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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 51. |
The figure shown a particle sliding on a frictionless track, which teminates in a straight horizontal section. If the particle starts slipping from the point `A`, how far away from the track will the particle hit the ground? |
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Answer» Applying the law of conservation of mechanical energy for the points `A` and `B`, `mgH = (1)/(2) mv^(2) + mgh` `g -(v^(2))/(2) = (g)/(2)` or `v^(2) = g rArr v = sqrt(g) = 3.1 ms^(-1)` After point `B` the particle exhibits projectile motion with `theta = 0^(0)` and `y = - 0.5 m` Horizontal distance travelled by the body `R = u sqrt((2h)/(g)) = 3.1 xx sqrt((2 xx 0.5)/(9.8)) = 1m`. |
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| 52. |
A body starts from rest and moves with uniform acceleration. What is the ratio of kinetic energies at the end of `1st , 2nd` and `3rd` seconds of its journey ?A. `1 : 8 : 27`B. `1 : 2: 3`C. `1 : 4: 9`D. `3 : 2 : 1` |
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Answer» Correct Answer - C `KE = (1)/(2) mv^(2) = (1)/(2) mg^(2)t^(2) (because v = g t)`. |
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| 53. |
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 charge of velocity at any momentD. is constant throughout |
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Answer» Correct Answer - A `K = (1)/(2) m v^(2)` `:. (dK)/(dt) =mv(dv)/(dt) =(m(dv)/(dt)) v = (ma) v`. |
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| 54. |
A bucket full of water is drawn up by a person. In this case the work done by gravitational force isA. negative because the force and displacement are in opposite directionsB. positive because the force and displacement are in the same directionC. negative because the force and displacement are in the same directionD. positive because the force and displacement are in opposite direction |
| Answer» Correct Answer - A | |
| 55. |
A body of mass `4kg` moving with a speed of `3 ms^(-1)` collides head on with a body of mass `3 kg` moving in the opposite direction at a speed of `2 ms^(-1)`. The first body stops after the collision. The final velocity of the second body isA. `3 ms^(-1)`B. `5 ms^(-1)`C. `-9 ms^(-1)`D. `30 ms^(-1)` |
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Answer» Correct Answer - A `m_(1)u_(1) - m_(2)u_(2) = m_(1)v_(1) + m_(1)v_(2)`. |
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| 56. |
A `6 kg` mass travelling at `2.5 ms^(-1)` collides head on with a stationary `4 kg` mass. After the collision the `6kg` mass travels in its original direction with a speed of `1 ms^(-1)`. The final velocity of `4 kg` mass isA. `1 ms^(-1)`B. `2.25 ms^(-1)`C. `2 ms^(-1)`D. `0 ms^(-1)` |
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Answer» Correct Answer - B According to law of conservation of linear momentum `m_(1)u_(1) + m_(2)u_(2) = m_(1)v_(1) +m_(2)v_(2)`. |
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| 57. |
A moving sphere `P` collides another sphere `Q` at rest. If the collision takes place along the line joining their centers of mass such that their total kinetic enegry is conserved and the fraction of `K.E` transferred by the colliding particle is `(8)/(9)`, then the mass of `P` and mass of `Q` bears a ratioA. `sqrt(8) : 3`B. `9 : 8`C. `2 : 3`D. `2 : 1` |
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Answer» Correct Answer - D `(4m_(1)m_(2))/((m_(1)+m_(2))^(2)) =(8)/(9)`, find `m_(1)` and `m_(2)`. |
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| 58. |
A body of mass `20 gm` is moving with a certain velocity. It collides with another body of mass `80 gm` at rest. The collision is perfectly inelastic. The ratio of the kinetic energies before and after collision of the system isA. `2 : 1`B. `4 : 1`C. `5 : 1`D. `3 : 2` |
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Answer» Correct Answer - C `m_(1) u_(2) = (m_(1) + m_(2)) v` `KE_(i) = (1)/(2) m_(1)u_(1)^(2) , KE_(f) = (1)/(2) (m_(1) +m_(2)) v^(2)`. |
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| 59. |
A ball of mass `M` moving with a velocity `v` collides perfectly inelastically with another ball of same mass but moving with a velocity `v` in the opposite direction. After collisionA. both the balls come to restB. the velocities are exchanged between the two ballsC. both of them move at right angles to the original line of motionD. one ball comes to rest and another ball travels back with velocity `2v` |
| Answer» Correct Answer - A | |
| 60. |
A body of mass 5 kg is acted upon by a variable force.the force varies with the distance covered by the body. Find the kinetic energy of the body when the body has covered 30 m distance ? Assume that the body starts from rest. A. 295 JB. 105JC. 300 JD. 80 J |
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Answer» Correct Answer - A Work done by force when body has covered 25 m =Area under curve upto distance 25 m Now `W=DeltaK=1/2 xx5(v^(2)-0^(2))=250` Area between 25 m to 30 m is (1/2)5x18=45 So total workd done by variable force,till 30 m is 250 +45=295 joule So change in kinetic energy =295 J. So final kinetic energy =295 J |
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| 61. |
An athlete in the Olympic gamed covers a distance of `100 m` in `10 s`. His kinetic energy can be estimated to be in range. (1) `200 J - 500 J` (2) `2 xx 10^(5) J - 3 xx 10^(5) J` (3) `20,000 J - 50,000 J` (4) `2,000 J - 5,000 J`.A. `200 J - 500 J`B. `2 xx 10^(5) J - 3 xx 10^(5) J`C. `20,000 J - 50,000 J`D. `2,000 J - 5,000 J`. |
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Answer» Correct Answer - D average velocity `v_(av) = (s)/(l) = (100)/(10) = 10 m//s` Assuming an athelete has mass about `50` to `100 kg` or average mass `75 kg`, His average kinetic energy. `k_(av) = (1)/(2) mv_(av)^(2) = (1)/(2) xx 75 xx 100 = 3750 J` Minimum possible kinetic energy, `(for 50 kg)` `k_(min) = (1)/(2) mv_(av)^(2) = (1)/(2) xx 50 xx 100 = 2500 J` Maximum possible kinetic energy `(for 100 kg)` `k_(max) = (1)/(2) mv_(av)^(2) = (1)/(2) xx100xx100 = 5000 J` Hence the kinetic energy could be in the range `2000` to `5000 J`. |
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| 62. |
Two bodies of unequal masses have same linear momentum. Which one has greater K.E. ?A. lighter bodyB. heavier bodyC. bothD. none |
| Answer» Correct Answer - A | |
| 63. |
A lorry and a car moving with the same `KE` are brought to rest by applying the same retarding force. ThenA. lorry will come to rest in a shorter distanceB. car will come to rest in a shorter distanceC. both come to rest in same distanceD. any of above |
| Answer» Correct Answer - C | |
| 64. |
Two bodies of masses `m_(1)` and `m_(2)` have equal `KE`. Their momenta is in the ratioA. `sqrt(m_(2)) : sqrt(m_(1))`B. `m_(1) : m_(2)`C. `sqrt(m_(1)) : sqrt(m_(2))`D. `m_(1)^(2) : m_(2)^(2)` |
| Answer» Correct Answer - C | |
| 65. |
Which of the following statements is correct?A. `KE` of a system cannot be changed without changing its momentumB. `KE` of a system can be changed without changing its momentumC. Momentum of a system cannot be changed with changing its `KE`D. A system cannot have energy without having momentum |
| Answer» Correct Answer - A | |
| 66. |
Which of the following graphs depicts the variation of `KE` of a ball bouncing on a horizontal floor with height ? (Neglect air resistances)A. B. C. D. None of these |
| Answer» Correct Answer - A | |
| 67. |
Internal forces can changeA. Linear momentum as well as kinetic energyB. Linear momentum but not the kinetic energyC. Kinetic energy but not linear momentumD. neither the linear momentum nor the kinetic energy |
| Answer» Correct Answer - C | |
| 68. |
The potential energy of a system increased if work is doneA. upon the system by nonconservative forceB. by the system against a conservative forceC. by the system against a nonconservative forceD. upon the system by a conservative force |
| Answer» Correct Answer - D | |
| 69. |
A chain is held on a frictionless table with L/4 hanging over. Knowing total mass of the chain is M and total length L, the minimum work required to pull hanging part back to the table is :A. `(MgL)/16`B. `(MgL)/8`C. `(MgL)/32`D. `(MgL)/24` |
| Answer» Correct Answer - C | |
| 70. |
Statement-1: Davisson-Germer experiment established the wave nature of electron Statement-2: If electrons have wave nature, they can interfere show differaction.A. Statement `1` is false, Statement `2` is trueB. Statement `1` is true, Statement `2` is falseC. Statement `1` is true, Statement is the correct explanation for statement `1`D. Statement `1` us true, Statement `2` is true, Statement `2` is not the correct explanation for statement `1`. |
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Answer» Correct Answer - A `k_(1)x_(1)=k_(2)x_(2) =F` `W_(1)=(1)/(2) k_(1)x_(1)^(2) =((k_(1)x_(1))^(2))/(2k_(1)) = (F_(2))/(2k_(1))` Similarly, `W_(2)=(F^(2))/(2k_(1)) rArr W prop (1)/(K)` `W_(1) gt W_(2) rArr k_(1) lt k_(2)` Statement II is true. Now, `W_(1) = (1)/(2) K_(1) X^(2)` `W_(2) = (1)/(2) K_(2) X^(2)` So, `W_(2) gt W_(1)`, Hence, statement `I` is false. |
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| 71. |
The total work done on a particle is equal to the change in its mechanical energyA. if the forces acting on are conservativeB. if gravitational force alone acts on itC. if elastic alone acts on it.D. none of these |
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Answer» Correct Answer - D Follows from energy conservation. |
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| 72. |
In which of the following, the work done by the mentioned force is negative ? The work done byA. the tension in the cable while the lift is ascendingB. the gravitational force when a body slides down an inclined planeC. the applied force to maintain uniform motion of a block on a rough horizontal surfaceD. the gravitational force when a boby is thrown up |
| Answer» Correct Answer - D | |
| 73. |
In the case of conservative forceA. work done is independent of the pathB. work done in a closed loop is zeroC. work done against conservative force is store is the form of potential energyD. all the above |
| Answer» Correct Answer - D | |
| 74. |
If `x,F` and `U` denote the dispalcement, force acting on and potential energy of a particle thenA. `U = F`B. `F = + (dU)/(dx)`C. `F = -(dU)/(dx)`D. `F = (1)/(x)((dU)/(dx))` |
| Answer» Correct Answer - C | |
| 75. |
A particle of mass `1 kg` is moving X-axis. Its velocity is `6 m//s` at `x = 0`. Acceleration-displacement curve and potential energy-dispalcement curve of the particle are shown : .A. the work done by all the forces is `704 J`B. the work done by external forces is `350 J`C. the work done by external forces is `384 J`D. the work done by conservation forces is `300 J`. |
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Answer» Correct Answer - A::C Area of `(a-t)` curve `= 32 ms^(-1) = V_(f) - V_(i)` `V_(f)=32 +V_(i)=32+6 = 38 ms^(-1)` work done by all forces `= Delta KE` =`(1)/(2)m(V_(f)^(2)-V_(i)^(2))=(1)/(2)(38^(2) -6^(2))=704 J` work done by external forces `= 704 -320 = 384 J`. |
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| 76. |
The spring constant of spring `A` is twice the spring constant of spring `B`. Each of the spring is cut into two pieces. First piece of spring `A` is `(4//5)` of the total length. Second piece of spring `B` is `(5//6)` of its total length. Both springs are of equal lengh initially :A. the ratio of force constant of first piece of spring `B` to the first piece of spring `A` is `(12//5)`B. the ratio of force constant of first piece of spring `B` to the first piece of spring `A` is `2`C. the ratio of force constant of second piece of spring `A` to the first piece of spring `B` is `5//3`D. the ratio of force constant of second piece of spring `A` to the first piece of spring `B` is `7//5`. |
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Answer» Correct Answer - A::C `K_(A)=2K,K_(B)=K` `(L_(A))_(1)=(4)/(5)L,(L_(A))_(2) =(L)/(5) ,(L_(B))_(1) =(L)/(6),(L_(B))_(2) =(5L)/(6)` force constant `prop(1//5 "length")` `(K_(A))_(1) =(5)/(4)(2K)=(5)/(2) K`, `(K_(A))_(2) =5(2K) = 10 K` `(K_(B))_(1) = 6(K)` and `(K_(B))_(2) = (6)/(5) (K)`. |
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| 77. |
A vehicle is moving with uniform speed along horizontal, concave and convex surface roads. The surface on which, the normal reaction on the vehicle is maximum isA. concaveB. convexC. horizontalD. same at all surfaces |
| Answer» Correct Answer - A | |
| 78. |
A car weighing `1000 kg` is going up an incline with a slope of `2` in `25` at a steady speed of `18 kmph`. If `g = 10 ms^(-2)`, the power of its engine isA. 4 kWB. 50 kWC. 625 kWD. 25 kW |
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Answer» Correct Answer - A `P = vecF.vec v = Fv = mg sin theta v`. |
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| 79. |
Two riffles fire the same number of bullets in a givem interval of time. The second fires bullets of mass twice that fired by the first and with a velocity that is half that of the first. The ratio of their powers isA. `1 : 4`B. `4 : 1`C. `1 : 2`D. `2 : 1` |
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Answer» Correct Answer - D `(P_(1))/(P_(2)) =((m_(1))/(m_(2))) ((v_(1)^(2))/(v_(2)^(2)))`. |
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| 80. |
In the figure shown upper block is given a velocity of `6 m//s` and lower block `3 m//s`. When relative motion between them is stopped. .A. Work done by friction on upper block is negativeB. Work done by friction on both blocks is positiveC. The magnitude of work done by friction on upper block is `10 J`D. Net work done by friction is zero. |
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Answer» Correct Answer - A::C From conservation of linear momentum `(1+2) v = (6 xx 1)+(2-3)` `:. v = 4m//s` (of both the blocks) from work energy theorm i.e., `W_("total") = Delta KE` on `1 kg` block, `W_(f) = (1)/(2) xx 1 xx (4^(2) - 6^(2)) = - 10 J` on `2 kg` block `W_(f) = (1)/(2) xx 2(4^(2) - 3^(2)) = + 7 J` `:.` Net work done by friction is ` - 3 J`. |
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| 81. |
Block `A` of mass `1 kg` is placed on the rough surface of block `B` of mass `3 kg`. Block `B` is placed on smooth horizontal surface. Blocks are given the velocities as shown. Find net work done by the frictional force. [in `(-) ve J`]. . |
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Answer» Correct Answer - `6` `(1+3)v = (1) (8) +(3)(4) = 20 , v = 5 m//sec` for block `A, W_(f) = (1)/(2)(1) (5^(2) -8^(2)) = -(39)/(2) J` for block `B, W_(f) =(1)/(2)(3)(5^(2) -4^(2)) = +(27)/(2) J` net work done by friction `= -6 J`. |
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| 82. |
If `W_(1) W_(2)` and `W_(3)` represent the work done in moving a particle from `A` to `B` along three different paths `1.2` and`3` respectively (asshown ) in the gravitational fieled of a point mass m, find the correct relation between ` `W_(1) W_(2)` and `W_(3)` A. `W_(1) gt W_(2) gt W_(3)`B. `W_(1) = W_(2) = W_(3)`C. `W_(1) lt W_(2) lt W_(3)`D. `W_(2) gt W_(1) gt W_(3)` |
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Answer» Correct Answer - B Work done will be same in all the cases because gravitational field is a conservative field Thus work done is independent of the point `B`, therefore. |
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| 83. |
A particle of mass `0.5kg` travels in a straight line with velocity `v=ax^(3//2)` where `a=5m^(-1//2)s^-1`. What is the work done by the net force during its displacement from `x=0` to `x=2m`?A. 50 JB. 20 JC. 80 JD. `45.5 J` |
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Answer» Correct Answer - A `W = Delta KE = (1)/(2) mv^(2) - (1)/(2) m u^(2)`. |
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| 84. |
A spring when compressed by `4 cm` has `2 J` energy stored in it. The force requried to extend it by `8 cm` will beA. 20 NB. 2 NC. 200 ND. 2000 N |
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Answer» Correct Answer - C `U = (1)/(2) Kx_(1)^(2) rArr K = (2U)/(x_(1)^(2))` and `F= Kx_(2)`. |
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| 85. |
Which of the following statements is not true?A. Work done by conservative force on an object depends only on the initial and final states and not on the path taken.B. The change in potential energy of a system corresponding to conservative intermal force is equal to negative of the work done by these forces.C. If some of the internal within a system are non-conservative, then the mechanical energy of the system is not constant.D. If the internal forces are conservative, the work done by the internal forces is equal to the change in mechanical energy. |
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Answer» Correct Answer - D Clear from definition |
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| 86. |
In the figure, block A is released from rest when the spring is its natural length for the block B of mass m to leave contact with the ground at some stage what should be the minimum mass of block A? .A. 2MB. MC. M/2D. A function of M and the force constant of the spring |
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Answer» Correct Answer - C Let m be minimum mass of ball. Let mass A moves downwards by x. From conservation of energy. `mgx=1/2kx^(2)` `x=((2mg)/k)` For mass M to leave contact with ground `kx=mg` `k((2mg)/k)=Mg` `m=M/2` |
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| 87. |
A particle of mass `m` moves along a circle of radius `R` with a normal acceleration varying with time as `a_(n) = bt^(2)`, where `b` is a constant. Find the time dependence of the power developed by all the forces acting on the particle, and the mean value of this power averaged over the first `2` seconds after the beginning of motion, `(m = 1,v = 2,r = 1)`. |
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Answer» Correct Answer - `2` `rArr v = sqrt(bR) t (dv)/(dt) = sqrt(bR)` For circular motion work done by normal force is zero. For tangential forces. `F_(t)=m(dv)/(dt)=m sqrt(bR) P=F_(t).v = F_(t) v cos theta` as `theta = 0^@`, `P = F_(t) v = mbRt` Average power `= (underset(0) overset(T) intP(t)dt)/(underset(0) overset(T)int dt) = underset(0) overset(T) int(mbRTdt)/(T)` =`(mbR(t^(2)//2)_(0)^(T))/(T) = (mbRt)/(2)`. |
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| 88. |
Block `A` has a weight of `300 N` and block `B` has weight `50 N`. Calculate the distance `A` must descent form rest before it obtains a speed of `4 m//s` (Neglect the mass of cord and pulleys). `(Take g =1 0 m//s^(2))`. . |
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Answer» Correct Answer - `2` `S_(A) = h` and `S_(B) = 2h , V_(A) = 4m//s , V_(B) = 8 m//s` From energy conservation decreases in `P.E` of `A` = Increase in `P.E.B+` Increase in `K.E` of both `A` and `B` `300h = 50 xx 2h +(1)/(2) xx30 xx 4^(2)+(1)/(2) xx 5xx 8^(2)` `300 h=100h + 240 + 160`, `200 h = 400 , h = 2m`. |
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| 89. |
In the above problem, the ratio of times of two consecutive rebounds isA. `1 : e`B. `e : 1`C. `1 : e^(2)`D. `e^(2) : 1` |
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Answer» Correct Answer - A `t_(n) = e^(n) t`. |
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| 90. |
In the above problem the ratio of distances travelled in two consecutive rebounds isA. `1 : e`B. `e : 1`C. `1 : e^(2)`D. `e^(2) : 1` |
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Answer» Correct Answer - C `h_(n) = e^(2n) h`. |
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| 91. |
In the figure the variations of componets of acceleration of particles of mass 1kg is shown w.r.t. time. The initial velocity of the particle is `vec(u)=(-3 hati+4hatj)`m/s. the total work done by the resultant force on the particles in time intervals from t=0 to t=4 seconds is : A. `22.5 J`B. 10 JC. 0D. None of these |
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Answer» Correct Answer - B From given graphs : `a_(x) =(3)/(4)t` and `a_(y) =-((3)/(4)t+1) rArr v_(x) =(3)/(8) t^(2) +C` At `t =0 ,v_(x) = -3 rArr C = -3` :. `v_(x)=(3)/(8)t^(2) -3 rArrdx =((3)/(8)t^(2)-3) dt` ….(1) Similarly , `dy =(-(3)/(8)t^(2)-t+4)dt` ….(2) As `dw =vec F. vecds = vec F.(dx i+xy j)` `:. underset(0) overset(w) int dw = underset(0) overset(4) int[(3)/(4) ti-((3)/(4) t+1)j].x` `[((3)/(8) t^(2)-3)i+(-(3)/(8)t^(2)-t+4) j]dt` `:. W = 10 J` Alternate Solution : Area of the graph , `int a_(x) dt =6=V_((x)f)-(-3) rArr V_((x)f) = 3` and `int a_(y) dt =-10=V_((y)f)-(-4)rArr V_((y)f) = -6` Now work done `= Delta KE = 10 J`. |
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| 92. |
A particle of mass `0.2 kg` is moving in one dimension under a force that delivers constant power `0.5 W` to the particle. If the initial speed (in `ms^(-1)`) after `5s` is. |
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Answer» Correct Answer - `5` `(1)/(2) m upsilon^(2) - (1)/(2) m u^(2) = W` `(1)/(2) mv^(2) = Pt` `upsilon = sqrt((2Pt)/(m)) = sqrt((2 xx 0.5 xx 5)/(0.2)) = 5ms^(-1)`. |
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| 93. |
Consider the following two statements: A. Linear momentum of a system of partcles is zero. B. Kinetic energ of a system of particles is zero.A. A does not imply `B & B` does not imply `A`B. `A` implies `B` but `B` does not imply `A`C. `A` does not imply `B` but `B` implies `A`D. `A` implies `B` and `B` implies `A` |
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Answer» Correct Answer - B Because linear momentum is vector quantity whereas kinetic energy is a scalar quantity. |
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| 94. |
An electric motor creates a tension of `4500` newton in a hoisting cable and reels it at the rate of `2 m//s`. What is the power of the motor ?A. 15 kWB. 9 kWC. 225 WD. 9000 kW |
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Answer» Correct Answer - B `P_("inst") = vec F.vecV = FV cos theta`. |
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| 95. |
A block of mass `1 kg` kept over a smooth surface is given velocity `2 m//s` towards a spring of spring constant `1 N//m` at a distance of `10 m`. Find after what time block will be passing through `P` again .A. `(20+2 pi)sec`B. `10 sec`C. `(10 + 2pi) sec`D. `(10 + pi) sec` |
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Answer» Correct Answer - D `t =(s)/(v) + pi sqrt((m)/(k))+(s)/(v)`. |
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| 96. |
Choose tha false statementA. In a perfect elastic collision the relative velocity of approach is equal to the relative velocity of separationB. In an inelastic collision the relative velocity of approach us less than the relative velocity of separationC. In an inelastic collision the relative velocity of separation is less than relative velocity of approachD. In perfect inelastic collision relative velocity of separation is zero] |
| Answer» Correct Answer - B | |
| 97. |
A particle of mass `m` has a velocity `-v_(0) i`, while a second particle of same mass has a velocity `v_(0) j`. After the particles collide, first particle is found to have a velocity `(-1)/(2) v_(0) overline i` then the velocity of othe particle isA. `(-1)/(2) v_(0) vec i+ v_(0) vec j`B. `(1)/(2) v_(0) vec i+ v_(0) vec j`C. `v_(0) vec i+ v_(0) vec j`D. `-v_(0) vec i+ v_(0) vec j` |
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Answer» Correct Answer - A `{:(("Total momentum",,,,),("before collision",,,,)):}={:(("Total momentum",,,,),("after collision",,,,)):}`. |
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| 98. |
Two particles of different masses collide head on. Then for the systemA. loss of `KE` is zero, if it was perfect elastic collisionB. If it was perfect inelastic collision, the loss of `KE` of the bodies moving in opposite directions is more than that of the bodies moving in the same directionC. loss of momentum is zero for both elastic and inelastic collisionD. `1,2` and `3` are correct |
| Answer» Correct Answer - D | |
| 99. |
In an elastic collisionA. The initial kinetic energy is equal to the final kinetic energyB. The final kinetic energy is less than the initial kinetic energyC. The kinetic energy remains constantD. the kinetic energy first increases then decreases. |
| Answer» Correct Answer - A | |
| 100. |
A `2 kg` mass moving on a smooth frictionless surface with a velocity of `10 ms^(-1)` hits another `2 kg` mass kept at rest, in a perfect inelastic collision. After collision, if they move togetherA. they travel with a velocity of `5 ms^(-1)` in the same directionB. they travel with a velocity of `10 ms^(-1)` in the same directionC. they travel with a velocity of `10 ms^(-1)` in opposite directionD. they travel with a velocity of `5 ms^(-1)` in opposite direction |
| Answer» Correct Answer - A | |