1582 + Interview Questions in WORK IN GENERAL AWARENESS Page 4 InterviewSolution

151.	A block of mass `0.50 kg` is moving with a speed of `2.00 cm^(-1)` can a smooth surface . It strikes another mass of `1.00 kg ` and then they move togather as a single body . The energy less charing the collision isA. `0.16 J`B. `1.00 J`C. `0.67 J`D. `0.34 J`
Answer» Correct Answer - C (c ) Initial kinetic energy of the system `K.E, = (1)/(2) mu^(2) + (1)/(2) M(0)^(2) = (1)/(2) xx 0.5 xx 2 xx 2 + 0 = 1J` For collision , uppying conservation of linear momentum `m xx u = (m+ M) xx nu ` `:. 0.5 xx 2 = (0.5 +1) xx nu rArr nu = (2)/(3) m//s ` Final kinetic energy of the system is `K.E_(f) = (1)/(2) (m+M) nu^(2) = (1)/(2) (0.5 + 1) xx(2)/(3) xx (2)/(3) = (1)/(3) j` :. Energy loss during collision `= (1 - (1)/(3)) J = 0.67 J`

Discussion

152.	The potential energy funtions for the force between two along in a distance molecule is approximatily given by `U(x) = (a)/(x^(12)) - b)/(x^(6)) ` where `a` and `b` are constant and `x` is the distance between the aloms , if the discision energy of the molecale is `D = [U(x = oo) - U` atequlibrium ] , D isA. `(b^(2))/(2a)`B. `(b^(2))/(12a)`C. `(b^(2))/(4a)`D. `(b^(2))/(6a)`
Answer» Correct Answer - C (c ) At equibrium : `(dU(x))/(dx) = 0` `rArr (-12a)/(x^(11) = (-6b)/(x^(5) rArr x = ((2a)/(b)) ^(1/6)` `:. U_(at equilibrium) = (a)/(((2a)/(b))^(2) - (b)/((2a)/(b)) = b^(2))/(4a) ` and `U_(x= 0) = 0` :. D= 0 - (-(b^(2))/(4a)) = (b^(2))/(4a)) `

Discussion

153.	A tanker filled with water starts at rest and then rolls, without any energy loss to friction, down a valley. Initial height of the tanker is `h_1`. The tanker, after coming down, climbs on the other side of the valley up to a height `h_2`. Throughout the journey, water leaks from the bottom of the tanker. How does `h_2` compare with `h_1`?
Answer» Correct Answer - `h_(2)=h_(1)`

Discussion

154.	(i) The cause of increases in kinetic energy when a man starts running without his feet slipping on ground is asked to two students. Their answers are– Harshit: Cause of increase in kinetic energy is work done by friction force. Without friction the man cannot run. Akanksha: Cause of increase in kinetic energy is work done by internal (muscle) forces of the body. Who is right? (ii) An inextensible rope is hanging from a tree. A monkey, having mass m, climbs to a height h grabbing the rope tightly. The monkey starts from rest and ends up hanging motionlessly on the rope at height h. (a) How much work is done by gravity on the monkey? (b) How much work is done by the rope on the monkey? (c) Using work – energy theorem, explain the increase in mechanical energy of the monkey.
Answer» Correct Answer - (i) Akanksha is right. (ii) a -mgh (b) 0 (c) Internal (muscle) forces of the body perfom work

Discussion

155.	A particle can move along a straight line. It is at rest when a force (F) starts acting on it directed along the line. Work done by the force on the particle changes with time(t) according to the graph shown in the fig. Can you say that the force acting on the particle remains constant with time?
Answer» Correct Answer - No, the forace is decreasing with time.

Discussion

156.	A man of mass M jumps from rest, straight up, from a flat concrete surface. Centre of mass of the man rises a distance h at the highest point of the motion. Find the work done by the normal contact force (between the man’s feet and the concrete floor) on the man.
Answer» Correct Answer - zero

Discussion

157.	In an industrial gun, when the trigger is pulled a gas under pressure is released into the barrel behind a ball of mass m. The ball slides smoothly inside the barrel and the force exerted by the gas on the ball varies as `F=F_(0)(1-x/L)` Where L is length of the end of the barrel from the initial position of the ball and x is instantaneous displacement of the ball from its initial position. Neglect any other force on the ball apart from that applied by the gas. Calculate the speed (V) of the ball with which it comes out of the gun.
Answer» Correct Answer - `v=sqrt((F_(0)L)/m) -`

Discussion

158.	A spring and block are placed on a fixed smooth wedge as shown. Following conclusions can be drawn for the block. (i) magnitude of its momentum will be maximum when `F_("net")` on block is zero (ii) its kinetic energy will be maximum when `F_("net")` on block is zero (iii) kinetic energy of block is maximum when block just touches the spring (iv) net force on block is maximum when KE = 0A. (i), (iii), (iv)B. (ii), (iii), (iv)C. (i), (ii), (iii)D. (i), (ii), (iv)
Answer» Correct Answer - D

Discussion

159.	A block of mass m = 10 kg is released from the top of the smooth inclined surface of a wedge which is moving horizontally toward right at a constant velocity of u = 10 m/s. Inclination of the wedge is `theta=37^(@)`. Calculate the work done by the force applied by the wedge on the block in two seconds in a reference frame attached to - (a) the ground (b) the wedge. [Take `g=10 m//s^(2)`]
Answer» Correct Answer - (a) 960 J (b) zero

Discussion

160.	A bus of mass 1000 kg has an engine which produces a constant power of 50 kW. If the resistance to motion, assumed constant is 1000 N. The maximum speed at which the bus can travel on level road and the acceleration when it is travelling at 25 m/s, will respectively be -
Answer» Correct Answer - `[50 m//s,1 m//s^(2)]`

Discussion

161.	A solid body starts rotating about a stationary axis with an angular acceleration `alpha=alpha_(0) cos theta`, where `alpha_(0)` is a constant and `theta` is an angle of rotation from the initial position. Find the angular velocity of the body as a function of `theta`.
Answer» Correct Answer - `[sqrt(2alpha_(0)sin theta]`

Discussion

162.	A ball is released from the top of a tower. The ratio of work done by force of gravity in 1st second, 2nd second and 3rd second of the motion of ball isA. `1:2:3`B. `1:4:16`C. `1:3:5`D. `1:9:25`
Answer» Correct Answer - C

Discussion

163.	A particle of mass `2 kg` starts moving in a straight line with an initial velocity of `2 m//s` at a constant acceleration of `2 m//s^(2)`. Then rate of change of kinetic energy.A. Is four times the velocity at any momentB. Is two times the displacement at any momentC. Is four times the rate of change of velocity at any momentD. Is constant through out
Answer» Correct Answer - A

Discussion

164.	Underline the correct alternative:In an inelastic collision of two bodies, the quantities which do not change after the collision are the total kinetic energy/total linear momentum/total energy of the system of two bodies.
Answer» Total linear momentum The total linear momentum always remains conserved whether it is an elastic collision or an inelastic collision.

164.

Underline the correct alternative:In an inelastic collision of two bodies, the quantities which do not change after the collision are the total kinetic energy/total linear momentum/total energy of the system of two bodies.

Answer»

Total linear momentum

The total linear momentum always remains conserved whether it is an elastic collision or an inelastic collision.

Discussion

165.	State if each of the following statements is true or false. Give reasons for your answer. (a) In an elastic collision of two bodies, the momentum and energy of each body is ocnserved. (b) Total energy of a system is always conserved, no matter what internal and external forces on the body are present. (c) Work done in the motion of a body over a closed loop is zero for every force in nature. (d) In an inelastic collision, the final kinetic energy is always less than the initial kinetic energy of the system.
Answer» (a) False. The total momentum and total energy of the system are conserved and not of each body. (b) False. The external forces on the body may change the total energy of the body. (c) False. Work done in the motion of a body over a closed loop is zero only when the body is moving under the action of conservative forces (like gravitational or electrostatic forces ). It is not zero when the forces are non-conservative e.g. frictional forces etc. (d) True, because in an inelastic collision, some kinetic energy usually changes into tsome other form of energy.

Discussion

166.	STATEMENT-l : In an elastic collision between two bodies, the relative speed of the bodies after collision is equal to the relative speed before the collision. STATEMENT-2 : In an elastic collision, the linear momentum of the system is conserved.A. If both, Assertion and Reason are true and Reason is correct explantion of Assertion.B. If both, Assertion and Reason are true but Reason is not a correct explanation of the Assertion.C. If Assertion is true but the Reason is false.D. If both, Assertion and Reason are false.
Answer» Correct Answer - D Here, Assertion is false, because in elastic collision the relative velocit of separation `(` not relative speed of separation `)` is equal to the relative velocity of approach `(` not relative speed of approach `)` of two bodies during collision . Hence reason is though true but cannot explain the Assertion. Hence false.

Discussion

167.	A collision in which there is absolutely no loss of `K.E.` is called an elastic collision. In such a collision, the linear momentum , total energy and kinetic energy, all are conserved.The coefficient of restitution `//` resilience of perfectly elastic collisions is unity. Read the above passage and answer the following questions `:` (i) When two bodies of equal masses undergo perfectly elastic collision in one dimension, what happens to their velocities ? (ii) How is this fact applied in a nuclear reactor ?
Answer» (i) The velocities of two bodies of equal masses are just interchanged after they undergo perfectly elastic collision in one dimesion. (ii) Fast moving neutrons in a nuclear reactor are slowed down by makin them collide against the nuclei of a moderator. As nuclei of a moderator, i.e., hydrogen atoms (vix. protons ) have roughtly the same mass as that of neutron, therefore, on collision, their velocities are interchanged, i.e., neutrons slow doen and protons acquire the speed of neutrons. Often hydrogenic materials (e.g., heavy water , paraffin, ets,) are used as moderators in nuclear reactors.

Discussion

168.	If two bodies stick together after collision. Will the collision be elastic or inelastic?
Answer» Inelastic collision.

Discussion

169.	Underline the correct alterntaive: a) when a conservative force does positivie work on a body, the potential energy of the body increase/decreases/remains unaltered. work done by a body against friction always results in a loss of its kinetic /potential energy. c) The rate of change of total momentum of a many-particle system is proportional to the external force/ sum of the internal forces on the system. d) In an inelastic collision of two bodies, the quantities which do not change after the collision are the total kinetic energy/total linear momentum/total enregy of the system of two bodies.
Answer» a) potential energy of the body decreases, the body in this case goes closer to the center of the force. Kinetic energy, because frictioin does its work against motion. c) internal forces can not change the total or net momentum of a system. Hence the rate of change of total momentum of many paticle system is proportional to the external force on the system. d) In an elastic collision of two bodies, the quantities which do not change after the collision are the total kinetic energy/total linear momentum total energy of the system of two bodies.

Discussion

170.	State if each of the following statements is true or false. Give reasons for your answer. a) In an elastic collision of two bodies, the momentum and energy of each body is conserved. b)Total energy of a systm is always conserved,no matter what internal and external forces on the body are present. Work done in the motion of a body over a closed loop is zero for every force in nature. d) In an inelastic collision, the final kinetic energy is always less than the initial kinetic energy of the sytem.
Answer» a) False, the total momentum and total energy of the system are conserved. False, the external force on the system may increase or decrease the total energy of the system. c) False, the work done during the motion of a body over a closed loop is zero only when body is moving under the action of a conservation force (such as gravitational or electrostatic force). Friction is not a conservative force hence work done by force of friction ( or work done on the body against friction) is not zero over a closed loop. d) True, usually in an elastic collision the final kinetic energy is always less than the initial kinetic energy of the sytem.

Discussion

171.	Momentum of a particle is increased by `50%`. By how much percentage kinetic energy of particle will increase?
Answer» When momentum is increased by `50%` velocity increases by `50%`, i.e., velcoity becomes `3//2` times, K.E. Becomes `9//4` times, i.e., `(9)/(4)xx100=225%`. Hence, increase in K.E. `=225-100=125%`

Discussion

172.	If linear momentum if increased by `50%` then kinetic energy will be increased byA. `50%`B. `100%`C. `125%`D. `25%`
Answer» Correct Answer - C

Discussion

173.	A block of mass `10 kg` is moving in x-direction with a constant speed of `10 m//s`. it is subjected to a retarding force `F=-0.1 x J//m`. During its travel from `x =20m` to `x =30 m`. Its final kinetic energy will be.A. `450J`B. `275J`C. `250J`D. `475 J`
Answer» Correct Answer - D Here, `m=10kg, u=10 m//s, F=-0.1x J//m` `[`as force is retarding `]` `x_(1)=20cm, x_(2)=30cm` As `W=int_(x_(1))^(x_(2))Fdx=int_(20)^(30)-0.1x dx =-0.1[(x^(2))/(2)]_(20)^(30)` `=-(0.1)/(2)[(30)^(2)-(20)^(2)]=-(0.1)/(2)[900-400]` `=-(0.1)/(2)[500]=-25J` `K.E_(i)=(1)/(2)m u^(2)=(1)/(2)xx10xx(10)^(2)=500J` According to work-energy theorem, `W=K.E_(f)-K.E_(i)` or `-25=K.E_(f)-500` `K.E_(f)=475J`

Discussion

174.	The momentum of a body is p and its kinetic energy is E. Its momentum becomes 2p. Its kinetic energy will beA. `(E )/(2)`B. 3EC. 2ED. 4E
Answer» Correct Answer - D (d) `K=(p^(2))/(2m)" or "K prop p^(2)` When P is doubled, kinetic energy will become four times.

Discussion

175.	A block of mass (m) slides along the track with kinetic friction mu. (A) man pulls the block through a rope which makes an angle `theta` with the horizontal as shown in the figure. The block moves with constant speed v. Power dlivered by man is .A. `Tv`B. `Tv cos theta`C. `(T cos theta-mumg)v`D. zero
Answer» Correct Answer - B `P=F. v Fv cos theta=TV cos theta`

Discussion

176.	A block of mass 10 kg, moving in x-direction with a constant speed of `10ms^(-1)`, is subjected to a retarding force `F=0.1xxJ//m` during its travel from x=20 m to 30 m. Its final KE will beA. 475JB. 450JC. 275JD. 250J
Answer» Correct Answer - a (a) From work-energy theorem, ` "Work done= Change in KE"` `implies W=K_(f)-K_(i)` `implies K_(f)=W+K_(i)=int_(x_(1))^(x_(2)) Fxdx+(1)/(2)mv^(2)` `=int_(20)^(30)-0.1.x.dx+1/(2)xx10xx10^(2)` `=0.1[(x^(2))/(2)]_(30)^(20)+500` `=-0.05[30^(2)-20^(2)]+500` `=-0.05[900-400]+500` `implies K_(f)=-25+500=475J`

Discussion

177.	A golf player hits a golf ball of mass 100 g and imparts it a speed of 50 m/s. If golf stick remains in contact with the ball for `0.01` s, then the average force exerted by the golf stick on the ball.
Answer» Initially, ball was at rest, then ball gains a momentum `=(100)/(1000)xx50implies5N-s` The momentum is gained in `0.01` s. Hence, the average force `=(5)/(0.01)=500N`

Discussion

178.	The pointer reading versus load graph for a spring balance is as shown in the figure. The spring constant isA. `15 kgf//cm`B. `5 kg//cm`C. `0.1 kgf//cm`D. `10 kgf//cm`
Answer» Correct Answer - C (c ) F=kx or k=slope of F-x grapha (F along Y-axis) Here, F is along X-axis. So, `k=(1)/(10)=0.1" kgf"//cm`

Discussion

179.	Assertion : Graph between potential energy of a spring versus the extension `//` compression `(x)` of the string is a straight ling. Reason : This is because potential energy is directly proportional to `x`.A. If both, Assertion and Reason are true and Reason is correct explantion of Assertion.B. If both, Assertion and Reason are true but Reason is not a correct explanation of the Assertion.C. If Assertion is true but the Reason is false.D. If both, Assertion and Reason are false.
Answer» Correct Answer - D `P.E.` of a spring, `U=(1)/(2) Kx^(2)` `:. U prop x^(2)`. The graph between `U` and `x` is a parabola. Choice `(d)` is correct.

Discussion

180.	In a vertical spring mass system, a block of mass m is initially at rest when there is no extension. Now if the mass is released suddenly, then the maximum elongation in the spring isA. `(mg)/(k)`B. `(2mg)/(k)`C. `(mg)/(2k)`D. `(mg)/(4k)`
Answer» Correct Answer - B

Discussion

181.	A body is projected from ground obliquely. During downward motion, power delivered by gravity of itA. IncreasesB. decreasesC. Remains constantD. First decreases and then becomes constant
Answer» Correct Answer - A

Discussion

182.	If a body of mass m suspended by a spring comed to rest after a downward displecemtn `y_(0)`, claculate (a) force constant of the spring, (b) loss in gravitational potential energy, (c) gain in elastic potential energy of spring.
Answer» (a) In equilibrium ,`mg=Ky_(0)` `K=(mg)/(y_(0))` (b) As mass descends through a distance `y_(0)`, `:.` loss in gravitational P.E. `=mg y_(0)`. (c) As the spring is stretched by `y_(0)`, therefore, gain in elastice potenntial energy `P.E. =(1)/(2)Ky_(0)^(2)=(1)/(2)((mg)/(y_(0)))y_(0)^(2) =(1)/(2)mg y_(0)`

Discussion

183.	If the potential energy of a gas molecule is `U=(M)/(r^(6))-(N)/(r^(12)),M ` and `N` being positive constants, then the potential energy at equlibrium must beA. zeroB. `M^(2)//4N`C. `N^(2)//4M`D. `MN^(2)//4`
Answer» Correct Answer - B Here, `U=(M)/(r^(6))-(N)/(r^(12)),` `F=-(dU)/(dr)=(-d)/(dr)((M)/(r^(6))-(N)/(r^(12)))` `=((-6M)/(r^(7))-(12N)/(r^(13)))` In equilibrium position, `F=0` `:. (6M)/(r^(7))=(12N)/(r^(13))` or ` r^(6)=(2N)/(r^(13))` Hence, `U=(M)/((2N//M))-(N)/((2N//M)^(2))=(M^(2))/(4N)`

Discussion

184.	A particle is acted upon by a force of constant magnitude which is always perpendiculr to the velocity of the particle. The motion of the particle takes place in a plane. It follows thatA. it moves in a circular pathB. its velcoity is constantC. its acceleration is constantD. its kinetic energy is constant
Answer» Correct Answer - A::D When force of constant magnitude is perpendicular to the velcoity of the particle, motion is along a circular path. Direction of velocity `//` acc. goes on changing , but `KE=(1)/(2) m upsilon^(2)=` constant. Choices `(a)` and `(d)` are correct.

Discussion

185.	A particle is acted upon by a force of constant magnitude which is always perpendiculr to the velocity of the particle. The motion of the particle takes place in a plane. It follows thatA. its velocity is constantB. its acceleration is constantC. its kinetic energy is constantD. it move in a circular path
Answer» Correct Answer - C::D

Discussion

186.	The block of mass m is released when the spring was in its natrual length. Spring constant is k. Find the maximum elongation of the spring. A. ` (mg)/(k)`B. `(2mg)/(k)`C. `(mg)/(2k)`D. Cannot be calculated
Answer» Correct Answer - B

Discussion

187.	A buttle weighting 10 g and moving with a velocity `800 ms^(-1)` strikes a 10 kg block resting on a frictionless surface. The speed of the block after the perfectly inelastic collision is approximatelyA. `8 cm s^(-1)`B. `80 cm s^(-1)`C. `8 ms ^(-1)`D. `0.8 cm s^(-1)`
Answer» Correct Answer - B

Discussion

188.	A block of mass 4 kg is resting on a horizontal table and a force of 10 N is applied on it in the horizontal direction. The coefficient of kinetic friction between the block and the table is 0.1. The work done in `sqrt(20)` sA. By the applied force is 150 JB. By the frictional force is 60 JC. By the applied force and the net force are 150 J and 90 J respectivelyD. All are correct
Answer» Correct Answer - D

Discussion

189.	A block of mass 2 kg is resting on a smooth surface. At what angle a force of 10 N be acting on the block so that it will acquire a kinetic energy of 10 J after moving 2 mA. `30^@`B. `45^@`C. `60^@`D. `90^@`
Answer» Correct Answer - C `triangleK=K_2-K_1=10-0=10J` `triangleK=W=Fscostheta` `10=10xx2xxcosthetaimpliescostheta=(1)/(2)` `theta=60^@`

Discussion

190.	Is it possible exert a force which does work on a body without changing its kinetic energy? If so, give example.
Answer» Yes, when a spring is compressed or when a body is pulled with a constant velocity on a rough horizontal surface.

Discussion

191.	(a) Can kinetic energy of a system be changed without changing its momentum ? (b) Can momentum of a system be changed without changing its kinetic energy?
Answer» (a) Yes, For example, when a bomb explodes linear momentum is conserved, but KE changes. (b) Yes. For example, in case of uniform circular motion. KE remains unchanged, but linear momentum changes because of change in the direction of motion.

Discussion

192.	A uniform chain of mass m & length L is kept on a smooth horizontal table such that `(L)/(n) `portion of the chaing hangs from the table. The work dione required to slowly bringsthe chain completely on the table is
Answer» Let `lambda=M//L`= mass per unit length of the chain and y is the length of the chain hanging over the edge. So the mass of the chain of length y will be `lambday` and the force acting on it due to gravity will be `mgy`. The work done in pulling the `dy` length of the chain on the table. `dW=F(-dy)` [dy is negative as y is decreasing] As the chain is pulled slowly, F=Weight of the hanging chain `=lambdayg` i.e., `dW=(lambdayg)(-dy)` So the work done in pulling the hanging portion on the table, `W=-underset(L//n)overset0intlambdagydy=-lambdag[y^2/2]_(L//n)^0=(lambdagL^2)/(2n^2)=(MgL)/(2n^2)` `[as lambda=M//L]`

Discussion

193.	Assertion:The work done by a conservative force such as gravity depends on the initial and final positions only Reason: The work done by a force can not be calculated if the exact nature of the force is not known.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If assertion and reason are false.
Answer» The work done by the force can be calculated sometimes when it is compresed or streched.

Discussion

194.	Asseration : If work by done conservative forces is positive, kinetic energy will increase. Reason : Because potential energy will decrease.A. both conservative and none conservative forceB. conservative forces onlyC. Non-conservative forces onlyD. nether conservative nor none conservative force
Answer» Correct Answer - B In case of non- conservative force the work done is dissipated its heat sound ect i.e. does not increase the potential energy but in case of conservative force work done is responsible for increasing the potential energy

Discussion

195.	Which are the following is not a conservative force?A. Force of frictionB. Magnetic forceC. Gravitational forceD. Electrostatic force
Answer» Correct Answer - A Among the given force , force of friction is a non conservative force whereas all other force are conservative

Discussion

196.	Which one of the following statement does not hold good when two balls of masses `m_(1)` and `m_(2)` undergo elastic collisionA. When `m_(1) lt lt m_(2)` and `m_(2)` at rest, there will be maximum transfer of momentumB. When `m_(1) gt gt m_(2) and m_(2)` at rest, after collision the ball of mass `m_(2)` moves with four times the velocity of `m_(1)`C. When `m_(1)=m_(2)` and `m_(2)` at rest, there will be maximum transfer of K.E.D. When collision is oblique and `m_(2)` at rest with `m_(1)=m_(2)` after collision the balls move in opposite direction.
Answer» Correct Answer - B::D

Discussion

197.	A body is moved from rest along a straight line by a machine delivering constant power. The ratio of displacement and velocity `(s//v)` varies with time `t` asA. B. C. D.
Answer» Correct Answer - A

Discussion

198.	What are the various types of equilibrium?
Answer» Different types of equilibrium are − Stable : (i) When a particle is displaced slightly from a position, then a force acing on it brings it back to the initial position, it is said to be in stable equilibrium position. (ii) Potential energy is minimum. (iii) \(\frac{d^2U}{dx^2 }= positive\) (iv) E.g. A marble placed at the bottom of a hemispherical bowl. Unstable : (i) When a particle is displaced slightly from a position, then a force acting on it tries to displace the particle further away from the equilibrium position, it is said to be in unstable equilibrium position. (ii) Potential energy is maximum (iii) \(\frac{d^2U}{dx^2 }= negative\) (iv) e.g. a marble balanced on top of a hemispherical bowl. Neutral : (i) When a particle is slightly displaced from a position it does not experience any force acting on it and continues to be in equilibrium in the displaced position, it is said be in neutral equilibrium (ii) Potential is constant (iii) \(\frac{d^2U}{dx^2 }= 0\) (iv) e.g. A marble placed on a horizontal table.

198.

What are the various types of equilibrium?

Answer»

Different types of equilibrium are −

Stable :

(i) When a particle is displaced slightly from a position, then a force acing on it brings it back to the initial position, it is said to be in stable equilibrium position.

(ii) Potential energy is minimum.

(iii) \(\frac{d^2U}{dx^2 }= positive\)

(iv) E.g. A marble placed at the bottom of a hemispherical bowl.

Unstable :

(i) When a particle is displaced slightly from a position, then a force acting on it tries to displace the particle further away from the equilibrium position, it is said to be in unstable equilibrium position.

(ii) Potential energy is maximum

(iii) \(\frac{d^2U}{dx^2 }= negative\)

(iv) e.g. a marble balanced on top of a hemispherical bowl.

Neutral :

(i) When a particle is slightly displaced from a position it does not experience any force acting on it and continues to be in equilibrium in the displaced position, it is said be in neutral equilibrium

(ii) Potential is constant

(iii) \(\frac{d^2U}{dx^2 }= 0\)

(iv) e.g. A marble placed on a horizontal table.

Discussion

199.	Prove that coefficient of restitution/resilience of perfectly elastic collision in one dimensions unity.
Answer» Suppose two balls A and B of masses m₁ and m₂ are moving initially along the same straight line with velocities u₁ and u₂ respectively. When u₁ > u₂ relative velocity of approach before collision, = u₁ - u₂ Hence the two balls collide, let the collision be perfectly elastic. after collision, suppose v₁ is velocity of A and v₂ is velocity of B along the same straight lien, (a) when v₂ > v₁, the bodies separate after collision. Relative velocity of separation after collision = v₂ - v₁ Linear momentum of the two balls before collision. = m₁u₁ + m₂u₂ Linear momentum of the two balls after collision. = m₁v₁ + m₂v₂ As linear momentum is conserved in an elastic collision, therefore m₁u₁ + m₂u₂= m₁v₁ + m₂v₂ .............(i) or m₂(v₂ − u₂) = m₁(u₁ − v₁) ….........(ii) Total K.E. of the two balls before collision = \(\frac{1}{2}m_1u^2_1+\frac{1}{2}m_1u^2_2..........(iii)\) Total K.E. of the two balls after collision = \(\frac{1}{2}m_1v^2_1+\frac{1}{2}m_1v^2_2..........(iv)\) As K.E. is also conserved in an elastic collision, there from (iii) and (iv), \(\frac{1}{2}m_1v^2_1+\frac{1}{2}m_1v^2_2=\frac{1}{2}m_1u^2_1+\frac{1}{2}m_1u^2_2\) \(\frac{1}{2}m_2(v^2_2-u^2_2)=\frac{1}{2}m_1(u^2_1-v^2_1)\) \(m_2(v^2_2-u^2_2)=m_1(u^2_1-v^2_1)......(v)\) Dividing (v) by (ii), \(\frac {m_2(v^2_2-u^2_2)}{m_2(v_2-u_2)}=\frac{m_1(u^2_1-v^2_1)}{m_1(u_1-v_1)}\) \(\frac{(v_2+u_2)(v_2-u_2)}{(v_2-u_2)}=\frac{(u_1+v_1)(u_1-v_1)}{(u_1-v_1)}\) or v₂ + u₂ = u₁ + v₁ or v₂ − v₁ = u₁ − u₂ …(iv) Hence is one dimensional elastic collision relative velocity of separation after collision is equal to relative velocity of approach before collision. From \(\frac{v_2-v_1}{u_2-u_1}=1\) by definition, \(\frac{v_2-v_1}{u_2-u_1}=1\) = e =1

199.

Prove that coefficient of restitution/resilience of perfectly elastic collision in one dimensions unity.

Answer»

Suppose two balls A and B of masses m₁ and m₂ are moving initially along the same straight line with velocities u₁ and u₂ respectively.

When u₁ > u₂ relative velocity of approach before collision,

= u₁ - u₂

Hence the two balls collide, let the collision be perfectly elastic. after collision, suppose v₁ is velocity of A and v₂ is velocity of B along the same straight lien,

(a) when v₂ > v₁, the bodies separate after collision.

Relative velocity of separation after collision

= v₂ - v₁

Linear momentum of the two balls before collision.

= m₁u₁ + m₂u₂

Linear momentum of the two balls after collision.

= m₁v₁ + m₂v₂

As linear momentum is conserved in an elastic collision, therefore

m₁u₁ + m₂u₂= m₁v₁ + m₂v₂ .............(i)

or m₂(v₂ − u₂) = m₁(u₁ − v₁) ….........(ii)

Total K.E. of the two balls before collision

= \(\frac{1}{2}m_1u^2_1+\frac{1}{2}m_1u^2_2..........(iii)\)

Total K.E. of the two balls after collision

= \(\frac{1}{2}m_1v^2_1+\frac{1}{2}m_1v^2_2..........(iv)\)

As K.E. is also conserved in an elastic collision, there from (iii) and (iv),

\(\frac{1}{2}m_1v^2_1+\frac{1}{2}m_1v^2_2=\frac{1}{2}m_1u^2_1+\frac{1}{2}m_1u^2_2\)

\(\frac{1}{2}m_2(v^2_2-u^2_2)=\frac{1}{2}m_1(u^2_1-v^2_1)\)

\(m_2(v^2_2-u^2_2)=m_1(u^2_1-v^2_1)......(v)\)

Dividing (v) by (ii),

\(\frac {m_2(v^2_2-u^2_2)}{m_2(v_2-u_2)}=\frac{m_1(u^2_1-v^2_1)}{m_1(u_1-v_1)}\)

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

or v₂ + u₂ = u₁ + v₁

or v₂ − v₁ = u₁ − u₂ …(iv)

Hence is one dimensional elastic collision relative velocity of separation after collision is equal to relative velocity of approach before collision.

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

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

Discussion

200.	For inelastic collision co-efficient of restitution is (a) 0 (b) 1 (c) 0 < e < 1 (d) ∞
Answer» (c) 0 < e < 1

Discussion

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