Solutions-Class-11-Science-Physics-Chapter-4-Laws of Motion-Maharashtra Board

(A) Only P

(D) All four.

(C) Mass 3m experiences maximum force.

(A) Component of F₂ adds up to weight to increase the normal reaction.

(B) 3/2

(B) 1.5

(D) While balancing an object on a pivot, the line of action of the gravitational force of the earth passes through the centre of mass of the object.

(D) All, (I), (II) and (III).

A real object on a horizontal surface stops moving because of the presence of resistive forces of friction or viscous drag. For its continued motion, an external force is required to overcome these resistive forces.

Newton's first law of motion is not dismissed for two reasons :

• Before Newton synthesized and firmly founded that branch of mechanics called dynamics, Artistotelians had a different view of the role of force. They held that force was the cause of uniform (unaccelerated) motion. Newton's first law of motion implies that a net force is the cause of change of velocity. Thus, the first law was an accepted emphasis on the changing point of view in mechanics.
• Newton's first law defines inertial frames of reference and limits the scope of the second law to such frames. In deriving the first law from the second law, \(\vec{v}\) is assumed to be measured in an inertial frame; i.e., Newton's first law is built into the argument and not truly ‘derived’ from the second law. The first law is independent and necessary to understand Newton's concept of force. The argument that the first law is redundant, as some sceptics would like to impress upon us, is a deceptive fallacy.

Limitation of Newton’s laws of motion

• Newton’s laws are applicable only in the inertial frames of reference which they define. If the body is in a frame of reference of acceleration (a), we need to use a pseudo force (−m ) in addition to all the other forces while writing the force equations.
• Newton’s laws are applicable for point objects.
• Newton’s laws are applicable to rigid bodies. A body is said to be rigid if the relative distances between its particles do not change for any deforming force.
• For objects moving with speeds comparable to that of light, Newton’s laws of motion do not give results that match with the experimental results and Einstein's special theory of relativity has to be used.
• Behaviour and interaction of objects having atomic or molecular sizes cannot be explained using Newton’s laws of motion, and quantum mechanics has to be used.

The push to the right is due to the acceleration of the observer's frame of reference – the capsule — to the left. This could be due to an acceleration of the capsule in a straight line towards the left or due to the circular motion of the capsule, in which case it is the centripetal acceleration.

The force to the right is a reality in the observer's reference frame.

Similar pseudo force is experienced in a lift which accelerates up or down, and in an  accelerating vehicle.

Majority of forces experienced in our daily life, such as force of friction, normal reaction, tension in strings, collision forces, elastic forces, viscosity (fluid friction), etc. are EM in nature.

Thus, the electromagnetic force, the second strongest of the four fundamental forces, almost governs our daily life.

The attraction of gravity is universal. The gravitational forces between ordinary uncharged objects, on the other hand, are hidden by frictional forces of one type or another, or by the enormously higher gravitational force owing to the Earth. As a result, we do not perceive all regular objects colliding owing to mutual pulls. Because of their reciprocal gravitational pulls, the friction forces between the friends and the earth keep them from colliding.

(a) The work done by a force in moving a body is defined to be the scalar product of the force and the displacement of the body.

Let  be the force at a point along the path, and let θ be the angle between  and  at this point. Then, the work done during the displacement  is

dW = (F cos θ) ds = \(\vec{F}.\vec{ds}\)

Then, W = \(\int_A^B{dW} = \int_A^B \vec{F}.\vec{ds}\)

This is the formula for calculating work done by variable force.

(b) The formula W = (\vec{F}.\vec{s}\) = (F cos θ)s  is applicable only if (\vec{F}\) is constant, both in magnitude and direction, over the displacement

If the force varies in magnitude and/or direction, then we must consider dW = \(\vec{F}.\vec{ds}\)  during an infinitesimal displacement \(\vec{ds}\)  in an infinitesimal time interval dt→0. The total work done by the variable force in moving the body from point A to point B is given by the integrating dW over the path from A to B.

W = \(\int_A^B{dW} = \int_A^B \vec{F}.\vec{ds}\)

(c) Graphically, we plot the component of \(\vec{F}\)  parallel to the direction of the motion at a point as a function of the displacement. Then, the total work is given by the area under the curve between the initial and final points.

Work is considered to be done when a body moves in response to an external force.

The effect of another body or bodies is referred to as external force. Thus, if body 2 executes work on body 1, body 2 transfers energy to body 1, whose change in kinetic energy is equal to the work done on it according to the work-energy theorem. As a result, work and energy are considered to be two sides of the same coin.

• Let the ball dropped from the terrace of a building of height h have mass m. during free fail, the ball is acted upon by gravity (accelerating conservative force).
• While coming down, the work that is done is equal to the decrease in potential energy.
• This work done however is not entirely converted into kinetic energy but some part of it is used in overcoming the air resistance (retarding non-conservative force). This part of energy appears in some other forms such as heat, sound, etc.
• Thus, in this case of an accelerating conservative force along with a retarding non—conservative force, the work—energy theorem is given as, Decrease in the gravitational, PE. = increase in the kinetic energy + work done against non-conservative forces
• Thus, the relation mgh = ½ mv² is not applicable when non-conservative forces are considered. The part of the energy converted to heat, sound, etc. also needs to be added to the equation.
• The ball will not bounce to the same height from where it was dropped due to the loss in kinetic energy during the collision making it an inelastic collision.

Principle or law of conservation of linear momentum : When the resultant external force acting on a system is zero, the total linear momentum of the system is constant i.e., it is conserved. OR The total linear momentum of an isolated system, i.e., a system subject only to internal forces, is constant i.e., it is conserved.

The principle is a consequence of Newton's third law of motion.

Examples :

• As a person jumps out of a stationary boat, the boat moves in the backward direction. Initially, the momentum of the system was zero because both of them were at rest. While jumping, the forward momentum of the person must be equal in magnitude to the backward momentum of the boat such that their sum is zero and the total linear momentum of the person and boat is conserved.
• When a bomb explodes in mid-air, the fragments fly off in different directions However, it is found that the vector sum of the momenta of all the fragments at any instant always remains equal to the initial momentum of the bomb in both magnitude and direction. Here, there is an increase in KE and production of heat, sound and light because the chemical PE of its explosive ingredients is partly converted to these forms.

Linear momentum conservation during the burst of a cracker:

• The law of conservation of linear momentum holds good during the bursting of a cracker.
• When a cracker is at rest before the explosion, the linear momentum of the cracker is zero.
• When the cracker explodes into number of pieces, scattered in different directions, the vector sum of the linear momentum of these pieces is also zero. This is as per the law of conservation of linear momentum.

Coefficient of Restitution e: When two bodies collide with each other, the negative ratio of their relative velocity after collision to their relative velocity before collision, is called the coefficient of restitution (e).

Coefficient of restitution during a head-on, elastic collision:

Consider the head-on collision of two particles of masses m₁ and m₂, moving before the collision with constant velocities \(\vec{u_1}\) and \(\vec{u_2}\) in the laboratory frame of reference, see below Fig. After the collision they have velocities \(\vec{v_1}\) and \(\vec{v_2}\), again considered to be constant.

Since the collision is head-on, i.e., the motion is confined to one dimension, the velocity vectors all lie in the same line and hence we can write the equations in scalar form with the usual sign convention. Therefore, by the principle of conservation of linear momentum,

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

or m₁(u₁ − v₁) = m₂(v₂ − u₂)  …..(2)

Also, since the collision is elastic, total kinetic energy before collision = total kinetic energy after collision :

½ m₁u₁² + ½ m₂u₂² = ½ m₁v₁² + ½ m₂v₂²

∴ m₁(u₁² − v₁²)=m₂(v₂² − u₂²)

Or m₁(u₁ − v₁) (u_{1 +} v₁)  = m₂(v₂ − u₂) (v₂ + u₂)    …….(3)

Dividing Eq. (3) by Eq. (2 ), we get,

u_{1 +} v₁ = v₂ + u₂      …….(4)

Rearranging the terms in Eq. (4),

u₁ − u₂ = v₂ − v₁   …….(5)

Here, (u₁ − u₂) is the velocity with which particle 1 approaches particle 2 before collision and (v₂ − v₁) is the velocity with which particle 2 separates away from particle 1 after collision.

Hence, in a head-on elastic collision, velocity of separation = velocity of approach.

From Eq. (5), the coefficient of restitution,

e = \(\frac{v_2-v_1}{u_1-u_2}\) = 1

In a perfectly inelastic collision, the colliding particles stick together and move with a common velocity. That is, their velocity of separation after the collision is zero. Therefore, the coefficient of restitution,

e = \(\frac{\text{velocity of separation}}{\text{velocity of approach}}\) = 0

Thus, for an elastic collision, coefficient of restitution, e =1. For a perfectly inelastic collision, e =0 (by definition). Thus, for any collision, the coefficient of restitution lies between 1 and 0.

Two particles of masses m₁ and m₂ move with initial velocities u₁ and u₂ such that particle 1 collides head-on elastically with particle 2.

Their respective velocities after the collision are

v₁ = \((\frac{m_1-m_2}{m_1+m_2})u_1+(\frac{2m_2}{m_1+m_2})u_2\)

v₂ = \((\frac{2m_1}{m_1+m_2})u_1+(\frac{m_2-m_1}{m_1+m_2})u_2\)

Special cases :

 (a) If the Colliding bodies are identical (or equal masses), m₁ = m₂

Substituting m₁ = m₂ in the above equations,

v₁ = \((0)u_1+(\frac{2m_2}{2m_2})u_2\) = u₂

and v₂ = \((\frac{2m_1}{2m_1})u_1+(0)u_2\) = u₁

This shows that in a head-on elastic collision of two particles of equal masses, the particles exchange their velocities.

Further, if the second particle is at rest before the collision, u₂ = 0, so that after the collision, v₁ = 0 and v₂  = u₁  that is, the first particle initially moving stops and the second one takes off with the initial velocity of the first.

(b) If colliding body is much heavier and the struck body is initially at rest, i.e., m₁ >> m₂ and u₂= 0

Then,

v₁ ≅ \((\frac{m_1}{m_1})u_1\)  ≅ u₁

and v₂ ≅ \((\frac{2m_1}{m_1})u_1\) ≅ 2u₁

that is, if a massive particle makes an elastic head-on collision with a light one at rest, the massive one continues its motion at almost the same speed, and the light one takes off at nearly twice this speed.

(c) The body which is struck is much heavier than the colliding body and is initially at rest, i.e., m₁ << m₂ and u₂= 0

Then, ignoring m₁ in comparison to m₂, the final velocities are 

v₁ ≅ \((-\frac{m_1}{m_1})u_1\)  ≅ −u₁ and v₂ ≅ (0)u₁  ≅ 0

Thus, if a light particle makes an elastic collision with a massive one at rest, it rebounds with almost its initial speed; the massive one is almost unaffected.

Consider a perfectly inelastic, head on collision of two bodies of masses m₁ and m₂ with respective initial velocities u₁ and u₂. As the collision is perfectly inelastic, they move jointly after the collision, i.e., their final velocity is the same. Let it be v.

According to conservation of linear momentum,

m₁u₁ + m₂u₂ = (m₁ + m₂)v

∴ v = \(\frac{m_1u_1+m_2u_2}{m_1+m_2}\)

This is the common velocity after a perfectly inelastic collision

Loss in the kinetic energy :

Loss in K.E. = Δ (K.E.) = Total initial K.E. − Total final K.E.

= ½ (m₁+m₂)v² – (½ m₁u₁² + ½ m₂u₂²)

= ½ (m₁+m₂)\((\frac{m_1u_1+m_2u_2}{m_1+M-2})^2\) – (½ m₁u₁² + ½ m₂u₂²)

= \(\frac{1}{2}[\frac{(m_1u_1+m_2u_2)^2}{m_1+m_2}-(m_1u_1^2+m_2u_2^2)]\)

= \(\frac{1}{2}[\frac{m_1u_1^2+2m_1m_2u_1u_2+m_2^2u_2^2-m_1^2u_1^2-m_2^2u_2^2-m_1m_2u_1^2-m_1m_2u_2^2}{m_1+m_2}]\)

= \(\frac{1}{2}[-\frac{m_1m_2(u_1^2+u_2^2-2u_1u_2)}{m_1+m_2}]\) = \(-\frac{1}{2}(\frac{m_1m_2}{m_1+m_2})(u_1-u_2)^2\)

Since the bracketed terms are always positive, the minus sign in Eq. (2) shows that the final kinetic energy is less than the initial kinetic energy. That is, there is a loss in kinetic energy.

∴ Loss in KE = initial KE — final KE = \(\frac{1}{2}(\frac{m_1m_2}{m_1+m_2})(u_1-u_2)^2\)

The loss in KE of the bullet is partly utilised in doing work against the retarding force of the block, partly in deforming the wood, and the rest is converted into heat and vibrational energy. Total energy is conserved, and the principle of energy conservation is not violated.

(a) For a force \(vec{F}\) which is constant over the time interval Δt during which it acts, the impulse of the force is defined as the product \(\vec{F}\)Δt :

\(\vec{J}\) = \(\vec{F}\)Δt

Newton's second law of motion, which duplicates the first law with quantitative precision, states that the change in momentum of a body equals the impulse of the applied force and is made in the direction of that force.

Then,

\(\vec{J}=\vec{dp}=\vec{p_f}-\vec{p_i}=m(\vec{v}-\vec{u})\)

where m is the mass of the body \(\vec{p_i}\) and \(\vec{p_f}\) are the initial and final momenta of the body over the time interval of the impulse.

For a time-varying force \(\vec{F}(t)\), whose magnitude over the interval Δt is given by the curve in Fig. (a), the impulse of the force is given by the area under the curve :

\(\vec{J}=\int_{t_i}^{t_f}\vec{F}dt\)

For a time-varying force, the magnitude of the impulse can also be written as

J = F_av Δt

where F_av is the average force that would impart the same impulse as the variable force over the time interval, i.e., the area F_av Δt of the rectangle in Fig. (b) is equal to the area under the curve in Fig. (a).

(b) Force is an abstract concept while what we actually observe is the change in momentum. The concept of impulse becomes more useful when the force varies with time, as in a two-body collision.

(a)

• While opening a door fixed to a frame on hinges, turning a door requires the least force when it is applied as far away from the rotation axis of the hinges as possible. So, handles on doors are put as far away as possible from the hinges.
• Also, it is easier to turn a door if the force is applied perpendicular to the rotation axis.
• Thus, turning effect of a force or torque depends on the magnitude of the force, its direction and point of application.
• For a measure of the turning effect of a force lying in a plane perpendicular to the axis of rotation, we must consider the magnitude of the applied force and the distance of its line of action from the axis.

(b) Torque or moment of a force about a point : The moment of a force about a point or torque is a quantitative measure of the ability of a force to change the state of rotation of a body. The magnitude of the torque is the product of the magnitude of the force and the moment arm.

Consider a rigid body free to rotate about an axis as shown (Fig). Suppose a force \(\vec{F}\) acts on the body at point P. The position vector of the point of application of the force with respect to the axis is \(\vec{r}\)

The moment arm of the force is the perpendicular distance r_L from the axis to the line of action of the force. If θ is the smaller angle between \(\vec{r}\) and \(\vec{F}\), r_L = r sin θ.

The magnitude of the torque is τ = Fr_L = Fr sin θ  = rF sin θ

The rotation of a body about a single axis would be clockwise or anticlockwise. The torque is considered positive if it tends to rotate the body anticlockwise, and is negative if it tends to rotate the body clockwise.

In vector form,

\(\vec{τ}\) = \(\vec{r}×\vec{F}\)

The direction of \(\vec{τ}\) is thus given by the right-handed screw rule for a vector product.

(c) Distinguish between the torques of a force and a couple :

In many situations the word couple is used synonymous to moment of the couple or its torque, i.e., every time we may not say it as torque due to the couple, but say that a couple is acting. (***write any two points)

Let a couple of a pair of forces \(\vec{F_1}\) and \(\vec{F_2}\) in the plane of the page act on the rigid body as shown (Fig.). \(\vec{F_2}\) = \(-\vec{F_1}\)  Let F₁ = F₂ = F. The perpendicular distance r_L = AB is the arm of the couple.

(1) About the point O :

The torque due to \(\vec{F_1}\);

\(\vec{τ_1}\) = \(\vec{OA}×\vec{F_1}\)

∴ τ ₁ = F(OA)  (anticlockwise)

The torque due to \(\vec{F_2}\)  is

\(\vec{τ_2}\) = \(\vec{OB}×\vec{F_2}\)

∴ τ ₂ = −F(OB) (clockwise)

Since the torques obey the superposition principle of forces, the resultant torque on the body is the sum \(\vec{τ_1}\) of \(\vec{τ_2}\)  and

τ = τ ₁ + τ ₂ = F(OA) − F(OB)

= F(OA − OB) = F(AB) = Fr_L (anticlockwise)  …..(1)

(2) About the point C :

τ ₁ = F(CA) and τ ₂ = F(CB) (both anticlockwise)

τ = τ ₁ + τ ₂ = F(CA) + F(CB)

= F(AB) = Fr_L (anticlockwise)  ……(2)

From Eqs. (1) and (2), the torque due to a couple about any point in its plane is the same, i.e., it is independent of the axis of rotation.

Balancing or mechanical equilibrium :

The momentum of a system is constant in the absence of an external unbalanced force. This state is called mechanical equilibrium.

• A rigid body is said to be in equilibrium if there is no change in its state of motion.
• If there is no change in its velocity, the body is said to be in translational equilibrium.
• If there is no change in the rotational state of the body, it is said to be in rotational equilibrium.

Condition for translational equilibrium:

A rigid body is in translational equilibrium when the resultant force acting on it is zero. By Newton's first law of motion, a body remains in its natural state of constant velocity (including zero) if the net force on the body is zero.

For translational equilibrium, the forces acting on the body must be concurrent. Thus, if F₁, F₂, F₃,..  are concurrent forces acting on a rigid body the first condition of equilibrium is

\(\vec{F_1}+\vec{F_2}+\vec{F_3}\) = 0 or ∑ \(\vec{F_i}\) = 0

Condition for rotational equilibrium : A rigid body is in rotational equilibrium when the resultant torque acting on it is zero. If the forces acting on a rigid body are not concurrent, then their resultant may form a couple. Since the sum of the forces constituting a couple is zero, the first condition for equilibrium is satisfied, But the action of the couple is to change the state of rotation of the body. A body at rest will start rotating, or if it was initially rotating, it will speed up or slow down.

Thus, if \(\vec{F_1}\), \(\vec{F_2}\), \(\vec{F_3}\) ,..   are nonconcurrent forces acting on a rigid body and \(\vec{τ_1}\), \(\vec{τ_2}\), \(\vec{τ_3}\),  are the respective torques produced by them about any point, the second condition for equilibrium is

\(\vec{τ_1}\) + \(\vec{τ_2}\) + \(\vec{τ_3}\)…. = 0 or ∑ \(\vec{τ_i}\)  = 0

Rotating fan in mechanical equilibrium :

• When a ceiling/table fan is stationary, it satisfies the condition for translational equilibrium.
• When the blades of the fan rotate with constant angular velocity, their angular acceleration is zero. Hence, it also satisfies the condition for rotational equilibrium. Therefore, a fan rotating with constant angular velocity is in mechanical equilibrium.

In real life, we always come across finite objects (objects of measurable sizes). Concept of centre of mass helps us in considering these objects to be point objects at a particular location, thereby allowing us to apply Newton’s laws of motion.

The centre of gravity of a body does not coincide with its centre of mass when the body is in a nonuniform gravitational field or, equivalently, when the extent of the body is very large.

Given : m = 5 ton = 5000 kg, u = 36 km/h = 36000 m/3600 s = 10 m/s, v = 0,

s = 1 km = 1000 m, g = 10 m/s²

v² − u² = 2as  ∴ 0 − u² = 2as

∴ a = \(-\frac{u^2}{2s}\) = \(-\frac{10^2}{2×1000}\) = \(-\frac{100}{2000}\)

= \(-\frac{1}{20}\) m/s²

The frictional force exerted by the road on the truck = |ma|

\(\frac{|ma|}{|mg|}\) = \(\frac{|a|}{|g|}\) = \(\frac{1/20}{10}\) = \(\frac{1}{200}\)

This gives the required fraction. It has the same value for the truck and the car.

Let m₁ and m₂ be me masses of lighy object A and heavy objects B and v₁ and v₂ be their respective velocities.

Since both are imparted with the same linear momentum,

m₁ v₁ = m₂ v₂

The kinetic energy of the lghter object A

K.E._A = \(\frac{1}{2}\)m₁ v₁²

The kinetic energy of the heavier object B

K.E._B = \(\frac{1}{2}\)m₂ v₂²

…(‘.’ m₁ v₁ = m₂ v₂)

m_1.K.E._A = m_2.K.E._B

As m₁ < m₂ therefore K.E._A > K.E._B lighter object A has more kinetic energy.

Given : mg = 50 kg wt, mg − ma =45 kg wt as the reading recorded by the weighing machine is less than mg, g = 10 m/s²

∴ \(\frac{mg-ma}{mg}\) = \(\frac{45}{50}\)

∴ \(1-\frac{a}{g}=\frac{9}{10}\)

∴ \(\frac{a}{g}=1-\frac{9}{10}=\frac{1}{10}\)

∴ a = g/10 = 10/10 = 1 m/s²

The lift must be coming down with this acceleration.

Let m_B and m_L be the masses of the block and the hanging load, respectively. Let m_o be the mass of the pan and m₁ the mass added to the pan. Let f_s be the friction (self-adjusting resistive) force on the block.

Given : m_B = 35 kg, m_L = 20 kg, m_o = 2 kg, f= 10% of m_Bg = 0.1m_Bg

The total mass on the right hand side is

m = m_o + m₁

From the free-body diagrams for the hanging load and the pan, shown in Figs. 4.7 (a) and (b), for the system to be at rest,

T₁ − m_Lg = 0  T₁ = m_Lg  ……(1)

and T₂ — mg = 0  T₂ = mg    ……..(2)

If the total mass on the right hand side, m, is less than a certain minimum, the block will tend to slide to the left, so that f_s will be to the right, Fig. (c). Then,

T₁ − T₂ − f_S = m_Ba (a is to the left)

m_Lg – mg − f_S = m_Ba      …....(3)

If m is more than a certain maximum, the block will tend to slide to the right, so that f_s will be to the left, Fig. (d). Then,

T₂ − T₁ − f_s = m_Ba (a is to the left)

mg − m_Lg − f_s = m_Ba   …...(4)

For the block to be at rest, i.e., a = 0, Eqs. (3) and (4) respectively sets the minimum and maximum values of m.

From Eq. (3),

(m_L − m − 0.1m_B) g = 0

∴ m = m_L − 0.1m_B = [20 − 0.1(35)]kg

m_o + m₁ = 20 − 3.5 = 16.5 kg

m₁ = 16.5 kg − m_o = 16.5 − 2 = 14.5 kg

Hence, minimum 15 weights, each of 1 kg wt, can be added to the pan.

From Eq. (4),

(m − m_L − 0.1m_B) g = 0

m = mL + 0.1m_B = [20 + 0.1(35)]kg

m_o + m₁ = 20 + 3.5 = 23.5 kg

m₁ = 23.5 kg − m_o = 23.5 − 2 = 21.5 kg

Hence, maximum 21 weights, each of 1 kg wt, can be added to the pan.

The power P delivered by a force is the time rate of work done by the force :

P = dW/dt

Under the action of a constant force  if the displacement of the body in an infinitesimal time interval dt is , the infinitesimal work done by the force in this time dt is

dW = \(\vec{F}\vec{ds}\)

∴ P = \(\vec{F}\frac{\vec{ds}}{dt}\) = \(\vec{F}.\vec{v}\) = Fv cos θ

Since \(\vec{F}\) is constant, the acceleration \(\vec{a}=\frac{\vec{F}}{m}\) is also constant.

∴ \(\vec{u}=\vec{u}+\vec{a}t=\vec{a}t\)  ('.'u = 0)

∴ \(\vec{u}\) = \(\frac{\vec{F}}{m}t\) = \(\frac{F^2t}{m}\)

∴ P = \(\vec{F}.\frac{\vec{F}}{m}t\)   is the required expression

. ∴ P ∝ t

Given : V = 40000, L = 40 m³, ρ = 0.9 g/cm³ = 900 kg/m³, h = 10 m,

g = 10 m/s², t = 30 min = 1800 s

The power of the pump, P =  work done/tlme

∴ P = \(\frac{mgh}{t}\) = \(\frac{(Vρ)gh}{t}\)

= \(\frac{40×900×10×10}{1800}\)

= 20 x 100 = 2000 W = 2kW

For the mass at the top see below Fig.,

T₁ − (T₂ + mg) = ma (upward) .... (1)

For the subsequent masses,

T₂ − (T₃ + mg) = ma  ……(2)

T₃ − (T₄ + mg) = ma  …..(3)

T₄ − (T₅ + mg) = ma  …..(4)

T₅ − (T₆ + mg) = ma  ……(5)

Adding all these equations, we get,

T₁ − T₆ − 5mg = 5ma

T₆ = T₁ − 5 mg − 5ma

Now, T₁ − 10 mg = 10 ma

.'.T₁ = 10 mg + 10 ma

.'. T₆ = 10mg + 10ma − 5mg − 5ma

.'.T₆ = 5mg + 5ma = 5m(g+a)

Now, a = g/2

The tension in the 6th string,

T₆ = 5m(g + \(\frac{g}{2}\)) = 5m\((\frac{3g}{2})\)  = \(\frac{15}{2}\)mg

= 7.5 mg

[Note : If we take a = 0.2g, then  T₆ = 5mg + 5m(0.2g) =5mg + mg = 6mg which is the answer given in the textbook]

Data: r = r_A + r_B = 5 x 10⁶ ly

m_A = 9 x 10⁹ M_s, m_B = 4 x 10⁹ M_s

M_s = solar mass

For the net gravitational force on the Sun due to the two galaxies to be zero, the path of the Sun must be perpendicular to the line joining the two galaxies. Let r_A and r_B be the distances of the galaxies A and B from the Sun as it crosses the line joining the galaxies. At this point, the magnitudes of the gravitational force on the Sun due to the galaxies A and B are, respectively,

F_SA = \(\frac{Gm_AM_S}{r^2_A}\) and  F_SB = \(\frac{Gm_BM_S}{r^2_B}\)

Since, F_SA = F_SB

\(\frac{m_A}{r^2_A}=\frac{m_B}{r^2_B}\)

∴ \(\frac{r^2_A}{r^2_B}=\frac{m_A}{m_B}\) = \(\frac{9×10^9}{4×10^9}\) = \(\frac{9}{4}\)

∴ \(\frac{r_A}{r_B}=\frac{3}{2}\)

∴ r_A = \(\frac{3}{2}\)r_B

∴ r_A + r_B = 5 x 10⁶ ly

∴ \(\frac{3}{2}\)r_B + r_B = 5 x 10⁶ ly

∴ \(\frac{5}{2}\)r_B = 5 x 10⁶ ly

∴ r_B = 2 x 10⁶ ly and r_A = 3 x 10⁶ ly

The neutral point on the line joining the two galaxies, where the sun experiences no net force due to the galaxies, is 3 million light years from the 9 billion solar massive galaxy.

From the graph (Fig.), the work done by the linearly varying force

= area of the trapezium PRST

= \(\frac{1}{2}\)(PR+ST) x RS

= \(\frac{1}{2}\)(5N + 3N) x (5m − 3m)

= \(\frac{1}{2}\) x 8 x 2 = 8 J

Given : F = (6x² − 4x − 8)N, x_i = 1 m, x_f=2 m

The work done by the force,

W = \(\int_{x_i}^{x_f} Fdx = \int_{x_i}^{x_f}(6x^2-4x-8)dx\) 

= \([6(\frac{x^3}{3})-4(\frac{x^2}{2})-8x]-{x_i=1}^{x_i=2}\)

= \([2x^3 − 2x^2 −8x]_1^2\)

=(2 x 8 − 2 x 4 − 8 x 2) − (2 – 2 − 8)

=(16 – 8 −16) − (−8)

= −8 + 8 = 0 J

Given: m = 100 g = 0.1 kg, g = 10 m/s², h=5 m, conversion of 64% of PE into KE during every bounce, t = 250 ms = 0.25 s, A = 0.5 cm² = 0.5 x 10^-4 m²

(a) Just before the ball strikes the ground for the first time, \(\frac{1}{2}\)mv₁² = mgh

Just after the first bounce,  \(\frac{1}{2}\)mv₂² = 0.64 mgh

∴ \(\frac{v_2^2}{v_1^2}\) = 0.64

∴ \(\frac{v_2}{v_1}\) = 0.8

The coefficient of restitution

e = [velocity of separation]/[velocity of approach] = \(\frac{v_2}{v_1}\) = 0.8

(b) Initial PE = mgh = 0.1 x 10 x 5 = 5 J

After the first bounce,

KE = 0.64 x 5 J

After the second bounce,

KE = 0.64 x 0.64 x 5 J

After the third bounce,

KE = 0.64 x 0.64 x 0.64 x 5 J

= (0.64)³ x 5 J

If v is the corresponding speed of the ball,

\(\frac{1}{2}\)mv² = (0.64)³ x 5

mv² = (0.64)³ x 10

Now, m = 0.1 kg  ∴ 0.1 v²  = (0.64)³ x 10

∴ v²  = (0.64)³ x 100

∴ v  = (0.64)^3/2 x 10 = [(0.64)^1/2]³ x 10

= 10 x (0.8)³ = 10 x \(\frac{512}{1000}\)  = 5.12 m/s

(c) After the second bounce \(\frac{1}{2}\)mv₂² = 0.64 x 0.64 x 5 J

∴ \(\frac{1}{2}\) x 0.1 v² = (0.64)² x 5

∴ v² = (0.64)² x 10 x 10

v = 0.64 x 10 = 6.4 m/S

Impulse = change in momentum

= 0.1 x 6.4 kg.m/s − (− 0.1 x 5.12 kg.m/s)

(considering the direction of the velocity)

= 0.1 x (6.4 + 5.12) = 0.1 x 11.52

= 1.152 kg.m/s

= 1.152 N.S

(d) Impulse \(\vec{F}\)t = 1.152 N.s, t = 0.25 s

∴ Average force, \(\vec{F}\) = \(\frac{1.152}{0.25}\) = 4 x 1.152

∴  = 4.608 N

 (e) Average pressure = \(\frac{\vec{F}}{A}\) = \(\frac{4.608}{0.5×10^4}\)

= 2 x 4.608 x 10⁴

= 9.216 x 10⁴ N/m²

= 9.216 x 10⁴ Pa

Given : m = 0.5 kg, g = 10 m/s², free fall for 10 S, \(\frac{m_1}{m_2}\) = \(\frac{1}{2}\) , m₁ + m₂ = 0.5 kg, v of m₁ = 60 m/s

∴ m₁ = \(\frac{m_2}{2}\)

Also, m₁ + m₂= 0.5 kg

∴ \(\frac{m_2}{2}\) + m₂ = 0.5

∴ \(\frac{3}{2}\)m₂ = 0.5 = \(\frac{1}{2}\)  ∴ m₂ = \(\frac{1}{3}\) kg

∴ m₁ = \(\frac{m_2}{2}\) = \(\frac{1}{6}\)kg

Free fall of m : v = gt = 10 x 10 = 100 m/s

The corresponding KE of m = \(\frac{1}{2}\)mv²

= \(\frac{1}{2}\) x 0.5 x (100)² = 2500 J

KE of m₁ = \(\frac{1}{2}\)m₁v₁² = \(\frac{1}{2}×\frac{1}{6}×(60)^2\) = 300 J

By momentum conservation,

mv = m₁v₁ + m₂v₂

∴ 0.5 x 100 = \(\frac{1}{6}×60+\frac{1}{3}v_2\) = 10 + \(\frac{v_2}{3}\)

∴ \(\frac{v_2}{3}\) = 50 − 10 = 40

∴ v₂ = 120 m/s

KE of m₂ = \(\frac{1}{2}\)m₂v₂² = \(\frac{1}{2}×\frac{1}{3}×(120)^2\) = 2400 J

Total KE of the fragments = 300 + 2400

= 2700 J

KE of the ball just before the explosion = 2500 J

Hence, the KE supplied during the explosion = 2700 J — 2500 J = 200 J

Mass of marble 1 m₁ = m, u₁ = 12 cm/s,

Mass of marble 2 m₂ = 2m, u₂ = 6 cm/s,

After collision speed of m₂ = average speed of two = (12+6)/2 = 9 cm/s = v₂

By momentum conservation,

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

∴ mu₁ + 2mu₂ = mv₁ + 2mv₂   (..m₁ = m, m₂ = 2m)

∴ 12m + 12m = mv₁ + 18 m   (..u₁ = 12, u₂ = 6)

∴ v₁ = 24 − 18 = 6 cm/s

v₂ − v₁ = 9 − 6 = 3 cm/s

u₁ − u₂ = 12 − 6 = 6 cm/s

The coefficient of restitution,

e = \(\frac{v_2-v_1}{u_1-u_2}=\frac{3}{6}\) = 0.5

Given : W = 20 kg, W₁ = 40 kg

In Fig.  AB = 2 m, AP = PC = CQ = QB = 0.5m,

Suppose the person starts walking from pivot P towards Q. Also, suppose the plank begins to topple when the person is at point D, a distance x from P. At that instant, the upward reaction R₁ at P becomes zero.

Taking moments about pivot Q,

W x QC + W₁ x QD + (−R₂) x 0 = 0

∴ W₁ x QD = −W x QC

∴ QD = (−W x QC)/W₁ = −(20 x 0.5)/40 = − 0.25 m

|d| = 0.25 m

PD = x = PQ − |d| = 1 − 0.25 = 0.75 m

∴ The person can walk up to 0.75 m from pivot P, i.e., 1.25 m from end A.

Data : From Fig. Ladder AB = L = 2 m, W = 10 kg, OB = 1.2 m, W₁ = 20 kg,

f = 0.5R₂

Since Δ OAB is a right triangle, right-angled at O,

OA = \(\sqrt{AB^2-OB^2}\) = \(\sqrt{2^2-1.2^2}\)

= \(\sqrt{4-1.44}\) = = \(\sqrt{2.56}\) = 1.6 m

According to the first condition for equilibrium,

∑ F_y = 0 = R₂ − W − W₁ = 0    ……(1)

and ∑ F_x = 0 = R₁ − f     …..(2)

From Eq. (1),

R₂ = W + W₁ = 10  + 20 = 30 kg

From Eq. (2) and data,

R₁ = f = 0.5 R₂ = 0.5 (30) = 15 kg

The horizontal reaction at the wall,

R = 15 kg  = 15 kg x 10 m/s²  = 150 N

From the second condition for equilibrium, taking moments about B,

—R₁ x OA + W₁ x BQ + W x BP + (R₂) x 0 + (—f) x 0 = 0

∴ W₁ x BQ = R₁ x OA – W x BP

BP =  \(\frac{1}{2}\)BE = \(\frac{1}{2}\)(1.2 m) = 0.6 m

∴ BQ = \(\frac{(15×1.6)-(10×0.6)}{20}\) = \(\frac{24-6}{20}\) = 0.9 m

BM/BA = BQ/BE

∴ BM = 2(\(\frac{0.9}{1.2})\) = 1.5 m

∴ The monkey can climb up 1.5 m along the ladder.

Given : L₁ = 10 cm, L₂ = 20 cm, L₃ = 30 cm, L₄ = 40 cm.

L₂/L₁ = 20/10 = 2, ∴ (L₂/L₁)³ = 8

L₃/L₁ = 30/10 = 3, ∴ (L₃/L₁)³ = 27

L₄/L₁ = 40/10 = 4, ∴ (L₃/L₁)³ = 64

Mass = volume x density; volume = L³

Let m₁= m = mass of cube A

Mass of cube B, m₂ = 8m; mass of cube C, m₃ = 27m; mass of cube D, m₄ = 64m

∴ m₁ + m₂ + m₃ + m₄ = m + 8m + 27m + 64m = 100m

From the figure, x₁ = 5 cm, x₂ = 20 cm, x₃ = 45 cm, x₄ = 80 cm

∴ X_CM = \(\frac{m_1x_1+m_2x_2+m_3x_3+m_4x_4}{m_1+m_2+m_3+m_4}\)

= \(\frac{m×5+8m×20+27m×45+64m×80}{100m}=\frac{6500m}{100m}\)= 65 cm

Also y₁ = 5m,  y₂ = 10 cm, y₃ = 15 cm, y₄ = 20 cm

∴ Y_CM = \(\frac{\sum_im_iy_i}{\sum_im_i}\) = \(\frac{1×5+8×10+27×15+64×20}{100}=\frac{1770}{100}\) = 17.7 cm

Similarly z₁ = 5 cm,  z₂ = 10 cm, z₃ = 15 cm, z₄ = 20 cm

∴ Z_CM = 17.7 cm

Let M = mass of the sphere of radius R, ρ = density of the material of the sphere

ρ = mass/volume

∴ M = \((\frac{4}{3}πR^3)ρ\)

If we imagine the hole (spherical cavity) to be filled with the same material as that of the sphere, the mass of this (smaller) sphere

= \(\frac{4}{3}π(\frac{R}{2})^3ρ=\frac{M}{8}\)

M − M/8 = 7M/8

Treating the hole as a sphere with mass –M/8, the X coordinate of the centre of mass (C) of the remaining sphere will be

X_CM = \(\frac{\sum_im_ix_i}{\sum_im_i}\) = \(\frac{0-(M/8)(R/2)}{7(M/8)}\)

= \(-\frac{R}{16}×\frac{8}{7}=-\frac{R}{14}\)

(a) —(ii) : (b) -(i), (iii), (iv), (v), (vi) : (c) - (iii), (iv) : (d) - (iii), (iv).