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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.
| 1. |
The ballistic pendulum is an apparatus used to measure the speed of a fast moving projectile, such as bullet. `A` bullet of mass `m_(1)` is fired into a large block of wood of mass `m_(2)` suspended from light wires. The bullet embeds in the block and the entire system swings through a height `h`. Determine the initial speed of the bullet in terms of `h`?` |
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Answer» As there is no external force in x-direction, applying conservation of momentum during collision `m_(1)v_(1)=(m_(1)+m_(2))V_(B)` `therefore" "V_(B)=(m_(1)v_(1))/(m_(1)+m_(2))`…..(i) Applying conservation of mechanical energy after collision `(1)/(2)(m_(1)+m_(2))V_(B)^(2)=(m_(1)+m_(2))=gh` here `h` is the displacement of center of mass: `V_(B)=sqrt(2gh)`....(ii) Form (i) & (ii) `therefore" "V_(1)=(1+(m_(2))/(m_(1)))sqrt(2gh)` |
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| 2. |
Statement-1 : The momentum of a particle cannot be changed without changing its kinetic energy. Statement-2 : Kinetic energy (K) of a particle is related to momentum (p) as `K=(p^(2))/(2m)`A. Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.B. Statement -1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.C. Statement-1 is true, statement-2 is false.D. Statement-1 is false, statement-2 is true. |
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Answer» Correct Answer - D momentum is a vector equantity |
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| 3. |
A small ball `B` of mass `m` is suspended with light inelastic string of length `L` from a block `A` of same mass in which can move on smooth horizontal surface as shown in the figure. The ball is displaced by angle `theta` from equilibrium position and then released. The displacement of block when equilibrium position isA. `(Lsintheta)/(2)`B. `Lsintheta`C. LD. none of these |
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Answer» Correct Answer - A `m_(1)Deltax_(1)+m_(2)Deltax_(2)=0` `rArrmx+m{-(Lsintheta-x)}=0` `rArrx=(Lsintheta)/(2)` |
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| 4. |
Statement-1 : A man falling from a given height on hard ground is hurt more than falling on a soft bed from same height. Statement-2 : A man falling from a given height on hard ground experiences larger impulse than falling on a soft bed in coming to rest in both cases.A. Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.B. Statement -1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.C. Statement-1 is true, statement-2 is false.D. Statement-1 is false, statement-2 is true. |
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Answer» Correct Answer - C |
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| 5. |
A ball of mass `m` collides horizontally with a stationary wedge on a rough horizontal surface, in the two orientations as shown. Neglect friction between ball and wedge. Two student comment on system of ball and wedge in these situation Saurav: Momentum of system in x-direction will change by significant amount in both cases. Rahul : There are no impulsive external forces in y-direction in both cases hence the total momentum of system in y-direction can be treated as conserved in both cases.A. Saruav is wrong and Rahul is correctB. Saurav is correct and Rahul is wrongC. Both are correctD. Both are wrong |
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Answer» Correct Answer - D In the second coase normal force by ground will be impulsive hence frictional force is also impulsive. So we can neglect it |
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| 6. |
The system of the wedge and the block connected by a massless spring as shown in the figure is released with the spring in its natural length. Friction is absent. Maximum elongation in the spring will be A. `(3Mg)/(5k)`B. `(6Mg)/(5k)`C. `(4Mg)/(5k)`D. `(8Mg)/(5k)` |
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Answer» Correct Answer - B Using conservation of mechanical energy `Mg(x_(m)sintheta)=(1)/(2)"K x"_(m)^(2)` `x_(m)=(2Mgsintheta)/(K)` |
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| 7. |
Two massless string of length 5 m hang from the ceiling very near to each other as shown in the figure. Two balls `A` and `B` of masses `0.25kg` and `0.5kg` are attached to the string. The ball `A` is released from rest at a height `0.45m` as shown in the figure. The collision between two balls is completely elastic. Immediately after the collision, the kinetic energy of ball `B` is `1 J` The velocity of ball `A` just after the collision is A. `5ms^(-1)` to the rightB. `5ms^(-1)` to the leftC. `1ms^(-1)` to the rightD. `1ms^(-1)` to the left |
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Answer» Correct Answer - D `(1)/(2)m_(B)v_(B)^(2)=K_(B)" "rArr" " (1)/(2)(0.5)v_(B)^(2)rArrv_(B)=2` `v_(B)=(2m_(A))/(m_(A)+m_(B))u_(A)` and `v_(A)=(m_(A)-m_(B))/(m_(A)+m_(B))u_(A)` `v_(A)=(m_(A)-m_(B))/(2m_(A))v_(B)=(0.25-0.5)/(0.5)xx2=-1m//s` |
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| 8. |
Two identical blocks `A` and `B` connected by massless string, are placed on a frictionless horizontal plane. A bullet havig the same mass, moving with speed `u` strickes block `B` from behind as shown. If the bullet gets embedded into block `B` then finda. the velocity of `A,B,C` after collision. b. impulse on `A` due to tension in the string,c. impulse on `C` due to normal force of collision, d. impulse on `B` due to normal force of collision. |
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Answer» (a) After collision, the blocks & the bullet will move together with same velocity, say `v` By conservation of linear momentum `m u=3m u` `v=(u)/(3)` (b) Net impulse on `A` is due to tension force, Impulse on `A=P_(f)-P_(i)` `intTdt=(m u)/(3)-0` (c) On the bullet `C`, net impulse is due to `N` `-intNdt=P_(f)-P_(i)` `=(m u)/(3)-m u` `=(-2m)/(3)` (d) On `B` two impulsive forces act i.e. Normal & Tension. `vec"J"=int(N-T)dt=intN" "dt-intT" "dt=(m u)/(3)` `rArr" "intN" "dt=(2m u)/(3)` The impulse due to normal force on both the colliding bodies is equal. Thus we can directly say impulse on `B` due to normal is same as impulse on `C` due to normal. |
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| 9. |
A small block of mass `2m` initially rests at the bottom of a fixed circular, vertical track, which has a radius of `R`. The contact surface between the mass and the loop is frictionless. `A` bullet of mass `m` strikes the block horizontally with initial speed `v_(0)` and remain embedded in the block as the block and the bullect circle the loop. Determine each of the following in terms of `m,R` and `g`. (a) The speed of the masses immediately after the impact. (b) The minimum initial speed of the bullet if the block and the bullet are to successfully execute a complete ride on the loop |
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Answer» Correct Answer - (a)`v_(0)//3,(b)3sqrt(5gR)` Using conservaion of momentum `mv_(0)+2m(0)=3mv` `rArr" "v=(v_(0))/(3)` (b) `v=sqrt(5gR)" "rArr(v_(0))/(3)=sqrt(5gR)" "rArr" "v_(0)=3sqrt(5gR)` |
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| 10. |
Consider a rubber ball freely falling from a height ` h = 4.9 m` onto a horizontally elastic plate. Assume that the duration of collision is negligible and the collisions with the plate is totally elastic . Then the velocity as a function of time and the height as a function of time will be :A. B. C. D. |
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Answer» Correct Answer - C First velocity is negetive and increasing. At the time of collision it becomes `+ve` from `-ve` again and again hence graph `C` |
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| 11. |
Statement-1 : In a head on elastic collision between a moving body `A` with an initally stationary body `B`, the kinetic energy lost by `A` is equal to the kinetic energy gained by `B` till the instant of maximum deformation during their contact. Statement-2 : The total kinetic energy of system before and after collision is same in elastic collision.A. Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.B. Statement -1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.C. Statement-1 is true, statement-2 is false.D. Statement-1 is false, statement-2 is true. |
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Answer» Correct Answer - D At the time of maximum defermation some potential energy is stored |
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| 12. |
In Column-I, 4 situations are depicted and in columa-II, 4 possible kinds of collision are listed. Match the situation with type of collision. |
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Answer» Correct Answer - (A) Q (B) S (C) P Evaluate `e=(v_(2)-v_(1))/(u_(1)-u_(2))` for each case alongwith using conservation of linear momentum. |
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| 13. |
Four particles of mass `5,3,2,4 kg` are at the points `(1,6),(-1,5),(2,-3),(-1,-4)` . Find the coordinates of their centre of mass. |
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Answer» Correct Answer - 1/7,23/14 `x_(cm)=(Sigmam_(i)x_(i))/(Sigmam_(i))=(5(1)+3(-1)+2(2)+4(-1))/(5+3+2+4)=(2)/(14)=(1)/(7)m` `y_(cm)=(Sigmam_(i)y_(i))/(Sigmam_(i))=(5(6)+3(5)+2(-3)+4(-4))/(5+3+2+4)=(23)/(14)m` |
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| 14. |
A square pland of mass `m_(1)=100kg` and edge length `L=20sqrt2m` is placed on a smooth surface. Two person each mass `m_(2)=m_(3)=50 "kg" ` are at corner of a plank as shown in figure. Two person begin to walk on the plank along two different paths as shown in figure and reach nearest corners. What is magnitude of displacment of plank (in `m`) in the process. |
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Answer» Correct Answer - 10 Let displacement of plank be represented by `xhati+yhatj` For x-component `50[L-x]-100x-50x=0` `x=(L)/(4)` Similarly y-component `50[L-y]-100y-50y=0` `y=(L)/(4)` Thus `vecr=-(L)/(4)hati-(L)/(4)hatj` or `|vecr|=(L)/(4)sqrt2` |
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| 15. |
Two identical blocks `A` and `B` , each of mass `m` resting on smooth floor are connected by a light spring of natural length `L` and spring constant `k`, with the spring at its natural length. A third identical block `C` (mass `m`) moving with a speed `v` along the line joining `A` and `B` collides with `A`. The maximum compression in the spring is |
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Answer» Initially there will be collision between `C` and `A` which is elastic, therefore, by using conservation of momentum we obtain, `mv_(0)=mv_(A)+mv_(C)" "," "v_(0)=v_(A)+v_(C)` Since `e=1,v_(0)=v_(A)-v_(C)` Solving the above two equation, `v_(A)=v_(0)` and `v_(c)=0` Now `A` will move and compress the spring which in turn acceleration `B` and retard `A` and finally both `A` and `B` will move with same velocity `v`. (a) Since net external force is zero, therefore momentum of the system (A and B) is conserved. Hence `mv_(0)=(m+2m)v` `rArr" "v=v_(0)//3` (b) If `x_(0)` is the maximum compression, then using energy conservation `(1)/(2)"mv"_(0)^(2)=(1)/(2)("m+2m")"v"^(2)+(1)/(2)"kx"_(0)^(2)` `rArr" "(1)/(2)"mv"_(0)^(2)=(1)/(2)(3"m")("v"_(0)^(2))/(9)+(1)/(2)"kx"_(0)^(2)" "rArr" "x_(0)=v_(0)sqrt((2m)/(3k))` Hence minimum distance `D=l_(0)-x_(0)=l_(0)-v_(0)sqrt((2m)/(3k))` |
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| 16. |
A uniform chain of length 2L is hanging in equilibrium position, if end `B` is given a slightly downward displacement the imbalance causes an acceleration. Here pullley is small and smooth & string is inextensible The velocity `v` of the string when it slips out of the pulley (height of pulley from floor`gt2L`)A. `sqrt((gL)/(2))`B. `sqrt(2gL)`C. `sqrt(gL)`D. none of these |
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Answer» Correct Answer - C `v(dv)/(dx)=(x)/(L)g" "rArrint_(0)^(v)"v dv"=(g)/(L)int_(0)^(L)"x dx"rArrvsqrt(gL)` |
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| 17. |
The figure shows the position-time (x-t) graph of one-dimensional motion of a body of mass 0.4kg. The magnitude of each impulse is A. `0.4 Ns`B. `0.8 Ns`C. `1.6 Ns`D. `0.2 Ns` |
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Answer» Correct Answer - B Impulse `=Dp=mDV="m 2 V"` `=0.4xx2xx"slope"` `0.4xx2xx(2)/(2)=0.8` |
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| 18. |
In a one-dimensional collision between two particles, their relative velocity is `vecv_(1)` before the collision and `vecv_(2)` after the collisionA. `vecv_(1)=vecv_(2)` if the collision is elasticB. `vecv_(1)=-vecv_(2)` if the collision is elasticC. `|vecv_(2)|=|vecv_(1)|` in all casesD. `vecv_(1)=-kvecv_(2)` in all cases, where `kge1` |
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Answer» Correct Answer - B::D `vecv_(2)-vecv_(1)=e(vecu_(1)-vecu_(2))` for `1-D` collision. |
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| 19. |
A `45.0kg` girl is standing on plank that has a mass of `150kg`. The plank, originally at rest, is free to slide on a frozen lake, which is a flat, frictionless supporting surface. The girl begins to walk alnong the plank at a constant speed of `1.50m//s` relative to the plank. (a) What is her speed relative to the ice surface? (b) What is the speed of the plank relative to the ice surface? |
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Answer» Correct Answer - 15/13m/s (b) `-9//26m//s` `m_("girl")v_("girl")+m_("plank")v_("plank")=0` [Initially both at reat] `rArr" "45v_("girl")+150v_("plank")=0`…(i) `v_("girl/plank")=v_("girl")-v_("plank")=1.5`…(ii) Solving(i)and(ii),we get `v_("girl")=(1.5xx150)/(195)=(15)/(13)m//s` `v_("plank")=v_("girl")-1.5=(15)/(13)-(3)/(2)` `v_("plank")=-(9)/(26)m//s` |
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| 20. |
Two particles A and B of masses 1kg and 2kg respectively are projected in the directions shown in figure with speeds `u_A=200m//s` and `u_B=50m//s`. Initially they were `90m` apart. They collide in mid air and stick with each other. Find the maximum height attained by the centre of mass of the particles. Assume acceleration due to gravity to be constant. `(g=10m//s^2)` |
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Answer» `vecP_("sys")=M_(T)vecV_(CM)` `(dvecP_("sys"))/(dt)=M_(T)veca_(CM)` `vecF_(ext)=M_(T)veca_(CM)` Net external force is the gravitational force `F_(ext)=M_(T)xxg` `therefore" "veca_("cm")=gdarr`(dwonwards) `(vecV_(cm))=(m_(A)vecV_(A)+m_(B)vecV_(B))/(m_(A)+m_(B))=(1xx200-2xx50)/(3)=(100)/(3)m//suarr` (upwards) initial height of centre of mass from `A`,`h_(0)=(1xx0+2xx90)/(1+2)=60m` `h_(max)=`inital height `(h_(0))+(v_(cm)^(2))/(2g)=60+(((100)/(3))^(2))/(2xx10)=115.55m` |
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| 21. |
Two blocks of masses `M_(1)=1kg` and `M_(2)=2kg` kept on smooth surface, are connected to each other through a light spring `(k=100N//m)` as shown in the figure. When we push mass `M_(1) ` with a force `F=10N` find the acceleration of centre of mass of system. |
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Answer» `a_(CM)=(F_(ext))/(M_(Total))` `=(10)/(3)m//s^(2)` |
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| 22. |
Two cars intially at rest are free to move in the `x` direction. Car `A` has mass 4 kg and car `B` has mass 2kg. They are tied together, compressing a spring in between them. When the string holding them together is burned, car `A` moves off with a speed of `2//s` (a) with what speed does car `B` leave. (b) how much energy was stored in the spring before the string was burned. |
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Answer» Correct Answer - 4m/s, 24J (a) `m_(A)v_(A)+m_(B)+v_(B)=0` [Initially both of them were at rest] `rArrv_(B)=-(m_(A)v_(A))/(m_(B))=-((4)(2))/((2))=-4m//s` (b) `E=(1)/(2)m_(A)v_(B)^(2)+(1)/(2)m_(B)v_(B)^(2)` `rArrE=(1)/(2)(4)(2)^(2)+(1)/(2)(2)(-4)^(2)=24J` |
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| 23. |
The diagram to the right shows the velocity-time graph for two masses `R` and `S` that collided elastically. Which of the following statements is true? [I] `R` and `S` moved in the same direction after the collision. [II] Kinetic energy of the system (R & S) is minimum at `t=2` milli sec.[III] The mass of `R` was greater than mass of `S`. A. I onlyB. II onlyC. I and II onlyD. I,II,only III |
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Answer» Correct Answer - D (i) From graph final velocity of both body is positive. (ii) K.E. is minimum when velocity of each body becomes equal. (iii) Magnitude of change in velocity of `R` is less. |
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| 24. |
A system of `N` particles is free from any extenal forces. (a) Which of the following is true for the magnitude of the total momentum of the system?A. It must be zeroB. It could be non-zero, but it must be constantC. It could be non-zero, and it might not be constantD. The answer depends on the nature of the internal forces in the system |
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Answer» Correct Answer - B (a) since momentum is conserved as `F_(ext)=0` so, if body have some momentum then it will not change it will be constant. i.e. Momemtum could non-zero but it must be constant internal force have no affect on momentum of body |
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| 25. |
Figure shows a rod of mass 10 kg of length 100cm with some masses tied to it at different position. Find the point on the rod at which if the rod is picked over a knife edge, it will be in equilibrium about that knife edge. |
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Answer» Centre of mass of the system shown in figure, will be the point, at which if we place a knife edge, system will remain in equilibrium. To locate the centre of mass of the system, we consider origin at the left end of the rod. With respect to this origin the position of centre of mass of the system is `x_(c)=(m_(1)x_(1)+m_(2)x_(2)+m_(3)x_(3)+m_(4)x_(4)+m_(rod)x_(rod))/(m_(1)+m_(2)+m_(3)+m_(4)+m_(rod))` `x_(c)=(25xx0+8.5xx30+10xx50+5xx80+5xx90)/(53.5)=30cm` |
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| 26. |
Statement-1 : Internal forces can change total linear momentum of system. Statement-2 : Due to work of internal forces kinectic energy of system may change.A. Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.B. Statement -1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.C. Statement-1 is true, statement-2 is false.D. Statement-1 is false, statement-2 is true. |
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Answer» Correct Answer - D Linear momentum of the system is changed by net external force but internal force can do some non zero work |
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| 27. |
Statement-1 : The centre of mass of any uniform triangular plate is at its centroid. Statement-2 : The centre of mass of any symmetrical body lies on its axis of symmetry.A. Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.B. Statement -1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.C. Statement-1 is true, statement-2 is false.D. Statement-1 is false, statement-2 is true. |
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Answer» Correct Answer - B Centoid is not the point about which body is symmetric. |
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| 28. |
Two spacemen are floating together with zero speed in a gravity-free region of space. The mass of spaceman `A` is 120 kg and that of spaceman `B` is 90 kg. Spacenab `A` pushes `B` away from him with `B` attaining a final speed of `0.5m//s`. Find the final reciol speed of `A`?. |
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Answer» Correct Answer - `0.38m//s` `m_(A)v_(A)+m_(B)v_(B)=0+0`[conservation of momentum] `rArrv_(A)=-(m_(B)v_(B))/(m_(A))=-((90)(0.5))/(120)=-0.375m//s` |
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| 29. |
Bullets of mass 10 g each are fired from a machine gun at rate of 60 bullets/minute. The muzzle velocity of bullets is `100m//s`. The thrust force due to firing bullets experienced by the person holding the gun stationary is ________. |
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Answer» Correct Answer - 1 N `F=(dp)/(dt)=((dn)mv)/((dt))=(1)((10)/(1000))(100)=1N` `(dn)/(dt)=60`bullets/mintue=1bullet/sec |
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| 30. |
A spring block system is placed on a rough horizontal floor. The block is pulled towards right to give spring some elongation and released. A. The block may stop before the spring attains its natural lengthB. The block must stop with spring having some comprission.C. The block may stop with spring having some compression.D. It is not possible that the block stops at antural length of the spring. |
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Answer» Correct Answer - A::C Take `mu` high and low |
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