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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.
| 301. |
Assertion: Two blocks A and B are connected at the two ends of an ideal spring as shwon in figure. Initially, spring was released. Now block B is pressed. Linear momentum of the system will not remain constan till the spring reaches its initial natural length. Reason: An external force will act from the wall on block A.A. If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.B. If both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.C. If Assertion is true but Reason is false.D. If Assertion is false but Reason is true. |
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Answer» Correct Answer - A Linear momentum will not remain constant till spring will remain compressed. Therefore a force will act on block A from the wall. |
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| 302. |
An `80 kg` man is riding on a `40 kg` cart travelling at a speed of `2.5 m//s` on a frictionless horizontal plane jumps off the cart, such that, his velocity just after is zero with respect to ground. The work done by him on th esystem during his jump is given as `(A)/(4) KJ` (Ain integer). Find the value of A. |
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Answer» Correct Answer - 3 By conservation of linear momentum `(80 + 40) (2.5) = 80 (0) + 40 (v) rArr = v = 7.5 m//s` work done `= DeltaKE = (1)/(2) 40 (7.5)^(2) - (1)/(2) (80 + 40) (2.5)^(2) = 750 J` |
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| 303. |
Statement I. A particle strikes head-on with another stationary particle such that the first particle comes to rest after collision. The collision should necessarily be elastic. Statement II: In elastic collision, there is no loss of momentum of the system of the particles.A. Both assertion and reason are true and reason is the correct explanation of assertion.B. Both assertion and reason are true but reason is not the correct explanation of assertion.C. Assertion is true and reason is false.D. Assertion is false and reason is true. |
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Answer» Correct Answer - D If the collision is elastic, then masses of both particles should be same. But if masses are different, then collision is inelastic. Further, momentum remains conserved in any kind of collision. |
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| 304. |
A mass `m_(1)` moves with a great velocity. It strikes another mass `m_(2)` at rest in head-on collision. It comes back along its path with low speed after collision. ThenA. `m_(1)gtm_(2)`B. `m_(1)ltm_(2)`C. `m_(1)=m_(2)`D. there is no relation between `m_(1)` and `m_(2)` |
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Answer» Correct Answer - B Conservation of linear rmomentum just before and after yields. ltbr `m_(1)vi=m_(2)v_(2)i+m_(1)v_(1)(-i)`…..i `implies m_(1)v=m_(2)v_(2)-m_(1)v_(1)` `e=-(v_(1)+v_(2))/(0-v)=1` therefore `v=v_(1)+v_(2)` ………ii [for elastic collision `e=1`] Elaminating `v_(2)` form i and ii we obtain `m_(1)v=m_(2)(v-v_(1))-m_(1)v_(1)` `implies v_(1)=(m_(2)-m_(1))/(m_(1)=m_(2))v` Since `v_(1)gt0,(m_(2)-m_(1))/(m_(1)+m_(2))gt1` `impliesm_(2)-m_(1)gt0impliesm_(1)ltm_(2)`. |
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| 305. |
A mass `m_(1)` moves with a great velocity. It strikes another mass `m_(2)` at rest in head-on collision. It comes back along its path with low speed after collision. Then find out whether `m_(1)ltm _(2)` or `m_(1)gtm_(2)`. |
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Answer» Since `m_(2)` is at rest `(u_(2)=0)`, velocity of `m_(1)` after collision: `v_(1)=((m_(1)-em_(2))/(m_(1)+m_(2)))u_(1)` Given `v_(1)lt0impliesm_(1)-em_(2)lt0` `impliesm_(1)ltem_(2)impliesm_(1)ltm_(2)` |
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| 306. |
In an eleastic collisionA. Both momentum and KE are conservedB. only momentum is conservedC. only KE is conservedD. Neither KE nor momentum is conserved |
| Answer» Correct Answer - A | |
| 307. |
After falling from rest through a height `h`, a body of mass `m` begins to raise a body of mass `M (M gt m)` connected to it through a pulley. Determnethe time it will take for the body of mass `M` to return to its original positionA. `(2m)/(M_m)sqrt((2h)/g)`B. `(2m)/(M-m)sqrt((2h)/g)`C. `(2m)/(M-m)sqrt(h/g)`D. `m/(M-m)sqrt((2h)/g)` |
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Answer» Correct Answer - B `u=sqrt(2gh)` is the velocity of `m` just before collision, and `v` is the velocilty of system `M+m` just after collision, then `(M+m)v="mu"` `implies v=(msqrt(2gh))/(M+m)` `a=((M-m)/(M+m))g` `t=(2v)/a=((2m)/(M-m))sqrt((2h)/g)` |
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| 308. |
After falling from rest through a height `h`, a body of mass `m` begins to raise a body of mass `M (M gt m)` connected to it through a pulley. Find the fraction of kinetic energy lost when the body of mass `M` is jerked into motionA. `M/(M+m)`B. `M/(M-m)`C. `(2M)/(M+m)`D. `M/(2(M+m))` |
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Answer» Correct Answer - A Fraction of `KE` lost`=(1/2"mu"^(2)-1/2(M+m)v^(2))/(1/2"mu"^2)` `=1-(M+m)/m (v/u)^(2)=1-(M+m)/m(m/(M+m))^(2)=m/(M+m)` |
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| 309. |
A pulley fixed to the ceiling carried a thread with bodies of masses `m_(1)` and `m_(2)` attached to its ends. The mases of the pulley and the thread are negligible and friction is absent. Find the acceleration of the centre of mass of this system. |
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Answer» Let us assume that `m_(2)gtm_(1)`. We can see that the masses have equal and opposite acceleration of the same magnitude. `a=(m_(2)-m_(1))/(m_(2)+m_(1))g` and tension in string is `T=(2m_(1)m_(2)g)/(m_(2)+m_(1))` Taking downward direction as positive `a_(1)=-,a_(2)=+a` `a_(CM)=(m_(1)a_(1)+m_(2)a_(2))/(m_(1)+m_(2))=((m_(2)-m_(1))a)/(m_(2)+m_(1))` substituting for the value of a we have `a_(CM)=((m_(2)-m_(1))/(m_(2)+m_(1)))^(2) g` Alternative method: `a_(CM)=F_(ext)/(m_(1)+m_(2))=((m_(1)g+m_(2)g)-2T)/(m_(1)+m_(2))=g-(2T)/(m_(1)+m_(2))` `=g-(4m_(1)m_(2)g)/((m_(2)+m_(1))^(2))=((m_(2)-m_(1))/(m_(2)+m_(1)))^(2)g` (downwards) |
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| 310. |
A ball collides at `B` with velocity `10m//s` at `30^(@)` with vertical. There is a flag at `A` and a will at `C`. Collision of ball with groundis perfectly inelastic `(e=0)` and that with wall is elastic `(e=1)`. Given `AB = BC=10 m`. Find the time after which ball will collide with the flag. A. 4B. 5C. 6D. Ball will not collide with the flag |
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Answer» Correct Answer - C As horizontal velocity (i.e velocity along the surface) remains constant so required time `= (30)/(5) = 6s` |
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| 311. |
A block C of mass m is moving with velocity `v_(0)` and collides elastically with block A of mass m and connected to another block B of mass `2m` through spring constant k. What is k if `x_(0)` is compression of spring when velocity of A and B is same? A. `(mv_(0)^(2))/(x_(0)^(2))`B. `(mv_(0)^(2))/(2x_(0)^(2))`C. `(3mv_(0)^(2))/(2x_(0)^(2))`D. `(2mv_(0)^(2))/(3x_(0)^(2))` |
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Answer» Correct Answer - D Their common velocity would be when C strikes A, A accquires the velocity of C and C beocmes stationary. A forces B to move and ultimately both acquires common velocity with spring compressed by `x_(0)`. `v=(mv_(0)/(m+2m)` = `v_(0)/3)` Now applying conservation of mechanical energy `1/2mv_(0)^(2)=1/2kx_(0)^(2) + 1/2(3m)(v_(0)/3)^(2) rArr k = 2/3(mv_(0)^(2)/(x_(0)^(2)` |
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| 312. |
Three objects `A` ,` B` and `C` are kept in a straight line on a smooth horizontal surface. These have masses `m `, `2m` and `3 m` , respectively . The head - on elastic collision takes place between `A` and `B` and then `B` makes completely inelastic collision with `C`. All motions occur on the same straight line. The final speed of `C` will be |
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Answer» Correct Answer - `4 m//s` For collision between A & B `v_(A) = ((m - 2m))/((m + 2m)) (9) = -3 ms^(-1)` `v_(B) = (2(m))/((m + 2m))(9) = +6ms^(-1)` For collision between B and C `v_(c ) = ((2m)/(2m + m)) (6) = 4ms^(-1)` |
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| 313. |
A particle of mass `m` moving with a speed `v` hits elastically another staionary particle of mass `2m` on a smooth horizontal circular tube of radius `r`. Find the time when the next collision will take place?A. `(2pir)/(v)`B. `(4pir)/(v)`C. `(3pir)/(2v)`D. `pir)/(v)` |
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Answer» Correct Answer - A Relative speed of seperation = relative speed of approach `therefore` Time of next collision = `(2pir)/v` |
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| 314. |
Two masses m and `2m` are placed in fixed horizontal circular smooth hollow tube of radius r as shown. The mass m is moving with speed u and the mass `2m` is stationary. After their first collision, the time elapsed for next collision. (coefficient of restituation `e = 1//2` ) A. `(2pir)/(u)`B. `(4pir)/(u)`C. `(3pir)/(u)`D. `(12pir)/(u)` |
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Answer» Correct Answer - (B) Let the speeds of balls of mass m and `2m` after collision be `v_(1)` and `v_(2)` as shown in figure. Applying conservation of momentum `mv_(1) + 2mv_(2) = m u` & `-v_(1) + v_(2) = (u)/(2)`. Solving we get `v_(1) = 0` and `v_(2) = (u)/(2)` Hence the ball of mass m comes to rest and ball of mass `2m` moves with speed `(u)/(2). t = (2pir)/(u//2) = (4pir)/(u)` |
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| 315. |
A sphere of mass `m`, impinges obliquely on a sphere, of mass M, which is at rest. Show that, if `m=eM`, the directions of motion of the sphere after impact are at right angles. |
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Answer» Let CT stands for common tangent direction and CN for common normal direcitons. In the common tangent directions velocity components remain unchanged. In common normal direction applying conservation of linear momentum and definition of `e`, `eMv_2=eMv_3+Mv_4` ...(i) From the definition of coefficient of restitution `e=(v_4-v_3)/(v_2)` ...(ii) Solving these two equations we get, `v_3=0` but `v_4!=0` So, after collision velocity of m is along CT while that of M along CN or they are moving at right angles. |
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| 316. |
The uniform solid sphere shown in the figure has a spherical hole in it. Find the position of its centre of mass. |
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Answer» Correct Answer - A::B::C `x_(CM)=(V_1x_1-V_2x_2)/(V_1-V_2)` `=((4/3piR^3)(0)-((4pia^3)/(3))(b))/((4/3piR^3)-4((pia^3)/(3)))` `=(-a^3b)/(R^3-a^3)` |
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| 317. |
A particle moves in the `xy` plane under the influence of a force such that its linear momentum is `vecP(t) = A [haticos(kt)-hatjsin(kt)]`, where `A` and `k` are constants. The angle between the force and momentum isA. `0^(@)`B. `30^(@)`C. `45^(@)`D. `90^(@)` |
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Answer» Correct Answer - D `vecF=(vec(dP))/(dt)=kAsin(kt)hati-kAcos(kt)hatj` `vecP=cos(kt)hati-Asin(kt)hatj` Now `vecF.vecP=0` Therefore angle betweenn `vecF` and `vecP `should be `90^(@)` |
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| 318. |
A `30 kg` projectile moving horizontally with a velocity `vecv_(0)=(120m//s)hati` explodes into two fragments `A` and `B` of masses `12 kg` and `8 kg`, respectively. Taking point of explosion as origin and knowing that `3s` later position of fragment a is (`300m, 24m, -48m`), determine the position of fragment `B` at the instant. |
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Answer» Initial corrdinates of `CM=(0,0,0)` `x` coordinates of `CM` after `3 s` `x_(CM)=120xx3=360m` `y` coordinates of `CM` after `3s`, `y_(CM)=-1/2"gt"^(2)=-45m` After `3s`, `(m_(1)+m_(2))x_(CM)=m_(1)x_(1)+m_(2)x_(2)` `x_(2)=(30xx360-12xx300)/18=400m` similarly `(m_(1)+m_(2))y_(CM)=m_(1)y_(1)+m_(2)y_(2)` `=-30xx45=12xx24+18xy_(2)` `y_(2)=-91m` and in `z` coordinate `m_(1)z_(1)+m_(2)z_(2)=0implies z_(2)=32m` |
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| 319. |
Assertion: If a projectile explodes in mid air, linear momentum of center of mass of different fragments remains constant.A. If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.B. If both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.C. If Assertion is true but Reason is false.D. If Assertion is false but Reason is true. |
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Answer» Correct Answer - A Path of center of mass remains unchanged because during explosion no external force acts on COM. |
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| 320. |
A shell fired vertically up, when reaches its highest point, explodes into three fragments A,B and C of masses `m_(A) = 4kg, m_(a) = 2_kg` and `m_(c ) = 3kg`. Immediately after the explosion, A is observed moving with velocity `v_(A) = 3m//s` towards north and B with a velocity `v_(b) = 4.5 m//s` towards east as shown in the figure. Find the velocity `v_(c )` of the piece C. |
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Answer» Explosion takes negligible duration, therefore, impulse of gravity, which is a finite external force, can be neglelted. The pieces fly off acquiring above-mentioned velocities due to internal forces developed due to expanded gases produced during explosion. The forces aplied by the expanding gases are internal forces, hence, momentum of the system of the three pieces remains conserved during the explosion and total momentum before and after the explosion are equal. Assuming the east as positive x-direction and the north as positive y-direction, the momentum vectors `vecp_A` and `vecp_B` of pieces A and B become `vec(p)_(A) = m_(A)v_(A)hat(j) = 12 kg-m//s` and `vec(p)_(B) = m_(B)v_(B)hat(i) = 9 kg-m//s` Before the explosion, momentum of the shell was zero, therefore from the principle of conservation of momentum, the total momentum of the fragments also remains zero. `vec(p)_(A) + vec(P)_(B)+vec(P)_(c ) = vec(0) rarr " " vec(p)_(c) = -(9hat(i) + 12hat(j))` From the above equation, velocity of the piece C is `vec(v)_(c ) = (vec(p)_(C ))/(m_(C )) = -(3hati+4hatj)=5m//s, 53`.south west. |
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| 321. |
A mud ball at rest explodes into three fragmennts of masses in the ragio `1:2:1`. the two equal masses move with velocities `2hati+5hatj-6hatk` and `-4hati+3hatj+2hatk`. Find the velocity of the find mass. |
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Answer» Apply conservation of momentum: `implies m_(1)vecv_(1)+m_(2)vecv_(2)=m_(3)vecv_(3)=0` `m(2hati+5hatj-6hatk)+2mvecv_(2)+m(-4hati+3hatj+2hatk)=0` `vecv_(2)=hati-4hatj+2hatk` |
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| 322. |
Assertion: The relative velocity of the two particles in head-on elastic collision is unchanged both in magnitude and direction. Reason: The relative velocity is unchanged in magnitude but gets reversed in direction.A. If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.B. If both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.C. If Assertion is true but Reason is false.D. If Assertion is false but Reason is true. |
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Answer» Correct Answer - D e= (|RVOS i.e., Relative velocity of separation|)/(|RVOA i.e., Relative velocity of approach|) In elastic collision e=1 `therefore |RVOS|=|RVOA|` |
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| 323. |
A gun which fires small balls of mass `20 g` is firing `20 balls` per second on the smooth horizontal table surface `ABCD`. If the collision is perfectly elastic and balls are striking at the centre of table with a speed of `5 m//s` at an angle of `60^(@)` with the vertical just before collision, then force exerted by one of the legs on ground is (assume total weight of the table is `0.2 kg`) A. `0.5 N`B. `1 N`C. `0.25 N`D. `0.75 N` |
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Answer» Correct Answer - B Force exerted by one leg on the ground `N = (1)/(4) xx ["Total force"]` `= (1)/(4) ["wt + rate of change of momentum"]` `= (1)/(4) [Mg + n(mv cos 60^(@)) xx 2] = 1N` |
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| 324. |
A body of mass `M_(1)` collides elastically with another mass `M_(2)` at rest. There is maximum transfer of energy when :A. `M_(1) gt M_(2)`B. `M_(1) lt M_(2)`C. `M_(1) = M_(2)`D. same for all values of `M_(1)` and `M_(2)` |
| Answer» Correct Answer - C | |
| 325. |
A particle of mass `m` travelling with velocity `v` and kinetic energy `E` collides elastically to another particle of mass `nm`, at rest. What is the fraction of total energy retained by the particle of mass `m`?A. `((n+1)/n)^(2)`B. `((n+1)/(n-1)^(2))`C. `((n-1)/(n+1))^(2)`D. none of these |
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Answer» Correct Answer - C `mv=mv_(1)+nmv_(2)` `v=v_(1)+mv_(2)=v_(2)v_(1)` `implies v_(2)=(2v)/((n+1)),v_(1)=((1-n)/(1+n))v` `(KE_(1))/(KE_(2))=(1/2mv_(1)^(2))/(1/2mv^(2))=((1-1)/(n+1))^(2)` |
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| 326. |
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» Correct Answer - A Using `m_Ar_A=m_Br_B` or `(1)(r_A)=(2)(r_B)` or `r_A=2r_B` and `r_A+r_B=90m` Solving these two equations, we get `r_A=60m` and `r_B=30m` i.e. COM is at height 60 m from the ground at time `t=0`. Further, `a_(COM)=(m_Aa_A+m_ga_B)/(m_A+m_B)` `=g=10m//s^2` (downwards) as `a_A=a_g=g` (downwards) `u_(COM)=(m_Au_A+m_Bu_B)/(m_A+m_B)` `=((1)(200)-(2)(50))/(1+2)=(100)/(3)m//s` (upwards) Let, h be the height attained by COM beyond 60m. Using, `v_(COM)^2=u_(COM)^2+2a_(COM^h)` or `0=(100/3)^2-(2)(10)h` or `h=(100)^2/180=55.55m` Therefore, maximum height attained by the centre of mass is `H=60+55.55=115.55m` |
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| 327. |
Figure 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. The velocities of `R` and `S` were equal at the mid time of the collision. iii. The mass of `R` was greater than mass of `S`. Which of the following is true?A. I onlyB. II onlyC. I and II onlyD. I, II and III |
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Answer» Correct Answer - D `COLM : m_(R ) (0.8) + m_(R ) (0.2) + m_(S)(1.0)` `rArr 0.6 m_(R ) = m_(S) rArr m_(R ) gt m_(S)` |
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| 328. |
Figure 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. The velocities of `R` and `S` were equal at the mid time of the collision. iii. The mass of `R` was greater than mass of `S`. Which of the following is true?A. ionlyB. ii onlyC. i and ii onlyD. i,ii, and iii |
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Answer» Correct Answer - D Since both have positive final velocities hene, both moved in the same direction after collision. ii At `t=2s`, both had equal velocities. iii By conservation of linear momentum, `m_(1)(0.8)=m_(1)(0.2)+m_(2)(1)=m_(1)/m_(2)=5/3impliesm_(1)gtm_(2)` |
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| 329. |
A uniform chain of mass M and length L is held verticallyi in such a way that its lower end just touches the horizontal floor. The chain is released from rest in this position. Any portion that strikes the floor comes to rest. Assuming that the chain does not form a heap on the floor, calculate the force exerted by it on the floor when a length x has reached the floor. |
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Answer» Let us consider a small element at distance x from the floor of length dy So, `dm=M/Ldx` so the velocity with which the element will strike the floor is `v=sqrt(2gx)` ltbr.`:.` so the momentum transferred to the floor is `M=(dm)v=M/L.dxsqrt(2gx)` [because the element comes to rest] So, the forec exerted on the floor changes in momentum ils givne by `F_1(dM)/(dt)=M/L.(dx)/(dt)sqrt(2g)` Because `v=((dx)/(dt))=sqrt(2gx)` for the chain element `F_=M/L.(sqrt(2gx))^` `=M/L.2gx=(2Mgx)/L` Againforce the force exerted due to x lengh of the chian on the floor due to its own weight is given by `W=M/L(x)xxg=(Mgx)/L` so the total force exerted is given by `F=F_1+W` `=(2Mgx)/L+(Mgx)/L` `=(3Mgx)/L` |
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| 330. |
A steel ball is suspended by a light in extensible string of length l from a fixed point `O`. When the ball is in equilibrium it just touches a vertical wall as shown in the figure. The ball is first taken aside such that string becomes horizontal and then released from rest. If coefficient of restitution is e, then find the maximum deflection of the string after `nth` collision. |
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Answer» Let `v_(0)` be the spee do fhte ball just before first collisiion and `v_(1)` be the speed just after first collision. From the definition of the coefficient of restitution. `v_(1)=ev_(0)`………..i Just before second collision seed of the ball will be `v_(1)` only. if `v_(2)` be the speed just after second collision then `v_(2)=ev_(1)=e^(2)v_(0)`...............ii Similarly `V_(n)=e^(n)v0`..............iii If after `nth` collisionmaximum deflection of the strig from veritcal is `theta` then from COE, we get `=1/2mv_(n)^(2)=mg(1-costheta)` `implies e^(2n)v_(0)^(2)=2gl(1-costheta)`.........iv also from COE `v_(0)^(2)=2gl....................v` frm iv and `v` we get `theta=cos^(-1)(1-e^(2n))` |
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| 331. |
If the net external forces acting on the system oif particles is zero, then which of the following may vary?A. Momentum of the systemB. Velocity of center of massC. Position of center of massD. None of the above |
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Answer» Correct Answer - D The center of mass under the given condition may be at rest or may be moving with constant velocity. |
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| 332. |
If net force on a system of particles is zero. Then `{:(,"Column I",,"Column II",),((A),"Acceleration of centre of mass",(p),"Constant",),((B),"Velocity of centre of mass",(q),"Zero",),((C ),"Momentum of centre of mass",(r ),"May be zero",),((D),"Velocity of an individual particle of the system",(s),"May be constant",):}` |
| Answer» Correct Answer - A -q, B-r, C`r`r,D`rightarrow`s | |
| 333. |
Shown in the figure is a system of three particles of mass `1 kg, 2 kg` and `4 kg` connected by two springs. The acceleration of `A. B` and `C` at any instant are `1 ms^(-2), 2 ms^(-2)` and `1//2 ms^(-2)` respectively directed as shown in the figure external force acting on the system isA. `1N`B. `7N`C. `3N`D. none of these |
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Answer» Correct Answer - C The aceeleration of centre of mass of the system `a_(cm)=(|m_(1)veca_(1)=vecm_(2)a_(2)+m_(3)veca_(3)|)/(m_(1)+m_(2)+m_(3))` `implies` the net force acting of the system `=|m_(1)veca_(1)+vecm_(2)a_(2)+m_(3)veca_(3)|` `implies F_(net)=(m_(1)a_(1)+m_(2)a_(2)-m_(3)a_(3))` `=[(1)(1)+(2)(2)-1/2(4)]N=3N` |
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| 334. |
A ping-pong ball of mass `m` is floating in air by a jet of water emerging out of a nozzle. If the water strikes the ping-gong ball with a speed `v` and just after collision water falls dead, the rate of flow of water in the nozzle is equal toA. `(2mg)/V`B. `(mV)/g`C. `(mg)/V`D. none of these |
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Answer» Correct Answer - C The impact force `F=(/_p)/(/_ )` where `/_p=` change of momentum of water of mass `/_m` striking the pendulum with a speed `v` during the `/_ `. Since water falls dead after collision with the ping pong ball. `/_p=/_mvimpliesF=v(/_m)/(/_ )` where `(/_m)/(/_ )=` rate of flow of wate in the nozzle. Since the ball is in equlibrium `F-mg=0impliesF=mg` ` implies(v/_m)/(/_ )=mgimplies(/_m)/(/_ )=(mg)/v` |
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| 335. |
The collision of two balls of equal mass takes place at the origin of coordinates. Before collision, the components of velocities are (`v_(x) = 50cm^(-1)` , `v_(y) = 0`) and `(v_(x) = -40cm^(-1)` and `v_(y) = 30cms^(-1)`. The first ball comes to rest after collision. The velocity (components `v_(x)` and `v_(y)` respectively) of the second ball areA. `10 and 30 cms^(-)`B. `30 and 10cms^(-1)`C. `5 and 15cms^(-1)`D. `15 and 5cms^(-1)` |
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Answer» Correct Answer - A x-direction `m xx 50-mxx40 = mxx0+mv_(x)` `rArr v_(x) = 10cm//s` y-direction, `mxx0 = mxx0 + mv_(y) rArr v_(y) = 30cm//s` |
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| 336. |
Two particles of mass `m_(A)` and `m_(B)` and their velocities are `V_(A)` and `V_(B)` respectively collides. After collision they interchanges their velocities, then ratio of `m_(A)/m_(B)` isA. `v_(A)/v_(B)`B. `v_(B)/v_(A)`C. `(v_(A) + v_(B))/(v_(B) - v_(A))`D. 1 |
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Answer» Correct Answer - D By law of conservation of momentum. `m_(A)v_(A) + m_(B)v_(B) = m_(A)v_(B) + m_(B)v_(A)` `m_(A)(v_(A)-v_(B))= m_(B)(v_(A)-v_(B))` `rArr (m_(A))/(m_(B)) = 1` |
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| 337. |
Two masses, `m_(1)` and `m_(2)` , are moving with velocities `v_(1)` and `v_(2)`. Find their total kinetic energy in the reference frame of centre of mass. |
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Answer» `K=1/2m_(1)v_(1c)^(2)+1/2m_(2)v_(2c)^(2)`………..i where `v_(1c)` and `v_(2c)` are velocities relatives to the `CM` `v_(1c)=v_(1)-v_(cm)=v_(1)-((m_(1)v_(1)+m_(2)v_(2))/(m_(1)+m_(2)))=(m_(2)(v_(1)-v_(2)))/(m_(1)+m_(2))` `v_(2c)=v_(2)-v_(CM)=v_(2)-((m_(1)v_(1)+m_(2)v_(2))/(m_(1)+m_(2)))=(m_(1)(v_(2)-v_(1)))/(m_(1)+m_(2))` putting these eqn i `K=1/2(m_(1)m_(2))/((m_(1)+m_(2)))(v_(1)-v_(2))^(2)` |
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| 338. |
A body of mass `3 kg` moving with a velocity of `4 m//s` towards left collides head on with a body of mass `4 kg` moving in opposite direction with a velocity of `3 m//s`. After collision the two bodies stick together and move with a common velocity which isA. zeroB. `12 m//s` towards leftC. `12 m//s` towards rightD. `12/7` m//s towards left |
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Answer» Correct Answer - A Initial momentum `3xx4=4xx(-3)=0` Final momentum will also be zero but both stick together, hence both come at rest. |
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| 339. |
Two masses `nm` and `m`, start simultaneously from the intersection of two straight lines with velocities `v` and `nv` respectively. It is observed that the path of their centre of mass is a straight line bisecting the angle between the given straight lines. Find the magnitude of the velocity of centre of mass. [Here `theta=`angle between the lines] |
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Answer» `vecv_(cm)=(m_(1)vecv_(1)+m_(1)vecv_(2))/(m_(1)+m_(2))` `vecv_(cm)=(nmvhati+mnvcosthetahati+nmv sin theta hatj)/(m+nm)` `v_(cm)=sqrt((nmv+mnvcostheta)^(2)+(nmvsintheta)^(2))/(m(1+n))` `v_(cm)=(nmvsqrt((1+costheta)^(2)+(sintheta)^(2)))/(m(1+n))` `=(nvsqrt(1+cos^(2)theta+2costheta+sin^(2)theta))/((1+n))` `=((nv)/(n+1))sqrt(2costheta/2)` `v_(cm)=(2nv"cos" theta/2)/(n+1)` |
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| 340. |
Total work of pseudo forces in center of mass frame. Show that total work done in center of mass frame on all the particles of a system by pseudo forces due to acceleration of mass center is zero. |
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Answer» Let acceleration of mass centre relative to an inertial is `vec(a)_(c )`. Pseudo force on `I^th` particle of mass `m_i` in center of mass frame is `(-m_(i) vec(a)_(c ))`. Let displacement of Ith particle in a time interval is `Delta vec(r )_(i)` relative to the center of mass frame. Total work of pseudo forces on all the particles in center of mass frame can now be expressed by the expression `Sigma(-m_(i) vec(a)_(c )).Delta vec(r )_(i) = -vec(a)_(c).Sigma(m_iDelta vec(r )_(i))=-vec(a)_(c ).vec(0) = 0` |
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| 341. |
A mass of `0.5 kg` moving with a speed of `1.5 m//s` on a horizontal smooth surface, collides with a nearly weightless spring of force constant `k =50 N//m` The maximum compression of the spring would be.A. 0.12mB. 1.5mC. 0.5mD. 0.15m |
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Answer» Correct Answer - D From law of conservation of energy, `1/2mv^(2) = 1/2 xxkx^(2) rArr (0.5)(1.5)^(2) = 50x^(2)` `rArr x^(2) = (1.25/50) rArr x = 0.15` |
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| 342. |
Two balls `A` and `B` each of mass `m` are placed on a smooth ground as shown in the figure. Another ball `C` of mass `M` arranged to the right of ball `B` as shown. If a velocity `v_(1)` is given to ball `A` in rightward direction, consider two cases. Case-I `M gt m` and case-II `M lt m`. Take all the collisions perfectly elastic (`e = 1`), find the number of collision in case-I and case-II. |
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Answer» Correct Answer - a. 2, b. 3` a. First collision will be between balls `A` and `B`. Since both the balls are of same mass, after collition `A` will come to rest and `B` will move with `v_(1)`, now it will collide to `C`. If after this collision, velocities of balls `B` and `C` are `v_(B)` and `v_(C)` respectively, we have `v_(R)=((m-M)/(m+M))v_(I),v_(C)=((2m)/(m+M))v_(1)` Here as we have `Mltm,ltv_(C)` and both are positive thus both the balls are moving forward and will not collide again, hence there are total two collision. b. If in second collision we have `Mgtm`, we get `v_(B)` negative, so ball `B` moves in backward direction after between collision and will strike again to ball `A` (which is at rest to the left of it) and come to rest and ball `A` will more to the left with speed `v_(B)`. Then now there are total three collision. |
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| 343. |
The helicopter has a mass `m` and maintains its height by imparting a downward momentum to a column of air defined by the slipstream boundary as shown in figure. The propeller blades can project a downward air speed `v_0`, where the pressure in the stream below the blades is atmospheric and the radius of the circular cross section of the slipstream is `r`. Neglect any rotational energy of the air, the temperature rise due to air friction and any change in air density `rho`. If the power is doubled, the acceleration of the helicopter is :- A. `(mg)/(2r) sqrt(mg)/(pi rho)`B. `(mg)/(r ) sqrt(mg)/(pi rho)`C. `(2 mg)/(r ) sqrt(mg)/(pi rho)`D. `(mg)/(r ) sqrt(mg)/(2pi rho)` |
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Answer» Correct Answer - A Thrust `= F = rhoAv^(2) " " [A = pir^(2)]` at equilibrium `rhoAv^(2) = mg` Power `P = Fv = rhoAv^(3)` `= (mg)/(r ) sqrt((mg)/(p pi))` |
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| 344. |
At one instant, the centre of mass of a system of two particles is located on the x-axis at `x=3.0m` and has a velocity of `(6.0m//s)hatj`. One of the particles is at the origin, the other particle has a mass of `0.10kg` and is at rest on the `x-`axis at `x=12.0m`. (a) What is the mass of the particle at the origin? (b) Calculate the total momentum of this system. (c) What is the velocity of the particle at the origin? |
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Answer» Correct Answer - A::B::C::D (a) `x_(CM)=(m_1x_1+m_2x_2)/(m_1+m_2)` `implies 3=(m_1(0)+(0.10)(12))/(m_1+0.1)` Solving this equation we get, `=m_1+0.3kg` (b) `P_(CM)=m_(CM)v_(CM)` `=(0.1+0.3)(6hatj)` `=(2.4hatj)kg*m//s` (c) `P_(CM)=P_1+P_2` `:. (2.4hatj)=(0.3)v_1+(0.1)(0)` `:. v_1=(8hatj)m//s` |
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| 345. |
A rocket with a lift-off mass `3.5xx10^4 kg` is blasted upwards with an initial acceleration of `10m//s^2`. Then the initial thrust of the blast isA. `3.5xx10^(5) N`B. `7.0xx10^(5)N`C. `14.0xx10^(5)N`D. `1.75xx10^(5)N` |
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Answer» Correct Answer - B Initial thrust of the blast =m(g+a) `=3.5xx10^(4) xx20` `=7xx10^(5)N` |
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| 346. |
A continuous stream of particles, of mass `m` and velocity `r`, is emitted from a source at a rate of `n` per second. The particles travel along a straight line, collide with a body of mass `M` and get embedded in the body. If the mass `M` was originally at rest, its velocity when it has received `N` particles will beA. `(mvn)/(Nm+n)`B. `(mvN)/(Nm+M)`C. `(mv)/(Nm+M)`D. `(Nm+M)/(mv)` |
| Answer» Correct Answer - B | |
| 347. |
A continuous stream of particles, of mass `m` and velocity `r`, is emitted from a source at a rate of `n` per second. The particles travel along a straight line, collide with a body of mass `M` and get embedded in the body. If the mass `M` was originally at rest, its velocity when it has received `N` particles will beA. `(mvn)/(Nm + n)`B. `(mvN)/(Nm + M)`C. `(mv)/(Nm + M)`D. `(Nm + M)/(mv)` |
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Answer» Correct Answer - B `Nmv = (M + Nm)v_(f) rArr v_(f) = (mvN)/(M + Nm)` |
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| 348. |
A bob of mass `10m` is suspended through an inextensible string of length `l`. When the bob is at rest in equilibrium position, two particles, each of mass `m`, strike it as shown in Fig. The particles stick after collision. Choose the correct statement from the following: A. Impulse in the string due to tension is `2"mu"`B. Velocity of the system just after collision is `v=(usqrt(3))/14`C. Loss of energuy is `137/28"mu"^(2)`D. Loss of energy is `137/56"mu"^(2)` |
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Answer» Correct Answer - A Momentum in vertical direction `"mu"cos60^(@)+3"mu"cos60^(@)=4"mu"cos60^(@)=2"mu"` This momentum becomes zero due to impulse in string. Hence, impulse in string `=2"mu"` Conservation of linear momentum in horizontal direction `m(3u)sin60^2-"mu"sin60^(@)=12mv` `impliesv=(sqrt(3)u)/12` Loss of energy `=1/2m(9u^(2))+1/2"mu"^(2)-1/2xx12mxx((sqrt(3)u)/12)^(2)` `=(39"mu"^(2))/8` |
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| 349. |
Two balls are thrown simultaneously in air. The acceleration of the centre of mass of the two balls while in airA. depends on the direction of the motion of the balls.B. depends on the masses of the two balls.C. depends on the speeds of two ballsD. is equal to g |
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Answer» Correct Answer - D Both the balls in airhve acceleration g downwards. Hence the acceleration of their center of mass will also be g downwards. |
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| 350. |
The center of mass of a system of two particles divides the distance between them.A. in inverse ratio of square of masses of particles.B. in direct ratio of square of masses of particles.C. in inverse ratio of masses of particles.D. in direct ratio of masses of particles. |
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Answer» Correct Answer - C `r alpha 1/m` |
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