solution: (D) Maximum height will not effected.
So Range ucos T 1/ 2(g/4)T2
R gT2/ 8 R g / 8 4u2sin2 / g2 = R + H
41. Given : . Which of the following is perpendicular to ?
(A) (B)
(C) (D)
Solution: (C) is in fourth quadrant.
is in the first quadrant.
Clearly, can be perpendicular to . For confirmation, let us check whether their dot product is zero.
This shows that is perpendicular to .
42. The angle between and is
(A) 0 (B) 90
(C) 45 (D) 60
Solution: (D)
43. In projectile motion maximum horizontal range get at which angle ?
(A) 0 (B) 90
(C) 45 (D) 60
Solution: (C)
44. The velocity vectors of a point A varies with time as , where a and b are positive constants. Find the equation of trajectory of the point.
(A) (B)
(C) (D)
Solution: (A)
Eliminating t from these to equations, we get
45. The velocity vectors of a point A varies with time as , where a and b are positive constants. Find the time dependence of the angle between acceleration and velocity vectors.
(A) (B)
(C) (D)
Solution: (B) acceleration
46. The velocity vectors of a point A varies with time as , where a and b are positive constants. Find
the time dependence of magnitude of acceleration and velocity vectors.
(A) (B)
(C) (D)
Solution: (C) acceleration is independent of time
47. A point moves along a circle with a velocity v = at where a = 0.50 m/s2. Find the total acceleration of the point at the moment when it has covered the nth (n = 0.10) fraction of the circle after beginning of the motion.
Solution: Let v be the velocity at the instant when the particle has covered the nth fraction of the circle
Tangential acceleration =
Radial acceleration = aN=
Total acceleration =
48. A projectile is fired from the surface of level ground at an angle 0 above the horizontal. Then elevation angle tan is
(A) (B)
(C) (D)
Solution: (C)
H = Maximum height
R = Maximum range
49. A projectile is fired from the surface of level ground at an angle 0 above the horizontal. Then calculate for 0 = 45.
(A) (B)
(C) (D)
Solution: (D) 0 = 45
50. A man is running with constant speed along a circular path of radius 22m. He completes 1 round in 10 second. Find average velocity in first 2.5 sec.
(A) (B)
(C) (D)
51. A man is running with constant speed along a circular path of radius 22m. He completes 1 round in 10 second. Find instantaneous speed at 2.5 sec.
(A) m/s (B) m/s
(C) m/s (D) m/s
52. A man is running with constant speed along a circular path of radius 22m. He completes 1 round in 10 second. Find average velocity in 20 seconds
(A) 2 (B) 5
(C) 0 (D) 3
Solutions :(50-52)
50(a) Let man starts from A, distance covered in first 2 seconds = Path ACB
Displacement = AB= =4m
Average velocity =
51(b) = r = = m/s
52(c) Net displacement = zero, so average velocity = 0
53. A rocket is fired vertically and ascends with constant vertical acceleration of 20m/s2 for 1 minute. Its fuel is then all burnt and it continues to move as a free body. Find the maximum height reached by the rocket
(A) 108000m (B) 106000m
(C) 108033m (D) 98000m
54. A rocket is fired vertically and ascends with constant vertical acceleration of 20m/s2 for 1 minute. Its fuel is then all burnt and it continues to move as a free body. Find the total time elapsed from the take off till the rocket strikes the earth.(g=10m/s2).
(A) 526.76 s (B) 326.96 s
(C) 400.96 s (D) 226.96 s
Solution : 53(a)For the time interval from 0 to 60 seconds rocket accelerates and after then, it moves freely under gravity. Distance moved by it in 60 seconds is given by
S1 = = 36000m.
v(60s) =
If H be the maximum height reached.
Then 0 = , (v2 = u2 + 2as)
H = 36000 + H = 108000m
(b) time taken to ascend is
t1 = 60s + = 180 s, [t = t1 + ]
Let time taken to descend is t2 then
108000 =
t2 =
Total time T = t1 + t2 = 180 + 146.96 = 326.96 s
55. If the velocity of projection of a projectile is trebled, then its maximum range will be
(A) quadrupled (B) nine times
(C) six times (D) eight times.
Solution : (B) Max. range .
56. The times of flight of a projectile is maximum when angle of projection is
(A) 30o (B) 45o
(C) 60o (D) 90o.
Solution: , For maximum T, sin should be maximum i.e. should be 90o.
57. If air resistance and air buoyancy are to be ignored, then which of the following factors determines the time of flight of the projectile?
(A) intial velocity
(B) horizontal component of initial velocity
(C) vertical component of initial velocity
(D) vertical component of acceleration.
Solution: (C)
58. A body is thrown with a velocity of 10 ms–1 at an angle of 60o with the horizontal. Its velocity at the highest point is
(A) 7 ms–1 (B) 9 ms–1
(C) 18.7 ms–1 (D) 5 ms–1
Solution: (D) cos = 10 cos 60o = 5 ms–1
59. A ball is projected upwards. Its acceleration at the highest point is
(A) infinite (B) zero
(C) directed upwards (D) directed downwards.
Solution: (D) Acceleration due to gravity is always directed vertically downward.
60. A body is projected at an angle of 30o with the horizontal with momental with moment p. At its highest point, the momentum is
(A) p (B)
(C) (D) .
Solution: (C) p = m
P’ mcos 30o =
35. Two particles start simultaneously from the same point and move along two straight lines, one with uniform velocity and other with a uniform acceleration a. If is the angle between the lines of motion of two particles then the least value of relative velocity will be at time given by
(A) (B)
(C) (D) .
Sol. : (B)
Velocity of 1st particle at time t is v1 = v
Velocity of 2nd particle at time t is v2 = at
Relative velocity
For vr to be least or to be least we must have
36. Two particles start simultaneously from the same point and move along two straight lines, one with uniform velocity and other with a uniform acceleration a. If is the angle between the lines of motion of two particles then, the least value of relative velocity is
(A) sin (B) cos
(C) tan (D) cot .
Sol. : (A)
39. A 2m wide car is moving with a uniform speed of 8 ms–1 along the edge of a straight horizontal road. A pedestrian starts to cross the road with a speed when the car is 12 m away from him. The time to cross the moving vehicle safely is
(A) (B)
(C) (D) .
Sol. : (B)
For pedestrian to cross the road safely, the time taken by the pedestrian to reach P equals the time taken by the edge A of the car to reach P. To reach P total distance (12 + 2 cos ) is travelled by the car with a velocity of 8 ms–1. Hence
40. The velocity of a boat in still water is times less than the velocity of flow of the river ( > 1). The angle with the stream direction at which the boat must move to minimise drifting is
(A) (B)
(C) (D) .
Sol. : (C)
vb = velocity of boat = v (say)
vr = velocity of river = v
To start with, we must observe that velocity of
river flow is times greater than the velocity of
boat. So, the boat has to drift and in this problem we are to minimise the drift. Let, the velocity of the boat make an angle with the river velocity. Then vx = vcos is the component of the velocity of boat along the river flow, whereas vy = v sin q is the component of the velocity of boat perpendicular to river flow. Due to this the vr + vx = v + v cos = V (say)
t =
During this time the drift equals (say x)
x = Vt
x =
To MINIMISE x, we must have
41. An armoured car 2 m long and 3 m wide is moving at 10 ms–1 when a bullet hits it in a direction making an angle with the length of the car as seen by a stationary observer. The bullet enters one edge of the car at the corner and passes out at the diagonally opposite corner. Neglecting any interaction between the car and the bullet, the time for the bullet to cross the car is
(A) 0.20 s (B) 0.15 s
(C) 0.10 s (D) 0.50 s.
Sol. : (A)
44. A boar which has a speed of 5 kmh–1 in still water crosses a river of width 1 km along the shortest possible path in 15 minute. The velocity of the river water in kmh–1 is
(A) (B) 3
(C) 4 (D) .
Sol. : (B)
45. A car A is going north east at 80 kmh–1 and another car B is going south east with a velocity of 60 kmh–1. The velocity of A relative to B makes an angle with the north equal to
(A) (B)
(C) (D) cot .
Sol. : (D)
80 kmh–1 NE
Since
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