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Saturday, May 12, 2018

Solutions to Problems on "ROTATIONAL MECHANICS" - H C Verma's Concepts of Physics, Part-I, Chapter-10, Questions for Short Answer

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ROTATIONAL MECHANICS:--
QUESTIONS FOR SHORT ANSWER

1. Can an object be in pure translation as well as in pure rotation?

ANSWER: No. In pure translation, each particle of the body moves in parallel lines while in a pure rotation the particles of a body move in concentric circles, Both cannot be true at the same time.

2. A simple pendulum is a point mass suspended by a light thread from a fixed point. The particle is displaced towards one side and then released. It makes small oscillations. Is the motion of such a simple pendulum a pure rotation? If yes where is the axis of rotation?

ANSWER: Since in a pure rotation all particles of a body move in concentric circles, hence this motion of a simple pendulum is a pure rotation, only the direction of motion reverses periodically. The axis of rotation is the line perpendicular to the plane of rotation and passing through the fixed point where the thread of the pendulum is tied.

Since the oscillations are small, for all practical purposes the movement of the ball can be assumed on a straight line and also the motion can be taken as periodic and Simple Harmonic motion can be assumed.

But these are the differences in theory and practice.

3. In a rotating body, a=ɑr, v=ɷr. Thus a/ɑ=v/ɷ. Can you use the theorems of ratio and proportion studied in algebra so as to write
(a+ɑ)/(a-ɑ) = (v+ɷ)/(v-ɷ)?

ANSWER: In the given relations there are five unknowns. When the first one is divided by the second, in fact, one unknown "r" is being eliminated. But if we use the theorems of ratio and proportion to write it as (a+ɑ)/(a-ɑ) = (v+ɷ)/(v-ɷ), we are not eliminating any unknown but complicating the relation a/ɑ=v/ɷ. Also, this equation is not dimensionally correct. So we can not write it like that.

4. A ball is whirled in a circle by attaching it to a fixed point with a string. Is there an angular rotation of the ball about its center? If yes, is this angular velocity equal to the angular velocity of the ball about the fixed point?

ANSWER: Yes the ball has an angular rotation about its center. If we notice the position of the point on the ball where it is tied with the string, we can see that it is moving about its center as its position changes. This point again comes to the same position when the ball makes a complete revolution around the fixed point. Thus the ball rotates about its center through 360° in the same time in which the ball revolves through 360° around the fixed point. Hence the angular velocity of the ball about its center is equal to the angular velocity of the ball about the fixed point.

5. The moon rotates about the earth in such a way that only one hemisphere of the moon faces the earth (figure-10Q1). Can we ever see the "other" face of the moon from the earth? Can a person on the moon ever see all the faces of the earth?

Figure for Question-5

ANSWER: Since only one hemisphere of the moon faces the earth, we can never see the "other" face of the moon from the earth.

But a person on the moon can see all the faces of the earth. The reason is that the time period and the direction of rotation of the moon is exactly the same as the revolution around the earth (about 28 days)-just like a ball being rotated in a circle around a fixed point. But the time period of the earth's rotation is not the same which is about 24 hours. So during one revolution of the moon, the earth rotates about 28 times. So a person on the facing side of the moon can see all faces of the earth.

6. The torque of the weight of a body about any vertical axis is zero. Is it always correct?

ANSWER: A force can only produce a torque around an axis if it has a component in the plane perpendicular to the axis and this component does not pass through the point of intersection of this plane and the axis. In the given case, the force of weight is parallel to any given vertical axis, hence it can not have a component in a plane perpendicular to this axis (Horizontal Plane). So it can not produce a torque about any vertical axis i.e. torque is zero. It is always correct.

7. The torque of a force F about a point is defined as Γ=rxF. Suppose r, F and Γ are all nonzero. Is rxΓ||F always true? Is it ever true?
ANSWER: The direction of the vector Γ (torque) is perpendicular to the plane in which r and F lie. The direction of the vector rxΓ will be perpendicular to the plane in which r and Γ lie. So it is not always true that rxΓ||F. It can only be true if r丄F but their directions will be opposite.

8. A heavy particle of mass m falls freely near the earth's surface. What is the torque acting on this particle about a point 50 cm east to the line of motion? Does this torque produce any angular acceleration in the particle?

ANSWER: The torque acting on the particle about 50 cm away from the line of motion = mg*(0.50) N-m. Where m = mass of the particle in kg, g=acceleration due to gravity in m/s² and r = 50 cm = 0.50 m.

This torque cannot produce any angular acceleration in the particle because there is no force in the opposite direction of the weight at 50 cm east of the line of motion. For a torque to be effective through a force, there must be equal and opposite non-linear force acting on the particle.

9. If several forces act on a particle, the torque on the particle may be obtained by first finding the resultant of the force and then taking torque of this resultant. Prove this. Is this result valid for the forces acting on different particles of a body in such a way that their lines of action intersect a common point?

ANSWER: Let F_1, F_2, F₃ .... be the forces acting on the particle and position vector of the particle from the point O (About which torque is being calculated) is r. Then the torque on the particle is equal to,
Γ= rxF₁ +rxF₂ + rxF₃ ....
= r x (F₁ +F₂ + F₃ ....) {Because vector cross product is distributive over addition}
= rxF
So, The Torque of the resultant force is equal to the sum of torques of individual forces on a particle.
Hence Proved.

Diagram for Q - 9

Second Part

If the forces act on different particles of a body such that their lines of actions intersect a common point then also this result is valid. Because we can shift the forces along their lines of action till their ends come to the point of intersection (Equivalent forces). This will not change the torque effect of a force (Since the magnitude of the torque of a force = Magnitude of the force x Shortest distance of the line of action of the force from the point about which torque is being calculated). Having done this we have a condition like first part where all the forces act at a point. So again
Γ= rxF₁ +rxF₂ + rxF₃ ....
= r x (F₁ +F₂ + F₃ ....) {Because vector cross product is distributive over addition}
= rxF

So, The Torque of the resultant force is equal to the sum of torques of individual forces on a body.

10. If the sum of all the forces acting on a body is zero, is it necessarily in equilibrium? If the sum of all the forces on a particle is zero, is it necessarily in equilibrium?

ANSWER: If the sum of all the forces acting on a body is zero then it is not necessarily in equilibrium. In fact, it will be in translatory equilibrium but may not be in rotational equilibrium. Suppose two equal and opposite forces having their line of actions at a certain distance apart act on a body, then the body will remain stationary at the point but will have rotation due to the torque produced.

If the sum of all the forces acting on a particle is zero then it is necessarily in equilibrium because there will not be a distance between antiparallel forces.

11. If the angular momentum of a body is found to be zero about a point, is it necessary that it will also be zero about a different point?

ANSWER: No.

We can express the angular momentum of a body about a point as mvr. Since it depends on 'r', it will be different about different points.

12. If the resultant torque of all the forces acting on a body is zero about a point, is it necessary that it will be zero about any other point?

ANSWER: Let us understand it with a simple example as in the figure below:-

Diagram for Q-12

There are two forces having each a magnitude F, one horizontal and other vertical. Point O is equidistant from each force = d. The torque of each force is Fd about O but equal and opposite. So total torque is zero about O.

Now consider a point P which is in the line of the horizontal force. The total torque of the forces about the point P =Fr (anticlockwise). So it is not zero.

So if the resultant torque of all the forces acting on a body is zero about a point, it is not necessary that it will be zero about any other point.

13. A body is in transitional equilibrium under the actions of coplanar forces. If the torque of these forces is zero about a point, is it necessary that it will also be zero about any other point?

ANSWER: Yes. In an inertial or equivalent frame of reference, It will be zero about any other point.

14. A rectangular brick is kept on a table with a part of its length projecting out. It remains at rest if the length projected is slightly less than slightly half the total length but it falls down if the length projected is slightly more than half the total length. Give reason.

ANSWER: The weight of the projected part produces a torque that will tend to overturn the brick about the edge of the table while the weight of the part on the table produces a torque that will tend to restore the brick from overturning. Assuming a uniform density of the brick, in the first case- the weight of projected part will be less than the weight of the part on the table. Also, the distance of the line of action of the weight of the projected part from the edge of the table is less in comparison to the other part. Hence the overturning torque, in this case, will be less than the restoring torque. So the brick remains at rest.

The situation is just reverse in the second case where the overturning torque is greater than the restoring torque, so the brick overturns and falls.

15. When a fat person tries to touch his toes keeping the legs straight, he generally falls. Explain with references to figure (10-Q2).

Figure for Q-15

ANSWER: We can understand the reason considering the body of the fat person in two parts - Legs up to waist as one part and rest of the body as the second part. (See diagram below).

Diagram for Q - 15

Center of the mass of the first part is at A (Somewhere near the knees) and the center of the mass of the second part is at B (Somewhere near the chest). So the center of mass of the whole body is at C on the line AB. The total weight of the fat person acts at C and the Normal force by the ground on his body (Which is also equal to W) acts opposite to the weight at his toes T. Though both weight and the Normal force are equal and opposite, their lines of action are not the same. Their lines of action are parallel having some distance d between them. So these antiparallel forces produce a torque equal to W*d on the fat person (Which is clockwise in the figure), and he generally falls.

16. A ladder is resting with one end on a vertical wall and the other end on a horizontal floor. Is it more likely to slip when a man stands near the bottom or near the top?

ANSWER: The ladder tends to slip more when the man stands near the wall.

Let us first draw a diagram to understand the problem. AB is the ladder with end A resting on the ground and end B against the wall. OA = d, OB = h, and W is the weight of the man on the ladder at distance x from end A horizontally. R and R' are normal forces at the ends A and B also F and F' are friction forces at A and B.

Diagram for Q - 16

Assuming µ same at both contact surfaces,
F =µR, F' = µR'
Equating forces vertically,
R+F' =W
R+µR' =W
Equating forces horizontally
F = R'
Taking Moments of all forces about B, we have
W(d-x)+Fh-Rd = 0
→W(d-x)+µRh-Rd = 0
→W(d-x) = R(d-µh)
→R=W(d-x)/(d-µh)
Overturning Torque (anticlockwise) about A
= Wx
Restoring torque (Clockwise)
= R'h+F'd
So net overturning torque T
=Wx-(R'h+F'd)
=Wx-(Fh+µR'd)
=Wx-(Fh+µFd)
=Wx-F(h+µd)
=Wx-µR(h+µd)
=Wx-µW(d-x)(h+µd)/(d-µh)
={Wxd-Wxµh-µW(dh+µd²-xh-µxd)}/(d-µh)
={Wxd-~~Wxµh~~-µWdh+µ²d²W+~~µxhW~~+µ²Wxd)}/(d-µh)
={Wxd(1+µ²)+Wµd(µd-h)}/(d-µh)
In this expression all except x are constant and we can see that as x increases T also increases.
So, the ladder tends to slip more when the man stands near the wall.

17. When a body is weighed on an ordinary balance we demand that the arm should be horizontal if the weights on the two pans are equal. Suppose equal weights are put on the two pans, the arm is kept at an angle with the horizontal and released. Is the torque of the two weights about the middle point (Point of support) zero? Is the total torque zero? If so, why does the arm rotate and finally become horizontal?

ANSWER: In fact, middle point (Point of support) is not in a straight line which joins the hanging point of the panes. This point of support (Middle point, pivot) is slightly above the center of the arms. Thus these three points make a triangle with a wide base. See diagram below,

Diagram for Q- 17

When the arms are horizontal, the resultant weight of the pans 2W (Downard) acts in the middle and the balancing normal force 2W also acts upwards a bit above but in the same line. So net torque is zero.
When the arm is kept at an angle with the horizontal, the lines of actions of resultant weight and the normal force are not the same but are at some distance d. (See the second diagram). These two equal and opposite forces each equal to 2W but at distance d produce a net restoring torque =2Wd which acts till the arms are horizontal because only then d=0 i.e. the torque is zero. Now it is in stable equilibrium.

18. The density of the rod AB continuously increases from A to B. Is it easier to set it in rotation by clamping it at A and applying perpendicular force at B or by clamping it at B and applying the force at A?

ANSWER: By clamping at B and applying the force at A.

It is due to the fact that in this position the center of mass is comparatively closer to the clamp. Hence moment of inertia I of the rod about the clamp will be less. So for the same angular acceleration, the torque needed (=Iα) will be less.

19. When tall buildings are constructed on earth, the duration of day-night slightly increases. Is it true?

ANSWER: Yes, True.

When tall buildings are constructed, the effective radius of the earth slightly increases. Though the mass remains constant, the moment of inertia of the earth I slightly increase. To conserve the angular momentum L of the earth the angular velocity ω (=L/I) slightly decrease. This means that the earth rotates slower now and takes more time to complete one rotation. So the duration of day-night slightly increases.

20. If the ice at the poles melts and flows towards the equator, how will it affect the duration of day-night?

ANSWER: The duration of day-night slightly increases.

The reason is same as above problem. The moment of inertia I increases.

21. A hollow sphere, a solid sphere, a disc and a ring all having same mass and radius are rolled down on an inclined plane. If no slipping takes place, which one will take the smallest time to cover a given length?

ANSWER: Since each object has the same radius, the total torque on each of them will be same. So angular acceleration α =Γ/I. Thus the object with minimum I will have maximum angular acceleration. Let us see the value of I for given objects:-

Hollow sphere = (2/3)MR²
Solid Sphere = (2/5)MR²
Disc = ½MR²
Ring = MR²
Obviously, minimum I of them is for the solid sphere. So the solid sphere will take the smallest time to cover a given length.

22. A sphere rolls on a horizontal surface. Is there any point on the sphere which has a vertical velocity?

ANSWER: Theoretically, no point on the sphere has velocity vector pointing in the vertical direction.
Practically, at the points very near to the point which is in contact at the instant has a nearly vertical velocity with a very small magnitude. Just at the contact point, when the velocity becomes true vertical, its magnitude becomes zero. So we can not say that a vertical velocity.

===<<<O>>>===

Links to the chapters -

ALL LINKS
CHAPTER- 10 - Rotational Mechanics

Questions for Short Answers

OBJECTIVE - I

OBJECTIVE - II

EXERCISES - Q-1 TO Q-15

EXERCISES - Q-16 TO Q-30

EXERCISES Q-31 TO Q-45

EXERCISES Q-46 TO Q-60

EXERCISES Q-61 TO Q-75

EXERCISES Q-76 TO Q-86

CHAPTER- 9 - Center of Mass, Linear Momentum, Collision

Questions for Short Answers

Objective - I

Objective - II

EXERCISES Q-1 TO Q-10

EXERCISES Q-11TO Q-20

EXERCISES Q-21 TO Q-30

EXERCISES Q-31 TO Q-42

EXERCISES Q-43 TO Q-54

HC Verma's Concepts of Physics, Chapter-8, WORK AND ENERGY

Click here for → Question for Short Answers

Click here for → OBJECTIVE-I

Click here for → OBJECTIVE-II

Click here for → Exercises (1-10)

Click here for → Exercises (11-20)

Click here for → Exercises (21-30)
Click here for → Exercises (31-42)
Click here for → Exercise(43-54)

HC Verma's Concepts of Physics, Chapter-7, Circular Motion

Click here for → Questions for Short Answer
Click here for → OBJECTIVE-I
Click here for → OBJECTIVE-II

Click here for → EXERCISES (1-10)
Click here for → EXERCISES (11-20)
Click here for → EXERCISES (21-30)

HC Verma's Concepts of Physics, Chapter-6, Friction

Click here for → Friction OBJECTIVE-I
Click here for → Friction - OBJECTIVE-II

Click here for → Questions for Short Answer

Click here for → OBJECTIVE-I
Click here for → Friction - OBJECTIVE-II

Click here for → EXERCISES (1-10)

Click here for → Exercises (11-20)

Click here for → EXERCISES (21-31)
For more practice on problems on friction solve these "New Questions on Friction" .

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HC Verma's Concepts of Physics, Chapter-5, Newton's Law's of Motion

Click here for → QUESTIONS FOR SHORT ANSWER

Click here for→ Newton's laws of motion - Objective - I

Click here for → Newton's Laws of Motion - Objective -II

Click here for → Newton's Laws of Motion-Exercises(Q. No. 1 to 12)

Click here for→Newton's Laws of Motion,Exercises(Q.No. 13 to 27)

Click here for→Newton's Laws of Motion,Exercises(Q.No. 28 to 42)

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HC Verma's Concepts of Physics, Chapter-4, The Forces

The Forces-

"Questions for short Answers"

Click here for "The Forces" - OBJECTIVE-I

Click here for "The Forces" - OBJECTIVE-II

Click here for "The Forces" - Exercises

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HC Verma's Concepts of Physics, Chapter-3, Kinematics-Rest and Motion:---

Click here for "Questions for short Answers"

Click here for "OBJECTIVE-I"

Click here for EXERCISES (Question number 1 to 10)

Click here for EXERCISES (Question number 11 to 20)

Click here for EXERCISES (Question number 21 to 30)

Click here for EXERCISES (Question number 31 to 40)

Click here for EXERCISES (Question number 41 to 52)

===============================================

Chapter -2, "Vector related Problems"

Click here for "Questions for Short Answers"

Click here for "OBJECTIVE-I

Click here for "OBJECTIVE-II"

Click here for "Exercises"

Wednesday, March 28, 2018

Solutions to Problems on "CENTER OF MASS, LINEAR MOMENTUM, COLLISION" - H C Verma's Concepts of Physics, Part-I, Chapter-9, EXERCISES-Q55-Q64

My Channel on YouTube → SimplePhysics with KK

For links to

other chapters - See bottom of the page

Or click here → kktutor.blogspot.com

CENTER OF MASS, LINEAR MOMENTUM, COLLISION:--
EXERCISES Q-55 to Q-64

55. A bullet of mass 10 g moving horizontally at a speed of 50√7 m/s strikes a block of mass 490 g kept on a frictionless track as shown in figure (9-E17). The bullet remains inside of the block and system proceeds towards the semicircular track of radius 0.2 m. Where will the block strike the horizontal part after leaving the semicircular track?

Figure for Q-55

ANSWER: Mass of the bullet, m = 10g = 0.01 kg

The speed of the bullet, v = 50√7 m/s.

The mass of the block, m' = 490 g = 0.49 kg

The speed of the block, v' = 0

Combined mass of the system, M = 10+490 = 500 g = 0.50 kg

Let the speed of the system after the collision is V. From the principle of conservation of linear momentum,

MV = mv + m'v'

→0.50V = 0.01*50√7 + 0.49*0

→V = 0.50√7/0.50 =√7 m/s

The K.E. of the system = ½MV² = ½*0.50*7 =0.25*7 = 1.75 J

Let the speed of the system when it separates from the semicircular part be u. At this point, the force applied by the system on the surface due to the circular motion would be equal to the component of weight along the radial direction. i.e.

Mu²/r = Mg.cosθ {θ = The angle that u makes with horizontal}

→u² = gr.cosθ = 10*0.2cosθ =2cosθ

Figure for problem 55

The total energy of the system at this point

= K.E. + P.E.

= ½Mu² + Mgh

= ½*0.50u² + 0.50*10*h

{h = height of the point where it separates

=0.2+0.2cosθ}

= 0.25u² + 5*.2(1+cosθ) = 0.25*2cosθ+1+cosθ

= 1+1.5cosθ

Taking horizontal portion as a reference for P.E., the total energy of the system just before it enters the semicircular part = K.E.

Hence from the conservation law of energy,

1+1.5cosθ = 1.75

→ Cosθ = (1.75-1)/1.5 =0.75/1.5 =0.5

→θ = 60°

Hence, u = √(2*0.5) = √1 = 1 m/s and

h = 0.2+0.2*cosθ =0.2+0.2*½ =0.3 m

If the time taken by the system to fall through the height h to the ground is t, then from

h=ut+½gt² we have

0.3 = -1*sin60°*t+½*10*t² =-√3t/2+5t²

→10t²-√3t-0.6 =0

t = {√3土√(3+4*10*0.6)}/20 = {√3土√27}/20

= (1.732土5.196)/20

= 0.34 s or -0.17 s

Since the time cannot be negative, hence t = 0.34 s. The distance traveled by the system in this time = u.cosθ*t

=1*0.5*0.34 = 0.17 m

Its horizontal distance from the outermost point O at the time of separation = 0.2-0.2*sinθ =0.2-0.2*√3/2

=0.2-0.1*1.73 =0.2-0.17 =0.03 m

Hence the horizontal distance of the point where it falls from O

=0.03+0.17 = 0.20 m

Hence it falls at the junction of the straight and curved path.

56. Two balls having masses m and 2m are fastened to two light strings of same length l (Figure 9-E18). The other ends of the strings are fixed at O. The strings are kept in the same horizontal line and the system is released from rest. The collision between the balls is elastic. (a) Find the velocities of the balls just after the collision. (b) How high will the balls rise after the collision?

Figure for Q-56

ANSWER: When the balls collide their speeds v will be equal and opposite and v = √(2gl), if after the collision the speed of m is u and of 2m is u', then from conservation law of linear momentum,

mu+2mu' = mv-2mv = -mv

→u+2u' = -v

→u = -(v+2u')

K.E. before collision =½mv²+½*2m*v² = 3mv²/2

K.E. after the collision = ½mu²+½*2mu'² =½mu²+mu'²

If the reference line for P.E. is assumed at the collision level then these are the total energies of the balls before and after the collision. From the energy conservation principle,

½mu²+mu'² = 3mv²/2

→u²+2u'² = 3v²

→(v+2u')²+2u'² = 3v²

→v²+2v*2u'+4u'²+2u'² = 3v²

→6u'²+4vu'-2v² =0

→3u'²+2vu'-v² =0

→3u'²+3vu'-vu'-v² = 0

→3u'(u'+v)-v(u'+v) =0

→(u'+v)(3u'-v) = 0

If u'+v = 0 → u' = -v and u = -(v+2u') =-u' =v. Since we have taken speed towards the right as positive, this suggests that after the collision speed of m (i.e. u) is towards right and speed of 2m (i.e. u') is towards the left which is not possible because after the collision they cannot cross each other.

Hence 3u'-v = 0 →u' =v/3 = √(2gl)/3

So the speed of the ball with mass 2m is √(2gl)/3 towards the right.

And u = -(v+2u') = -{√(2gl) + 2√(2gl)/3}

=-5√(2gl)/3 = -√5²√(2gl)/3 -√(50gl)/3

Hence the speed of the ball with mass m is √(50gl)/3 towards left.

ALTERNATELY

In an elastic collision,

the speed of separation =speed of approach

Hence u'-(-u) = v-(-v) =2v

u+u' = 2v --------(A)

Since here we have taken u' positive and u negative, from conservation law of linear momentum,

2mu'-mu=mv-2mv

2u'-u=-v ---------(B)

Adding (A) and (B), we get,

3u'=v

→u' = v/3 =√(2gl)/3.

And u = 2v-u' =2v-v/3 =5v/3

→u = 5√(2gl)/3 =√(50gl)/3.

(b) Total energy just after the collision, For m = ½mu²

= ½m*50gl/9 =50mgl/18

If it can rise a maximum height h, then

mgh = 50mgl/18

→h=50l/18 = 2.78l

But the strings are fixed at O, hence maximum height mass m can rise is 2l.

For 2m, Total energy =½*2m*(2gl)/9 =4mgl/9 = 2mgl/9

If maximum height attained be h', then

2mgh' = 2mgl/9

→h' = l/9

So the maximum height the mass 2m can rise is l/9.

57. A uniform chain of mass M and length L is held vertically in such a way that its lower end just touches the horizontal floor. The chain is released from rest in this position. Any portion that strikes the floor comes to rest. Assuming that The chain does not form a heap on the floor, calculate the force exerted by it on the floor when a length x has reached the floor.

ANSWER: The mass per unit length of the chain =M/L
Consider an infinitesimally small length dx at length x above the floor initially.
Mass of this length = dm = (M/L)dx
The velocity of this mass when it reaches the floor,
v=√(2gx)
The linear momentum of this mass when it reaches the floor =v.dm
After the strike, its velocity is zero. Hence the change of momentum = v.dm-0 =v.dm = √(2gx)(M/L)dx
And the force = Rate of change of momentum
=v.dm/dt = √(2gx)(M/L)dx/dt {But dx/dt =v}
=√(2gx)(M/L)v
=√(2gx)(M/L)*√(2gx)
=(2gx)(M/L)
=2Mgx/L
Force due to weight of already reached length x =(M/L)xg =Mgx/L
Hence total force when x length has reached the floor = Force due to change of momentum+Force of weight
=2Mgx/L+Mgx/L
=3Mgx/L

58. The blocks shown in figure (9-E19) have equal masses. The surface of A is smooth but that of B has a friction coefficient of 0.10 with the floor. Block A is moving at a speed of 10 m/s towards B which is kept at rest. Find the distance traveled by B if (a) the collision is perfectly elastic and (b) the collision is perfectly inelastic. Take g=10 m/s².

Figure for Q-58

ANSWER: (a) When the collision is perfectly elastic and the blocks have equal masses, the velocities just interchange. That means the block A will come to rest and the block B get a velocity (v) of 10 m/s.
Force of friction on B =µN = 0.10*mg
Retardation of the block due to friction, a=Force/mass =0.10mg/m
=0.10g =0.10*10 =1 m/s²
Hence 0² = v²-2as (s= distance traveled)
→s =v²/2a =100/2 =50 m

(b) When the collision is perfectly inelastic.
After the strike, both will stick together and move with the same velocity, say u. From the conservation law of linear momentum,
2mu = m*10+m*0 =10m
→u = 5 m/s
Force of friction = µN = 0.10*mg

Retardation of the block due to friction, a'=Force/mass =0.10mg/2m
=0.10*10/2 = 0.50 m/s²
Hence, 0²=u²-2a's
→s=5²/2*0.5 =25/1 = 25 m

59. The friction coefficient between the horizontal surface and each of the blocks shown in figure (9-E20) is 0.20. The collision between the blocks is perfectly elastic. Find the separation between the two blocks when they come to rest. Take g = 10 m/s².

Figure for Q-59

ANSWER: The force of friction on 2 kg block = µN =0.20*2g

=0.40g

Retardation of 2 kg block,a = 0.40g/2 =0.40*10/2 =2 m/s²

Let the velocity of the 2 kg block jst before the collision = v

d =16 cm =0.16 m, u = 1.0 m/s

Hence, v² = u²-2ad =1²-2*2*0.16 =1-0.64 =0.36

→v = 0.60 m/s

Since the second block is at rest, the velocity of approach = 0.60-0 =0.60 m/s.

The collision is elastic, hence the velocity of separation = velocity of approach. If after the strike, the velocity of 2 kg block is u' (towards left) and 4 kg block is v' (towards right),
then velocity of separation= v'-(-u') =0.60
v'+u' = 0.60 --------------------------------------(A)
And from conservation principle of linear momentum,
4v'-2u' = 2*0.60 =1.20
→2v'-u'=0.60 -------------------------------------(B)
Adding (A) and (B),
3v' = 1.20
→v' = 0.40 m/s
u' = 0.60-0.40 =0.20 m/s
Since retardation of 2 kg block = 2 m/s²,
Distance traveled by it = u'²/2a =0.20²/2*2
0.04/4 =0.01 m = 1cm (towards left from collision pont)
Retardation of 4 kg block, a' = Friction force/ mass
=µ*4g/4 = 0.20*10 =2 m/s²
Hence distance traveled by this block =v'²/2a'
=0.40²/2*2 =0.16/4 =0.04 m =4 cm (towards right from the collision point)
Hence total separation when they come to rest
= 1 cm + 4 cm = 5 cm

60. A block of mass m is placed on a triangular block of mass M, which in turn is placed on a horizontal surface as shown in figure (9-E21). Assuming frictionless surfaces find the velocity of the triangular block when the smaller block reaches the bottom end.

Figure for Q-60

ANSWER: Let the horizontal velocity of m with respect to ground = v'

the velocity of the triangular block with mass M =V

If v is the relative velocity of block m w.r.t. triangular block M along the slope, then in the horizontal direction

v.cosα = v'+V {Since the directions of v' and V are opposite}

→v' = v.cosα-V

Figure for Q-60

Now, there is no movement of block M in the vertical direction, the vertical component of the velocity of m (v.sinα) will remain unaffected by the motion of M and the velocity of m w.r.t. ground (say V') will have components v' and v.sinα.

Since there is no external force in the horizontal direction, the linear momentum will be conserved, hence

mv' = MV

→m(v.cosα-V) =MV

→mv.cosα-mV = MV

→mv.cosα = (m+M)V

→v = (m+M)V/m.cosα

When the smaller block reaches the bottom end of the larger block, it loses the P.E. of mgh and gains the K.E. in both the blocks. From the conservation law of enrgy,

½MV²+½mV'² =mgh

→MV²+m{(v')²+(v.sinα)²} = 2mgh

→MV²+m{(vcosα-V)²+ (m+M)²V²sin²α/m²cos²α}=2mgh

→MV²+[m{(m+M)V/m-V}²+ m(m+M)²V²sin²α/m²cos²α}=2mgh

→MV²+[m{(m+M)V-mV}²/m²+(m+M)²V²sin²α/mcos²α] =2mgh

→MV²+V²[M²/m+(m+M)²sin²α/mcos²α] =2mgh

→V²[M+M²/m+(m+M)²sin²α/mcos²α]=2mgh

→V² = 2m²gh.cos²α/[mMcos²α+M²cos²α+(m+M)²sin²α]

→V² = 2m²gh.cos²α/[Mcos²α(m+M)+(m+M)²sin²α]

→V² = 2m²gh.cos²α/[(m+M){Mcos²α+(m+M)sin²α]

= 2m²gh.cos²α/[(m+M){Mcos²α+msin²α+Msin²α}]

= 2m²gh.cos²α/[(m+M){M(cos²α+sin²α)+msin²α}]

= 2m²gh.cos²α/{(m+M)(M+msin²α)}

Hence V = [ 2m²gh.cos²α/{(m+M)(M+msin²α)}]½

61. Figure (9-E22) shows a small body of mass m placed over a larger mass M whose surface is horizontal near the smaller mass and gradually curves to become vertical. The smaller mass is pushed on the longer one at a speed v and the system is left to itself. Assume that all the surfaces are frictionless. (a) Find the speed of the larger block when the smaller block is sliding on the vertical part. (b) Find the speed of the smaller mass when it breaks off the larger mass at height h. (c) Find the maximum height (from the ground) that the smaller mass ascends. (d) Show that the smaller mass will again land on the bigger one. Find the distance traveled by the bigger block during the time when the smaller block was in its flight under gravity.

Figure for Q-61

ANSWER: Let the speed of the larger block be V. Momentum of the system just before the contact in the horizontal direction = Mx0+mv = mv

(a) The momentum of the system when the smaller block is sliding on the vertical part = (M+m)V. Since the linear momentum in the horizontal direction will be conserved in the absence of external force, so

(M+m)V = mv

→V = mv/(M+m)

(b) Let the speed of the smaller block at height h is v'. From the conservation of energy law,

½mv²=½mv'²+½MV²+mgh

→ v²=v'²+MV²/m+2gh

→v'²=v²-M*m²v²/m(M+m)² -2gh

=v²-Mmv²/(M+m)²+2gh

=v²{(M+m)²-Mm}/(M+m)² - 2gh

=v²{M²+m²+2Mm-Mm}/(M+m)² - 2gh

=v²{M²+Mm+m²}/(M+m)² - 2gh

Hence v' = [v²{M²+Mm+m²}/(M+m)² - 2gh]½

(c) When the smaller mass breaks off from the larger mass at height h, its speed v' is not vertical but at an angle, let it be α from the horizontal. Since just before the break-off, it was moving with the system with a uniform speed V, V will be the horizontal component of v'. Hence,

V=v'.cosα, →cosα=V/v'

And, sinα=√{1-cos²α}=√{1-V²/v'²}

The vertical component of v' will be u=v'.sinα

→u=v'√{1-V²/v'²} = √{v'²-V²}

Hence the maximum height achieved from the larger block,

=u²/2g

=(v'²-V²)/2g

=[v²{M²+Mm+m²}/(M+m)² - 2gh -m²v²/(M+m)²]/2g

=[(v²M²+Mmv²+m²v²-m²v²)/2g(M+m)²-h]

=[(v²M²+Mmv²)/2g(M+m)²-h]

=[v²M(M+m)/2g(M+m)²-h]

=[Mv²/2g(M+m) - h]

So the maximum height achieved from the ground

= h + [Mv²/2g(M+m) - h]

= Mv²/2g(M+m)

(d) The smaller block has a uniform horizontal component of the speed i.e. V which is same as the larger block. Hence the smaller block will again land on the larger block at the same point. The flight of the smaller block will be a projectile motion when seen from the ground.

If the time of flight of the smaller block is T, then from v=u-gt we have,

-u = u-gT

→T = 2u/g = 2[√{v'²-V²}]/g

The distance traveled by the larger block when the smaller block was in flight under gravity = VT

= V*2[√{v'²-V²}]/g

=(2V/g)*[v²{M²+Mm+m²}/(M+m)² - 2gh -m²v²/(M+m)²]½

={2mv/g(M+m)}*[(v²M²+Mmv²+m²v²-m²v²)/(M+m)²-2gh]½

={2mv/g(M+m)}*[(v²M(M+m)/(M+m)²-2gh]½

={2mv/g(M+m)}*[(v²M/(M+m)-2gh]½

={2mv/g(M+m)}*[(Mv²-2gh(M+m)]½/(M+m)½

= 2mv[Mv²-2(M+m)gh]½/g(M+m)^3/2

62. A small block of superdense material has a mass of 3x10²⁴ kg. It is situated at height h (much smaller than the earth's radius) from where it falls on the earth's surface. Find its speed when its height from the earth's surface has reduced to h/2. The mass of the earth is 6x10²⁴ kg.

ANSWER: Since the mass of the small block is half that of the earth, not only the block will move towards the earth but the earth will also move towards the ball. Also, there is no external force on the system, the center of mass of the system will remain unchanged and the linear momentum conserved. Hence,
MV+mv = 0
So. MV = -mv
→V = -(m/M)v
The negative sign shows that v and V have opposite direction.
{Here M & V are the mass and speed of the earth and m & v are the mass and speed of the block, at h/2 above earth}

Initial P.E. of the block = mgh

P.E. at height h/2 = mgh/2

Reduction of P.E. = mgh-mgh/2 = mgh/2

This must be gain in K.E. which is = ½MV²+½mv²

So, ½MV²+½mv² = mgh/2

→M(m/M)²v²+mv² = mgh

→m²v²/M+mv² = mgh

→v²(m/M+1) = gh

→v²(m+M)/M = gh
→v² = {M/(m+M)}gh = {6x10²⁴ /(3x10²⁴ +6x10²⁴ )}gh
→v² = 6gh/9 =2gh/3
→v = √(2gh/3)

63. A body of mass m makes an elastic collision with another identical body at rest. Show that if the collision is not head-on the bodies go at right angles to each other after the collision.

ANSWER: Let the speed of block A just before the collision be V at P. Since both the blocks have equal mass m and the other block B is at rest, Total momentum before collision = mV+m*0 =mV

If after the collision block A goes in PR direction at an angle α with the original direction and speed v; Block B goes in QS direction making an angle β with the original direction and speed v', then the linear momentum along the original direction after collision

= mv.cosα+mv'.cosβ

From the conservation law of linear momentum,

mv.cosα+mv'.cosβ = mV

→V = v.cosα+v'.cosβ

Figure for Q-63

In the direction perpendicular to the original direction:-

Linear momentum before collision = 0

Linear momentum after collision = v.sinα-v'.sinβ

Hence v.sinα-v'.sinβ =0

→ v.sinα = v'.sinβ

K.E. before collision = ½mV²

K.E. after collision = ½mv²+½mv'²

From the energy conservation law,
½mv²+½mv'² = ½mV²
→v²+v'²=V²
→v²+v'²= (v.cosα+v'.cosβ)²

→v²+v'² = v²cos²α+v'²cos²β+2.vv'cosα.cosβ

→v²(1-cos²α)+v'²(1-cos²β) = 2.vv'cosα.cosβ

→v².sin²α+v'².sin²β = 2.vv'cosα.cosβ

→v².sin²α+v².sin²α = 2.vv'cosα.cosβ

→2v²sin²α-2.vv'cosα.cosβ = 0

→2v(v.sin²α-v'.cosα.cosβ) = 0

Since v is not zero,

v.sin²α-v'.cosα.cosβ = 0

→v'sinβ.sinα-v'.cosα.cosβ = 0

→v'(cosα.cosβ-sinα.sinβ) = 0

→v'.cos(α+β)=0

Since v' is also not zero, hence

→cos(α+β) =0 = cos90°

So, α+β=90°⍴

Hence the bodies go at right angles after the collision when the collision is not head-on.

64. A small particle traveling with a velocity v collides elastically with a spherical body of equal mass and of radius r initially kept at rest. The center of this spherical body is located a distance ρ(<r) away from the direction of motion of the particle (figure 9-E23). Find the final velocities of the two particles.

Figure for Q-64

ANSWER: Let us the figure as below:-

Diagram for Q-64

Let the mass of the particle and the body be m. If the particle strikes the body at A making the angle between radius AO and direction of motion α, then the velocity v can be resolved along the radius AO and tangent at A as v.cosα and v.sinα respectively.
The tangential velocity v.sinα will have no effect on the body after the impact while the radial velocity v.cosα will have a head-on collision effect on the body. Since both the particle and the body have equal mass m and the body is at rest so the velocities will be exchanged. It means that after the collision the particle will have no radial velocity component while the body will move with the velocity v.cosα along the radial direction AO. So after the collision the velocity of the body = v.cosα
=v.(BO/AO)
=v.√(r²-⍴²)/r {Given that AB=⍴}
Since the smaller particle has no component along AO, it will have only the tangential component v.sinα =v.(AB/AO) =v⍴/r.

Hence the small particle goes along the tangent with a speed of v⍴/r and the spherical body goes perpendicular to the smaller particle with a speed of v.√(r²-⍴²)/r.