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This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
Assertion : Any object thrown at an angle of 90^(@) with an initial velocity is called projectile Reason : When any object is thrown at an angle of 45^(@), it reaches the maximum (horizontal) distance ex : Javelin throw, shot put throw etc. |
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Answer» ASSERTION and are correctand REASONIS CORRECTEXPLANATION of Assertion |
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| 2. |
Why a small bubble rises slowly through a liquid whereas the bigger bubble rises rapidly? |
| Answer» Solution :SMALL BUBBLES of liquid do not rise RAPIDLY because of their terminal veocity is very low. WHEREAS the bigger bubble rises rapidly due to their very high terminal VELOCITY. So, a small bubble rises slowly through a liquid whereas the bigger bubble rises rapidly. | |
| 3. |
In a HCl molecule, we may treat Cl to be of infinite mass and H alone oscillating. If the oscillations of HCI molecule shows frequency 9 xx 10^(13)s^(-1) , deduce the force constant. Given Avogadro's number 6 xx 10^(26)per kg mole. |
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| 4. |
One end of a spring of spring constant 80 Nm^(-1) and unstretched length of 30 cm is fixed at point A and the other end of the spring is fitted with a smooth ring of mass 300g as shown in the figure. The rings is allowed to slide on a horizontal rod fixed at a height of 40 cm. Initially the spring makes an angle of 60^(@) with the vertical and the system of spring and ring is released from rest. The speedof the ring when the spring becomes verticcal is _______ ms^(-1). |
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Answer» 32 |
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| 5. |
Three identical bodies, each of mass m, are located at the vertices of an equilateral triangle with side a. At what speed must they move if they all revolve under the influence of one another.s gravitation in a circular orbit circumscribing the triangle while still preserving the equilateral triangle ? |
Answer» Solution : The force of ATTRACTION on body C due to bodies at A and B are `F_1=(Gm^2)/a^2` along CA and `F_2=(Gm^2)/a^2` along CB The resultant force on the body at C is `F=sqrt(F_1^2+F_2^2+2F_1F_2cos60^@)=sqrt3F_1=sqrt3(Gm^2)/a^2` `(because F_1=F_2=(Gm^2)/a^2)` The resultant force acting along CD, Here `OC=a/sqrt3` When each body is DESCRIBING a CIRCULAR orbit with centre of orbit at O, the force F provides the required CENTRIPETAL force. The radius of the circular orbit is OC = `a//sqrt3`, If V is the speed of the body in circular orbit , then Centripetal force = resultant gravitational force `(mV^2)/(a//sqrt3)=(sqrt3Gm^2)/a^2` or `V=sqrt((Gm)/a)` |
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| 6. |
If the blocks are connected as shown in fig, the relation between their velocities is |
Answer» SOLUTION :The length of the string is  `L=2sqrt(d^(2)+y^(2))+x` Since, L is constant, DIFFERENTIATING with respect to tiem t, we get `(DL)/(dt)=(2xx2y)/(2(d^(2)+y^(2))^((1)/(2)))((dx)/(dt))+(dx)/(dt)=0` `rArr 2v_(1)cos theta + v_(2)= 0, v_(2)=-2v_(1)cos theta` |
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| 7. |
The escape velocity for a body projected vertically upwards from the surface of earth is 11kms^(-1). If the same body is projected at an angle of 45^@ with the vertical, the escape velocity will be |
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Answer» `11sqrt2 KMS^(-1)` |
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| 8. |
The positions of two masses, x1 and x2. The position of center of mass is |
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Answer» `(m_(1)m_(2))/(m_(1)x_(1)+m_(2)x_(2))` |
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| 9. |
A body takes 8 minutes to cool from 90^(@) C to 80^(@)C in a surrounding of temperature 25^(@)C. The time taken by it to cool from 80^(@)C " to"70^(@)C in the same surroundings is |
| Answer» ANSWER :B | |
| 10. |
Ice starts freezing in a lake with water at 0^(0)C when the atmospheric temperature is -10^(0)C. If the time taken for 1 cm of ice to be formed is 12 minute, the time taken for the thickness of the ice to change from 1 cm to 2 cm will be |
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Answer» 12 MINUTE |
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| 11. |
A cylinder with a movable piston contains air under a pressure P_1 and a soap bubble of radius 'r'. The pressure P_2 to which the air should be compressed by slowly pushing the piston into the cylinder for the soap bubble to reduce its radius by half will be : x[P_1 +3S//r], where S is the surface tension |
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Answer» |
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| 12. |
Object fallen down on Earth from 10 m height lost 20% energy, then it will gain …... height. |
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Answer» 20% is LOST MEANS 80% is GAINED `:. U_2=80%" of "U_1=U_1xx80/100=0.8U_1` From LAW of conservation of energy, `U_1=0.8U_1` `:. mgh_2=0.8xxmgh_1` `:. h_2=0.8h_1=0.8xx10=8m` |
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| 13. |
A body of mass 0.01 kg executes S.H.M about x = 0 under the influence of a force shown below. The period of S.H.M is |
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Answer» 1.05 sec |
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| 14. |
A book with many printing errors contains four different expressions for the displacement 'y' of a particle executing simple harmonic motion. The wrong formula on dimensional basis (i) y=A sin (2pi t//T)(ii) y=A sin (Vt)(iii) y=A//T sin (t//A)(iv) y=(A)/(sqrt(2))(sin omega t+cos omega t) |
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Answer» II only |
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| 15. |
A gas may expand either adiabatically orisothermally. A number of p-V curves are drawn for the two process over different ranges of pressure and volume. It will be found that a) Two adiabatic curves do not intersect b) Two isothermal curves do not intersect c) An adiabatic curve and an isothermal curvemay intersect d) The magnitude of the slope of an adiabaticcurve is greater than the magnitude of the slope of an isothermal curve for the samevalues of pressure and volume |
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Answer» only a,B are CORRECT |
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| 16. |
In a test experiment on a model aeroplane in a wind tunnel, the flow speeds on the upper and lower surfaces of the wing are 70 ms^(-1) and 63 ms^(-1) respectively. What is the lift on the wing if its area is 2.5 m^(2)? Take the density of air to be 1.3 kg m^(-3) |
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Answer» FORCE on WING `= (P_2-P_1)A = 0.65 xx 931 xx 2.5 = 1512.9 N` |
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| 17. |
Mass of a ring is non-uniformly distributed around its geometric centre. If R is radius of the ring, then (a) Centre of mass does not coincide with geometric centre b) Position of centre of mass from the geometric centre will be x(0 lt x lt R) (c) Centre of mass will be nearer to the greater mass distribution (d) Centre of mass may lie out side the periphery |
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Answer» only a and B are CORRECT |
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| 18. |
Four particles, each of mass 1kg, are placed at the corneres of a square OABC of side 1m. .O. is at the origin of the coordinate system. OA and OC are aligned along positive x-axis and positive y-axis respectively. The position vector of the centre of mass is (in .m.) |
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Answer» `HAT(i)+hat(j)` |
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| 19. |
In above question 60 the velocity with which A colliides with C is : |
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Answer» `6 m//sec` |
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| 20. |
A balloon of mass M is descending at a constant acceleration alpha. When a mass m is released from the balloon it starts rising with the same acceleration alpha . Assuming that its volume does not change, what is the value of m? |
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Answer» `(ALPHA)/(alpha+g)M` While ascending`F_(B)-(M-m)g=(M-m)alpha` where `.F_(B).` is the bouyant force |
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| 21. |
Draw graphs showing the variation of acceleration due to gravity with (i) height above the earth's surface. |
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Answer» Solution :The value of G varies with depth d as `g = g(1 - d/R) i.e g ALPHA (R - d)` Thus the graph of g VERSUS depth d is the STRAIGHT line AC. |
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| 22. |
A sample of gas is at 0^@C. To what temperature it must be raised in order to double the r.m.s speed of the molecule |
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Answer» `270^0C` |
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| 23. |
A black body does not |
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Answer» EMIT RADIATION |
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| 24. |
A block of mass 100g is moved with a speed of 5.0 m/s at the highest point in a closed circular tube of radius 10 cm kept in a vertical plane. The cross sectiion of th tube is such that the block just fits in it. The block makes wseverl oscillations inside the tube and finally stops at the lowest point. find teh work done by the tube on the block during the process |
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Answer» `v=5m/sec, r=10 cm` work DONE by the BLOCK =total energy at A-total energy AST B `=(1/2mv^2+mgh-0)` `rarr 2=1/2xx(0.1)xx25+(0.1)xx10xx(0.2)` `[h=2r=0.2m]` `=1.25+0.2=1.45J` so, the work done by the TUBE on the BODY is `W_1=1.45J` |
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| 25. |
A ship in a sea sends SONAR waves straight down into the seawater from the bottom of the ship. The signal reflects from the deep bottom bed rock and returns to the ship after 3.5 s. After the ship moves to 100 Km it sends another signal which returns back after 2s. Calculate the depth of the sea in each case and also compute the difference in height between two cases. |
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Answer» Solution :Speed of SONAR waves in water c = `1500 ms^(-1)` Time taken to reflect from the bottom of the sea, 2t= 3.5 sec `t = 1.75` sec Distance covered in FORWARD and reflected backward `(d_(1)) = c XX t` `d_(2) = 1500 xx 1.75 = 2625` m After ship moves in a distance = 150 km Time taken to reflect by the waves 2t = 2s t = 1 s Distance covered by the waves `(d_2) = c xx t = 1500 xx 1 = 1500` m The different between the HEIGHT of two cases `= 2625 - 1500` `h_("difference") = 1124` m |
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| 26. |
(A) : The density of water is maximum at 4^(@)C. (R ) : The volume of water decreases while heating from 0^(@)C to 4^(@)C. |
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Answer» Both (A) and (R ) are true and (R ) is the CORRECT explanation of (A) |
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| 27. |
A rod of negligible heat capacity has length 20 cm, area of cross--section 1.0" cm"^(2) and thermal conductivity 200" W/m"-""^(0)C. The temperature of one end is maintained at 0^(0)C and that of the other end is slowly and linearly varied from 0^(0)C to 60^(0)C in 10 minutes. Assuming noloss of heat through the sides. The total heat transmitted through the rod in these 10 minutes is x xx600. Find x |
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| 28. |
A mass m kg is whirled in a vertical plane by tying it at the end of a flexible wire of length l and area of cross section 'A'. When the mass is at its lowest position the strain produced in the wire is, if Youngs modulus of the wire is 'Y' (if V= sqrt( 5 gl ) ) |
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Answer» `(AY)/(6 MG) ` |
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| 29. |
The pulley arrangements of figures (a) and (b) are identical . The mass of the rope is negligible . In figure (a), the mass m is lifted up by attaching a mass 2m to the other end of the rope . In figure (b) , m is lifted up by pulling the other end of the rope with a constant downward forceF=2mg . Calculate the accelerations in the two cases. |
Answer» SOLUTION :In figure (a), for the MOTION of mass m , T-mg=ma For motion of mass 2M figure (B) 2 mg -T (2m)a…(2) Adding equation (1) and (2) , we get `2mg-mg= 2ma+ma, mg=3ma, a = g//3` (b)In figure `T^(1)-mg=ma^(1)` But `T^(1)=2mg , :. 2mg-mg =a^(1), ` Therefore `a^(1)=g` |
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| 30. |
Show that the angular momentum of a particle is equal to twice the product of mass and areal velocity. |
| Answer» Solution :As the person is moving away from the edge of a CIRCULAR PLATFORM towards the centre, the moment of INTERTIA will decrease. According to law of conservation MOMENTUM. `IOMEGA`=constant, angular speed will increase. | |
| 31. |
A block is pushed up a rough inclined plane of 45^(@). If the time of descent is twice the time of ascent, the coefficient of friction is |
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Answer» `0.6` |
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| 32. |
Which of the following physical quantity is dimensionless? |
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Answer» STRAIN |
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| 33. |
A car travel starting form rest with constant acceleration alpha and reaches a maximum velocity V. It travels with maximum velocity for some time and retards uniformly at the rate of beta and comes to rest. If s is the total distance and t is the total time of journey then t= |
Answer» Solution : `V=0+ALPHA t, , t_(1)=v/alpha` `S_(1)=1/2 alpha t_(1)^(2)=v^(2)/(2alpha)` `O-V=beta t_(3), t_(3)=v/beta` `O^(2)-V^(2)=-2]`  `S_(3)=v_(2)/(2beta)=1/2 betat_(3)^(2)` `t_(2)=(S_(2))/V=(S-(S_(1)+S_(3)))/(V)=(S-V^(2)/2 (1/alpha+1/beta))/(V)` `t=t_(1)+t_(2)+t_(3)` `=v/alpha+s/v-v/(2alpha)-v/(2beta)+v/(beta), t=s/v+v/2 ((alpha+beta)/(alpha beta))` `s_(2)=s-(s_(1)+s_(2))=s-((v^(2))/(2alpha)+v^(2)/(2alpha))` `=s-v^(2)/(alpha)=s-(vt-s), s_(2)=2s-vt` `[S_(2)`= distance TRAVELLED with constant VELOCITY] |
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| 34. |
A simple pendulam completes 20 oscillations in 25 S . The time is measured with a stop watch of least count 0.1 S . The error in the measurement is |
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Answer» 0.4 |
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| 35. |
You may have seen in a circus a motorcyclist driving in a vertical loop inside a 'death well' chamber with holes, so the spectators can watch from outside. Explain clearly why the motorcyclist does not drop down when he is at the uppermost point, with no support from below. What is the minimum speed required to perform a vertical loop if the radius of the chamber is 25 m ? |
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| 36. |
Twoforces10 Nand 15Naer actingon actingon abody. Whatcanbe the maximumandminimumresultantforceonthebody ? |
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Answer» SOLUTION :Here : `A=10N,B=15N,R=sqrt(A^(2)+B^(2)+2ABcos0)` (a) R will be maximum if COS 0=1 Then `R=A+B = UNDERSET(to)(10)+15 underset (to)(=)25N` in the direction of A and B (b) R will be minimum if cos `0=-1` then `R=15-10=5underset(to)(N)` in the direction of B Vector addition OBEYS (1) commutative law `VECA +vecB=vecB+vecA` (2) distributive law m `(vecA+vecB)=mvecA+mvecB` (3) Associative law `vecA+(vecB+vecC)=(vecA+vecB)+vecC` |
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| 37. |
At which of the following temperature would the molecules of a gas have twice the average kinetic energy they have at 20^(@)C |
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Answer» `40^@C` |
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| 38. |
A guard on duty observes a fall of 25% in the pitch of the silencer of a motor bike as it crosses him. What will be the speed of car if velocity of sound is 340 m/s ? |
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Answer» Solution :Before crossing, the source of sound is MOVING towards the listener (guard) `(V_(L) = 0)` `therefore""v' = (V)/(V - V_(s)) v""...(i)` After crossing, the source of sound is moving away from the listener. `therefore""v'' = (V)/(V + V_(s))v""...(II)` As `(v'')/(v) = (75)/(100)` Dividing (i) by (ii) `(v'')/(v) = (V)/(V + V_(s)) v xx (V - V_(s))/(Vv)` `(75)/(100) = (V - V_(s))/(V + V_(s))` `(75)/(100) = (340 - V_(s))/(340 + V_(s))` `75 (340 + V_(s)) = 100 (340 - V_(s))` `75 xx 340 + 75 V_(s) = 100 xx 340 - 100 V_(s)` `75 xx 340 + 175 V_(s) = 100 xx 340` `{:(175 V_(s) = 100 xx 340 - 75 xx 340),(""ul(100 xx 340 - 75 xx 340)),(""175):}` `= (34000 - 25500)/(175) - 48.57` m/s |
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| 39. |
A light rod of length 2 m is suspended from the ceiling horizontally by means of two vertical wires of equal length tied to its ends. One of the wires is made of steel and is of cross-section 10^(-3) m^2and the other is of brass of cross-section 2 xx 10^(-3) m^2 .Find out the position along the rod at which a weight may be hung to produce, i) equal stress in both wires ii) equal strains in both wiresYoung's modulus of brass = 1 xx 10^11 N//m^2Young's modulus of steel = 2 xx 10^11 N//m^2 |
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Answer» Solution :Suppose `a_1` and `a_2`are the cross-sectional areas, and `Y_1` and `Y_2`are the Young.s moduli of steel and brass wire respectively. Let `T_1` and `T_2`are tensions in the steel and brass WIRES respectively. Let x is distance of the position of the hanging weight from the steel wire. i) FIRST case : For equal stress in both wires, we have `(T_1)/(a_1) = (T_2)/(a_2)` `(T_1)/(10^(-3)) = (T_2)/(2 xx 10^(-3)) " or " T_2 = 2T_1 `....(i) As the whole system is in equilibrium, so`sum vec tau = 0.`TAKING moment of all the forces acting on the rod about C, we have `T_1 x = T_2 (2-x) =0`....(II) Solving eqn (i) and (ii) we get`x = 4/3 m` ii) second case : For equal strain in both the wires `e_1 = e_2` `(T_1l)/(a_1Y_1) = (T_2l)/(a_2Y_2)` `(T_1)/(10^(-3) xx 2 xx 10^11)= (T_2)/(2 xx 10^(-3) xx 10^11)` `T_1 = T_2`...(iii) From equation (ii) and (iii) , we get x = 1m |
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| 40. |
A body describes simple harmonic motion with an amplitude of 5 cm and a period of 0.2 s. Find the acceleration and velocity of the body when the displacement is (a) 5 cm (b) 3 cm (c) 0 cm. |
| Answer» SOLUTION :`(a)-5pi^(2)MS^(-2),0(b)-3pi^(2)ms^(-2),0.4pims^(-1),(c) 0:0.5pims(-1)` | |
| 41. |
Figure shows a block A on a smooth surface attached with a spring of force constant k to the ceiling. In this state spring is in its natural length l . The block A is connected with a massless and frictionless string to another identical mass B hanging over a light and smooth pulley. Find the distance moved by A before if leaves contact with the ground. |
Answer» Solution : Due to weight of block B, it moves down and pulls the block A now as block A and B move, spring gets stretched and BECOMES inclined as its lower end is attached to the block A. It will break off from the GROUND below it when the vertical component of the spring FORCE on block A will balance its weight mg Let it happens when A moves by a distances as shown in figure. At this instant let the spring be inclined at an angle with the vertical If the STRETCH in the spring at this instant is x, then it is given as `x = l sec theta - l ` or ` x = l ( sec theta - 1)` ........... (1) If mass A breaks off from ground below it , we have ` kx cos theta = mg ` From equation (1) and (2) substituting the value ofx, we get `kl( sec theta - 1) cos theta = mg or kl(1- cos theta ) = mg or cos theta =1 -(mg)/(kl) or tan theta =([k^(2)l^(2) (kl - mg)^(2) ]^(1//2))/(kl-mg)` At this instant the distance TRAVELLED by mass A and B is given by ` s=l tan theta (or) s=l([k^(2)l^(2)(kl-mg)^(2)]^(1//2))/(kl-mg)` |
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| 42. |
(A) : For a perfectly incompressible substance, bulkmodulus is infinity (R) : Bulkmodulus of elasticity is inversely proportional to change in volume. |
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Answer» Both .A. and .R. are TRUE and .R. is the correct EXPLANATION of .A. |
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| 43. |
When a source of sound is in motion towards a stationary observer, the effect observed is ………… |
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Answer» Solution :WHENTHE source and theobserver arein relativemotionwithrespectto eachotherandto themediumin whichsoundpropagates, thefrequency of thesoundwaveobservedisdifferentfromthe FREQUENCY of thesource. Thisphenmenonis calledDopplerEffect (i )SOurcemovestowards the observer: supposea sourceS movesto theright( as shownin figure ) withavelocity`V_s` and letthe frequencyof thesoundwavesproducedby thesourcebe ` F_s `. Weassumethe VELOCITYOF SOUND in mediumis v.the compression( soundin thefigure. whenS is at`C_2` and similarlyfor `X_3` the compressionis at`C_1` . whenS isatposition`X_2` , thecompressionis at ` C_2`andsimilary for ` x_3` and `C_3`Assumethat if ` C_1`reachesthe pointC asshownin the figure. it is obviousto seethat thedistancebetweencompressions`C_2`and ` C_3`is shorterthan distancebetween`C_1and C_2`.thismeansthe wavelengthdecreases whenthe sourceS movestowardstheobserver O ( sincesoundtravels ) .ButFrequency isinverselyrelatedto wavelengthand therefore , frequency increases .  Let `lamda ` bethewavelengthof the sourceS asmeasured by theobderverwhen Ssi atposition`X_1 and lamda ` . bewavelengthof thesourceobserved by theobserver whenS movestopositionsourceto travelbetween`X_1and x_2`therefore, `lamda. = lamda -v_s t ` But`t=(lamda )/(v ) ` on substitutingequation(2 )in equation(3), we get `lamda.= lamda(1- (v_s) /(v) )` sincefrequencyis inversely proportionalto wavelength, wehave ` f. =(v_s )/(lamda. )andf=(v_s )/(lamda )` hence ` f. =(f )/( (1-(V_s )/(V)))` since`(v_s )/( v )lt lt1,`Weusethe binomialexpansion andretatiningonlyfirstorderin ` ( V_s)/( v ) `, weget ` f.= f (1+(v_s)/(v))v` (ii ) Sourcemovesawayfrom theobserver : sincethe velocityhereof thesourceis oppositein direction whencomparedto case(a) , therefore, changingthe signof thevelocityof thesourcein theabovecasei.e.,bysubstituting`( v_s to - V_s )` in equation(1) , we get ` f. =(f) /((1+(v_s)/(v)))` Usingbinomialexpansion again , we get , ` f. = f ( 1-(v_s)/(v))` |
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| 44. |
For a pure rotational motion , |
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Answer» `omega `is samefor all PARTICLES of BODY |
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| 45. |
Which one is not a dimensional number? |
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Answer» Acceleration DUE to gravity `R_(E)=(rhovD)/(n)` |
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| 46. |
A uniform chain of mass m and length l is on a smooth horizontal table with (1/n)^th part of its leght is hanging from one end of the table. The velocity of the chain, when it completely slips off the table is |
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Answer» Solution :With respect to the TOP of the table, the initial potential energy of the chain, `U_(1)`= PE of the chain lying on the table + PE of the hanging PART of the chain `=L(1-(1)/(n))mg(0)+(m)/(n)g""((-L)/(2N))=-(MGL)/(2n^(2))` P.E of the chain, when it just slips off the table, `U_(2)= mg((-L)/(2))=-(mgL)/(2)` From jaw of conservation of energy `Delta K = - Delta U K_(f) - K_(i) = - (U_(f) - U _(i))` `thereforeK_(i)=0,K_(f)=-[-(mgL)/(2)-(-(mgL)/(2n^(2)))]` `K_(f)= (mgL)/(2)[1-(1)/(n^(2))]` It .V. is the velocity of the chain, then, `(1)/(2)mv^(2)=(mgL)/(2)[1-(1)/(n^(2))]` `thereforev=sqrt(gL[1-(1)/(n^(2))])` |
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| 47. |
A ball of mass 0.2kg strikes an obstacle and moves at 60^(@) to its original direction. If its speed also changes from 20m//s" to "10m//s, the magnitude of the impulse received by the ball is |
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Answer» `2sqrt7Ns` `J=DeltaP=sqrt(Deltap_(x)^(2)+Deltap_(y)^(2))` |
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| 48. |
The equatorialand polarregions of the earthreceive unequalsolar heat. The convection current arisingdue to thisis called |
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Answer» LAND breeze |
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| 49. |
A LASER is a source of very intense, monochromatic, and unidirectional beam of light.These properties of a laser light can be exploited to measure long distance. The distance of the Moon from the Earth has been already determined very precisely using a laser as a source of light . A laser light beamed at the Moon takes 2.56 s to return after reflection at the Moon's surface. How much is the radius of the lunar orbit around the Earth ? |
| Answer» SOLUTION :`3.84xx10^(8) m` | |
| 50. |
Consider a one dimensional elastic collision between a given incoming body A and body B, initially at rest. How would you choose the mass of B in comparison to the mass of A in order that B should recoil with (a) greatest speed (b) greatest momentum? |
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Answer» Solution :As in a collision, momentum is always conserved `m_(A) u= m_(A) v_(A) + m_(B)v_(B)` `u= v_(A) + kv_(B)` (with `K= m_(B)//m_(A)`) `v_(A)= u - kv_(B)` …(1) Now as collision is ELASTIC `(1)/(2) m_(A) u^(2) = (1)/(2) m_(A) v_(A)^(2) + (1)/(2) m_(B) v_(B)^(2)` `u^(2) = v_(A)^(2) + kv_(B)^(2)` ....(2) Substituting the value of `v_(A)` from Eqn. (1) and (2) `u^(2)= (u-kv_(B))^(2)+ kv_(B)^(2)` Which on solving gives `v_(B)= (2u)/(1+k)` ...(3) (a) So for `v_(B)` to be maximum, k must be minimum, i.e., `k= (m_(B))/(m_(A))rarr 0 or m_(B) LT lt m_(A)` (b) Now `P_(B)= m_(B) v_(B)` In perfect elastic collisions, when MASSES are equal, the total momentum of A is completely transfered to B. i.e., `m_(A)= m_(B)`. |
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