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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.
| 101. |
Assertion: A given force applied in turn to a number of different masses may cause the same rate of change in momentum in each but not the same acceleration to all. Reason: `F=(dp)/(dt)` and `a=F/m`A. (a) If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.B. (b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.C. (c) If Assertion is true, but the Reason is false.D. (d) If Assertion is false but the Reason is true. |
| Answer» Correct Answer - A | |
| 102. |
A ball is dropped from height 10 m . Ball is embedded in sand 1 m and stops, thenA. (a) only momentum remains conservedB. (b) only kinetic energy remains conservedC. (c) both momenutm and kinetic energy are conservedD. (d) neither kinetic energy nor momentum is conserved |
| Answer» Correct Answer - A | |
| 103. |
The average resisting force that must act on `5kg` mass to reduce its speed from 65 to `15ms^-1` in `2s` isA. (a) `12.5N`B. (b) `125N`C. (c) `1250N`D. (d) None of these |
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Answer» Correct Answer - B `Impu lse=F_(av)(Deltat)=|DeltaP|` `:. F_(av)=(|DeltaP|)/(Deltat)=(m(v_i-v_f))/(Deltat)` `=((5)(65-15))/(2)` `=125N` |
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| 104. |
A trolley of initial mass `m_0` is kept over a smooth surface as shown in figure. A constant force F is applied on it. Sand kept inside the trolley drains out from its floor at a constant rate of `(mu)kg//s`. After time t find: (a) total mass of trolley and sand. (b) net force on the trolley. (c) velocity of trolley. |
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Answer» Correct Answer - A (a) Total mass of trolley and sand `m="initial mass-mass of sand drained out"` or `m=m_0-mut` (b) Let v is the velocity of trolley at time t. Then, velocity of sand drained out is also v or relative velocity is zero. Hence, no thrust force will act in this case. Therefore, `F_(n et)=F` (c) `F_(n et)=F` or `ma=F` or `m(dv)/(dt)=F` or `(m_0-mut)(dv)/(dt)=F` `:. int_0^vdv=int_0^t(Fdt)/(m_0-mut)` Solving this equation we get, `v=1/mulog_e((m_0)/(m_0-mut))` |
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| 105. |
Sand drops from a stationary hopper at the rate of `5kg//s` on to a conveyor belt moving with a constant speed of `2m//s`. What is the force required to keep the belt moving and what is the power delivered by the motor, moving the belt? |
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Answer» Correct Answer - A::B Relative velocity of sand is `2m//s` in backward direction. Since mass is increasing, therefore thrust force is in the direction of relative velocity (backwards). Thrust force `=v_r(+(dm)/(dt))` `=(5)(2)=10N` (backwards) `:.` Force needed `(F_(ext))` to move the belt with constant velocity `(F_(n et)=0)` is `F_(ext)=F=10N` (in forward direction) `P=F_(ext) v=20W` |
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| 106. |
A metal ball falls from a height of 32 metre on a steel plate. If the coefficient of restitution is 0.5, to what height will the ball rise after second bounceA. 2 mB. 4mC. 8mD. 16 m |
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Answer» Correct Answer - A `h_(n) = he^(2n) = 32(1/2)^(4) = 32/16 = 2m`(here , n=2, e=1//2)` |
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| 107. |
An object of mass `40kg` and having velocity `4 m//s` collides with another object of mass `60 kg` having velocity `2 m//s` . The loss of energy when the collision is perfectly inelastic is |
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Answer» `m_91) = 40kg, m_(2) = 60kg, v_91) = 4ms^(-1) , v_(2) = 2ms^(-1)` `therefore DeltaK = (1/2)(m_(1)m_(2))/(m_(1) + m_(2))(v_(1) - v_(2))^(2) = 1/2(40 xx 60)/(40 + 60)(4-2)^(2) = 48J` |
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| 108. |
A wedge having a vertical slot in it is placed on smooth horizontal surface as shown in the figure. Two blocks are arranged as shown in the figure. The system is released from rest calculate the speed of the wedge when block 1 comes down a distance `h`. |
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Answer» We define system wedge `+` block 1 and block 2 Observation: No external force acts on the system in horizontal direction: Conclusion: Both the linear momentum direction. Analysis of Problem: If block 1 moves down a distance h with will be conserved for the system. respect to wedge, the block 2 will also slide towards left by the same distance with respect to wedge. As wedge is not moving in vertical direction, the displacement of block 1 with respect to wedge will be same as with respect to the ground. The linear momentum of the system in horizontal direction is conserved. Initial linear momentum of the system is zero. As block 2 moves towards left, to conserve linear momentum the wedge will start moving towards right. As block 1 is moving the slot in the wedge, the velocity of block 1 horizontal direction will be same as the velocity of wedge. Block 1 will also move in vertical direction with respect to wedge as well as with respect to ground. The velocity of block 2 with respect to wedge will be same as velocity of block I in vertical direction as both are connected with same string. Final step: Let velocity of block 1 in vertical direction be `v`. Hence velocity of block 2 with respect to wedge will be `v` towards left. Let velocity of wedge towards right be `V`. Hence velocity of block 2 towards left (with respect to ground) will be (`v - V`). Now using conservation of linear momentum for the system in horizontal direction. `(M+m)V-m(v-V)=0` as `m=3mimpliesv=5V`..............i Now using conservation of mechanical energy for the system `/_K+/_U=0` `[(1/2(M+m)V^(2)+1/2m(v-V)^(2)+1/2mv^(2))-0]+[-mg]=0` `=1/2(3m+m)V^(2)+1/2m(5V-V)^(2)+1/2m(5V)^(2)=mgh` `V=sqrt((2gh)/45)` |
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| 109. |
Figure shows a fixed wedge on which two blocks of masses `2 kg` and `3 kg` are placed on its smooth inclined surfaces. When the two blocks are released from rest, find the acceleration of centre of mass of the two blocks. |
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Answer» Block `A` slides with aceleration `g//2` and block `B` slides wth acceleration `sqrt(3)g//2`. Now the acceleration of centre of mass of the system of blocks `A` and `B` can be given both in `x` and in `y` directions as: `a_(x)=(3xx(sqrt3g)/2cos60^(@)-2xxg/2cos30^(@))/5=sqrt(3g)/20` and `a_(y)=(3xx(sqrt(3)g)/2sin60^(@)+2xxg/2sin30^(@))/5=(11g)/20` Thus acceleration of the centre of mass of the system is given as `a_(CM)=sqrt(a_(x)^(2)+a_(r)^(2)=sqrt((sqrt(3))/(20)6+((11g)/(20)^(2))=sqrt(31)/1g` |
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| 110. |
A boy of mass 25 kg stands on a board of maas 10 kg which in turn is kept on a frictionless horizontal ice surface. The boy maks a jump with a velocity component 5m/s in a horizontal direction with respect to the ice. With what velocity does the board recoil? with what rate are tehboy and theboard separting from each other? |
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Answer» Consider the board boy as a system. The exterN/Al forces on this system are a. weight of the system and b. normal contact force by the ice surface. Both these force are vertical and there is no exterN/Al force in horizontal directio. The horizontal component of linear momentum of the board+boy system is therefore constant. If the board redcoils at a speed v `0=(25kg)xx(5m/s)-(10kg)v` or, `v=12.5m/s` The boy and the board are separating wilth a rate `5m/s+12.5m/s=17.5m/s` |
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| 111. |
A railroad car of mass M is at rest on frictionless rails when a man of mass m strts moving on the car towards the engine. If the car recoils with a speed v backward on the rails, with what velocity is the man aproaching the engine? |
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Answer» Correct Answer - A A railrod car of ma sM is at rest o frictionless rails when a man of mass m starts movng on the casr towards the engine. The car recoils with a speed v backward on the rails. Let the mass moving with a velocity x w.r.t the engine. `:.` The velocity of the mass w.r.t earth is (x-v) towards right. `V_(CM)=0(Initially at rest)` `:.-0=-Mv+m(x-v)` `rarr Mv=m(x-v)` `rarr mx=Mv+mv` `rarr x=((M+m)/m)v` `=(1+M/m)v` |
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| 112. |
Figure shows a flat car of mass `M` on a frictionless road. A small massless wedge is fitted on it as shown. A small ball of mass `m` is released from the top of the wedge, it slides over it and falls in the hole at distance `l` from the initial position of the ball. Find the distance the flat car moves till the ball gets into the hole. |
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Answer» When the bell falls into the hole, with respect to the flat car, the bail travels a horizontal distance `l`. During this motion, to conserve momentum and to maintain the position of centre of mass, the car moves towards left, say by a distance `x`. Thus, the total distance travelled by the ball towards right is `(l-x)`. As centre of mass remains at rest, the change in mass moments of the two (ball and car) about any point must be equal to zero. Hence `m(l-x)=MXimpliesm=(ml)/(M+m)` |
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| 113. |
Statement-1 : A particle of mass m strikes a smooth wedge of mass M as shown in the figure. Linear momentum of particles along the inclined surface of wedge is conserved during collision. and Statement-2 : Wege exerts a force on particle perpendicular to inclined face of wedge during collision.A. Statement -1 is True, Statement -2 si True , Statement -2 is a correct explanation for Statement -1.B. Statement -1 is True, Statement -2 si True , Statement -2 is not correct explanation for Statement -1.C. Satement-1 is True, Statement-2 is Satement-1 is True, Statement-2 is False.D. Satement-1 is False, Statement-2 is True. |
| Answer» Correct Answer - A | |
| 114. |
A block of mass M is placed on the top of a bigger block of mass 10 M as shown in figure. All the surfaces are frictionless. The system is released from rest, then the distance moved by the bigger block at the instant the smaller block reaches the ground : |
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Answer» I the bigger block moves towards right by a distnce X, the smaller block will move towasrds left by a distance (2.2m-X). Taking the two blocks together as the system, there is no horizontal exterN/Al force on it. The centre of mass which was ast rest initially, will remain at the sme horizontal position. thus, ` M(2.2m-X)=MX` or, `2.2m=11x` |
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| 115. |
Two identical blocks, each having mass M, are placed as shown in figure. These two blocks A and B are smoothly conjugated, so that when another block C of mass m passes from A to B there is no jerk. All the surfaces are frictionless, and all three blocks are free to move. Block C is released from rest, then A. when in is at the highest position on `B`B. when in is at the lowest position and moving left.C. when in is at `C`D. when in is at lowest position and moving right. |
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Answer» Correct Answer - B Till then m remains on `B, B` will continue to acelerate due to contact force between `m` and `B`. Hence, velocity of `B` will be maximum when finally `m` leaves `B`. |
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| 116. |
Two identical billiard balls undergo an oblique elastic collision. Initially, one of the balls is stationary. If the initially stationary ball after collision moves in a direction which makes an angle of `37^(@)` with direction of initial motion of the moving ball, then the angle through which initially moving ball will be deflected isA. `37^(@)`B. `60^(@)`C. `53^(@)`D. `gt53^(@)` |
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Answer» Correct Answer - C When two indentical balls undergo elastic oblique collisions with one of them stationary, the after collision they will move at right angles to each other. Alternative method I: Just like in previous question, you can proceed using centre of mass frame of reference. Alternative method II: Using basic equation of conservation of momentum, conservation of energy and coefficient of restitution, you can have the desired result. |
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| 117. |
A pendulum consists of a wooden bob of mass `M` and length `l`. A bullet of mass `m` is fired towards the pendulum with a speed `v`. The bullet emerges immediately out of the bob from the other side with a speed of `v//2` and the bob starts rising. Assume no loss of mass of bob takes place due to penetration. If the bob stops where the string becomes horizontal then `v` isA. `(2M)/m sqrt(3gl)`B. `(2M)/msqrt(5gl)`C. `(2M)/msqrt(gl)`D. `(2M)/msqrt(2gl)` |
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Answer» Correct Answer - D Let velocity of bob after collision be `v_(1)`. Then `Mv_(1)=(mv)/2impliesv_(1)(mv)/(2M)` Height rasised by bob `=l ` `l=v_(1)^(2)/(2g)impliesl=(m^(2)v^(2))/(8M^(2)g)impliesv=(2M)/msqrt(2gl)` |
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| 118. |
A pendulum consists of a wooden bob of mass `M` and length `l`. A bullet of mass `m` is fired towards the pendulum with a speed `v`. The bullet emerges immediately out of the bob from the other side with a speed of `v//2` and the bob starts rising. Assume no loss of mass of bob takes place due to penetration. If the bob is just able to complete the circular motion, then tension at the lowest point just when the bob starts rising will beA. `6Mg`B. `5Mg`C. `3Mg`D. `Mg` |
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Answer» Correct Answer - A Here `v_(1)=sqrt(5gl)` `T=Mg+(Mv_(1)^(2))/l=Mg+(M5gl)/l=6Mg` |
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| 119. |
A pendulum consists of a wooden bob of mass `m` and length `l`. A bullet of mass `m_(1)` is fired towards the pendulum with a speed `v_(1)`. The bullet emerges out of the bob with a speed of `(v_(1))//3` and the bob just completes motion along a vertical circle, then `v_(1)` isA. `m/m_(1)sqrt5gl`B. `(3m)/(2m_(1))sqrt(5gl)`C. `2/3(m/m_(1))sqrt(5gl)`D. `(m_(1)/m)sqrt(gl)` |
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Answer» Correct Answer - B From conservation of linear momentum, `m_(1)v_91) = msqrt5gl) + m_(1)(v_(1)/3)` or `v_(1) = 3/2m/m_(1)sqrt(5gl)` |
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| 120. |
A gun (mass `= M`) fires a bullet (mass `= m`) with speed `v_(r)` relative to barrel of the gun which is inclined at an angle of `60^(@)` with horizontal. The gun is placed over a smooth horizontal surface. Find the recoil speed of gun. |
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Answer» Let the recoil seed of gun is `v`. Taking gun`+`bullet as the system. Net external force on the system in horizontal directioin zero. initial the system was rest. Therfore, applying the principle of conservation of inear momentum in horizontal direction, we get `Mv-m(v_(r)cos60^(@)-v)+0` `:. v=(mV_(r)cos60^(@))/(M+m)` or `v=(mV_(r))/(2(M+m))` |
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| 121. |
Five identical elastic balls are so suspended with strings of equal length in a row that the distance between adjacent balls are very small. If the extreme right ball is moved aside and released then A. one extreme left hand ball will bounce offB. two extreme left hand balls will bounce offC. three extreme left hand balls will bounce offD. all the left hand four balls will bounce off |
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Answer» Correct Answer - A Clear from figure |
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| 122. |
Two identical balls `A` and `B`. each of mass `2 kg` and radius `R`, are suspended vertically from inextensible strings as shown in Fig. The third ball `C` of mass `1 kg` and radius `r=(sqrt(2)-1)R` falls and hits `A` and `B` symmetrically with `10 m//s`. Speed of both `A` and `B` just after the collision is `3 m//s`. Impulse provided by each sting during collision isA. `6sqrt(2)Ns`B. `12Ns`C. `3sqrt(2)Ns`D. `6Ns` |
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Answer» Correct Answer - D impulse provided by each string `I_(1)=Isintheta=6Ns` |
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| 123. |
A ball of mass in `m= 1 kg` is hung vertically by a thread of length `l = 1.50 m`. Upper end of the thread is attached to the ceiling of a trolley of mass `M= 4 kg`. Initially, the trolley is stationary and it is free to move along horizontal rails without friction. A shell of mass `m=1 kg`, moving horizontally with velocity `v_(0) = 6 m//s` collides with the ball and gets stuck with it. As a result, the thread starts to deflect towards right. The maximum deflection of the thread with the vertical isA. `cos^(-1)(4/5)`B. `cos^(-1)(3/5)`C. `cos^(-1)(2/3)`D. `cos^(-1)(3/4)` |
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Answer» Correct Answer - A When shell strikes the ball and gets stuck with it, combined body of mass `2m` starts to move to the right, Let velocity of the combined body (just after collision) be `v_(1)`. According to law of conservation of momentum, `(m+m) v_(1)=mv_(0)` `v_(1)=(v_(0))/2=3m//s` As soon as the combined body starts to move rig wards, thread becomes inclined to the vertical. Horizontal com tension retards the combined body while trolley accelerate rightwards due to the same component of tension. Inclination of thread with the vertical continues to increase till velocities of both (combined body and trolley) become identical or combined body comes to rest relative to the trolley. Let velocity at that instant of maximum inclination of thread be `v`. according to law of conservation of momentum, `(2m+M)v=2mv_(1)` or `v=1m//s` During collision of ball and shell, a part of energy is lost. But after that there is no loss of energy., Hence, after collision, kinetic energy lost is used up in increasing gravitational potential energy of the combined body. If maximum inclination of threads with the vertical is `theta`, then according to law of conservation of energy. `=1/2(2m)v_(1)^(2)-1/2(2m+M)v^(2)=2mg(l-lcostheta)` `costheta=0.8` or `theta=37^(@)` |
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| 124. |
The two balls shwon in figure are indentical the first moving at a speed v towards right and the second staying at rest. The wall at the extreme right is fixed. Assume all collisions to be elastic. Show that the speeds of the balls remain unchanged after all the collisions have taken place. |
| Answer» Correct Answer - A | |
| 125. |
When a cannon shell explodes in mid air, then identify the incorrect statementA. (a) the momentum of the system is conserved at the time of explosionB. (b) the kinetic energy of the system always increasesC. (c) the trajectory of centre of mass remains unchangedD. (d) None of the above |
| Answer» Correct Answer - B | |
| 126. |
In a carbon monoxide molecule, the carbon and the oxygen atoms are separated by a distance `1.2xx10^-10m`. The distance of the centre of mass from the carbon atom isA. `0.48 xx 10^(-10)`mB. `0.51 xx 10^(-10)`mC. `0.56 xx 10^(-10`mD. `0.69 xx 10^(-10)`m |
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Answer» Correct Answer - D Distance distributes in inverse ratio of masses. Here, `r_(c) = d(m_(0))/(m_(0)+m_(c)) = 1.2 xx 10^(-6)(16/(16+2))` `0.069 xx 10^(-10)`m |
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| 127. |
Statement I: In a two-body collision, the momenta of the particles are equal and opposite to one another, before as well as after the collision when measured in the centre of mass frame. Statement. II: The momentum of the system is zero from the centre of mass frame.A. Statement -1 is True, Statement -2 si True , Statement -2 is a correct explanation for Statement -1.B. Statement -1 is True, Statement -2 si True , Statement -2 is not correct explanation for Statement -1.C. Satement-1 is True, Statement-2 is False.D. Satement-1 is False, Statement-2 is True. |
| Answer» Correct Answer - A | |
| 128. |
Statement I: In a two-body collision, the momenta of the particles are equal and opposite to one another, before as well as after the collision when measured in the centre of mass frame. Statement. II: The momentum of the system is zero from the centre of mass frame.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 - A The momentum of a system of particles from any frame is given by `vecP=mvecv_(CM)` From the centre of mass frame, `vecv_(CM)=vecO` `implies vecP=0` So individual bodies should have momenta is opposite directions to make net momentum zero. |
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| 129. |
A uniform bar of length `12L` and mass `48m` is supported horizontally on two fixed smooth tables as shown in figure. A small moth (an insect) of mass `8m` is sitting on end A of the rod and a spider (an insect) of mass `16m` is sitting on the other end B. Both the insects moving towards each other along the rod with moth moving at speed `2v` and the spider at half this speed (absolute). They meet at a point P on the rod and the spider eats the moth. After this the spider moves with a velocity `v/2` relative to the rod towards the end A. The spider takes negligible time in eating on the other insect. Also, let `v=L/T` where T is a constant having value `4s`. By what distance the centre of mass of the rod shifts during this time?A. (a) `(8L)/(3)`B. (b) `(4L)/(3)`C. (c) `L`D. (d) `L/3` |
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Answer» Correct Answer - A Till `t_1`, rod is stationary. For time `t_2` rod is moving with absolute speed `u(=v//6)` `:.` Displacement of rod `=(v/6)t_2` `=(v/6)((16L)/(v))=(8L)/(3)` |
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| 130. |
In one-dimensional elastic collision, the relative velocity of approach before collision is equal toA. e times the relative velocity of separation after collisionB. relative velocity of separation after collisionC. sum of the velocities of two bodiesD. 1/e times the relative velocity of separation after collision |
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Answer» Correct Answer - A e times the relative velocity of seperation after collision |
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| 131. |
The coefficient of restitution `( e)` for a perfectly elastic collision isA. 1B. zeroC. `oo`D. -1 |
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Answer» Correct Answer - A e=1 |
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| 132. |
A small block of mass `M` move on a frictionless surface of an inclimed from as down is figure . The engle of the inclime suddenly change from `60^(@)` to `30^(@)` at point `B` . The block is initally at rest at `A` Assume the collsion between the block and the incline are totally inclassic `(g = 10m//s^(2)`) The speed of the block at point `C` immediately before is leaves the second incline isA. `sqrt120 m//s`B. `sqrt105 m//s`C. `sqrt30 m//s`D. `sqrt75 m//s` |
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Answer» Correct Answer - B `v_(c )^(2) = v_(B)^(2) + 2g(h_(2) -h _(1)` `rArr = v_(c ) = sqrt(45^(2) + 2 xx 10 xx 3) = sqrt(105) ms^(-1)` |
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| 133. |
The position vectors of three particles of mass `m_(1) = 1kg, m_(2) = 2kg and m_(3) = 3kg` are `r_(1) = ((hat(i) + 4hat(j) +hat(k))`m, `r_(2) = ((hat(i)+(hat(j)+hat(k))`m and `r_(3) = (2hat(i) - (hat(j) -(hat(2k))`m, respectively. Find the position vector of their center of mass. |
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Answer» The position vector of CM of the three particles will be given by `r_(CM) = (m_(1)r_(1) + m_(2)r_(2) + m_(3)r_(3))/(m_(1)+ m_(2) +m_(3))` substituting, the values, we get `r_(cm) = (1(i+4j+k) + 2(i+j+k)+3(2i-j-2k))/(1+2+3)` `=(9i + 3j -3k)/6 rArr r_(cm) =1/2(3i + j -k)m` |
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| 134. |
A particle of mass `1 kg` has velocity `vec(v)_(1) = (2t)hat(i)` and another particle of mass `2 kg` has velocity `vec(v)_(2) = (t^(2))hat(j)` `{:(,"Column I",,"Column II",),((A),"Netforce on centre of mass at 2 s",(p),(20)/(9) unit,),((B),"Velocity of centre of mass at 2 s",(q),sqrt68 " unit",),((C ),"Displacement of centre of mass in 2s",(r ),(sqrt80)/(3) "unit",),(,,(s),"None",):}` |
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Answer» Correct Answer - (A) `rarr` q; (B) `rarr` r; (C ) `rarr` p For `1 kg v_(1) = (2t) hat(j) = 4 hat(i) , a_(1) = 2 hat(i) = 2 hat(i)` for `2 kg v_(2) = t^(2) hat(j) = 4 hat(j) , a_(2) = 2t hat(j) = 4 hat(j)` (A) Acceleration of centre of mass `= (m_(1)a_(1) + m_(2)a_(2))/(m_(1) + m_(2))` `vec(a)_(cm) = (2)/(3) hat(i) + (8)/(3) hat(j) rArr a_(cm) = sqrt((4)/(9) + (64)/(9)) rArr (sqrt68)/(3) m//s^(2)` `f = ma_(cm) = sqrt68 N` (B) Velocity of centre of mass `vec(v)_(cm) = (m_(1)v_(1) + m_(2)v_(2))/(m_(1) + m_(2)) = ((4)/(3)hat(i) + (8)/(3) hat(j)) m//s` `rArr |vec(v)_(cm) | = sqrt((16)/(9) + (64)/(9)) rArr (sqrt80)/(3)` (C ) `vec(v)_(cm) = (1(2t) hat(i) + 2(t^(2)hat(j)))/(1 + 2) + (2)/(3) that(i) + (2)/(3) t^(2)hat(j)` Displacement `= int_(0)^(2) vec(v)_(cm) dt = [(3)/(2)((t^(2))/(2))^(2) hat(i) + (2)/(3) ((t^(3))/(3)) hat(j) ]_(0)^(2) = (4)/(3)hat(i) + (16)/(9) hat(j)` `rArr | "Dispalcement"| = sqrt(((4)/(3))^(2) + ((16)/(9))^(2)) = (20)/(9)` unit |
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| 135. |
A body of mass `m_(1) ` moving with uniform velocity of 40 m/s collides with another mass `m_(2)` at rest and then the two together begin to moe wit h uniform velocity of 30 m/s. the ratio of their masses `(m_(1))/(m_(2))` isA. `1:3`B. `3:1`C. `1:1.33`D. `1: 0.75` |
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Answer» Correct Answer - B (b) Initial momentum of the sytem `=m_(1) xx 40 + m_(2) xx 0 = 40m_(1)` Final momentum of the system. `=m_(1) xx 40 + m_(2) xx 0 = 40m_(1)` Final momentumof the system, `=(m_(1) + m_(2)) xx 30` ltbrlt By law of conservation of momentum, Initial momentum = Final momentum `40m_(1) = ((m_(1) + m_(2))30` `40m_(1) - 30m_(1) = 30m_(2)` `10m_(1) = 30m_(2)` `m_(1)/m_(2) = 3/` |
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| 136. |
These questions consists of two statements each printed as Assertion and Reason. While answering these question you are required to choose any one of the following five responses. Reason: For inelastic collision, `0leelt1`. Hence, the magnitude of relative velocity of separation after collision is less than relative velocity of approach before collision.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. |
| Answer» Correct Answer - D | |
| 137. |
These questions consists of two statements each printed as Assertion and Reason. While answering these question you are required to choose any one of the following five responses. Reason: In an eleastic collision, the linear momentum of the system is conserved.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 - B In elastic collision magnitude of relative of separation = magnitude of relative velocity of approach. |
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| 138. |
Comprehension # 2 When two bodies collide normally they exert equal and opposite impulses on each other. Impulse `=` change in linear momentum. Coefficient of restitution between two bodies is given by :- `e = |"Relative velocity of separation"|/|"Relative velocity of approach"| = 1`, for elastic collision Two bodies collide as shown in figure. During collision they exert impulse of magnitude J on each other. If the collision is elastic, the value of J is ............. `N-s` :-A. `10//3`B. `5//4`C. `8//3`D. `3//2` |
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Answer» Correct Answer - C By applying conservation of momentum `2(6) + 1(4) = 1v_(2) + 2v_(1), 16 = v_(2) + 2v_(1)` ….(i) By applying newton law of collision `1 = (v_(2) - v_(1))/(2) rArr v_(2) - v_(1) = 2` ……(ii) `rArr v_(2) = (20)/(3), v_(1) (14)/(3)` Impulse `=` change in momentum `= (20)/(3) -4 = (8)/(3) N-s` |
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| 139. |
An insulated particle of mass `m` is moving in a horizontal plane (x - y) along X-axis. At a certain height above the ground, it suddenly explodes into two fragments of masses `m//4 and 3 m//4`. An instant later, the smaller fragment is at `Y = + 15`. The larger fragment at this instant is at :A. `y = -5 cm`B. `y = +20 cm`C. `y = + cm`D. `y = -20 cm` |
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Answer» Correct Answer - A `Deltabar(y) = (m_(1)Deltay_(1) + m_(2)Deltay_(2))/(m_(1) + m_(2))` `rArr 0 = (m)/(4) (15) + (3m)/(4) (y_(2))` `rArr = y_(2) = -5 cm` |
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| 140. |
Two identical smooth balls are projected from points `O` and `A` on the horizontal ground with same speed of projection. The angle of projection in each case is `30^(@)`.The distance between `O` and `A` is `100 m`. The balls collide in mid-air and return to their respective points of projection. If the coefficient of restitution is `0.7`, find the speed of projection of either ball (in `m//s`) correct to nearest integer. (Take `g = 10 ms^(-2)` and `sqrt(3)=1.7)` |
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Answer» Correct Answer - `38m//s` The collision does not alter the vertical component of the velocity of either ball so `t_(1)+t_(2)=T` `implies50/(ucos30^(@))+50/(0.7ucos30^(@))=(2usin30^(@))/10impliesu~~38m//s` |
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| 141. |
Three particles A, B and C of equal mass move with equal speed V along the medians of an equilateral triangle as shown in hgure. They collide at the centroid G of the triangle. After the collision, A comes to test, B retraces its path with the speed V. What is the velocity of C ? |
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Answer» Correct Answer - `vec(v)_(c ) = -vec(v)_(B)` COLM : implies that `vec(v)_(C )` & `vec(v)_(B)` are opposite to each other. |
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| 142. |
A bullet of mass 50 g is fired rom below into the bob of mass 450 g of a long simple pendulum as hown in figure. The bullet premains inside the bob and the bob rises thrugh a height of 1.8 m. Find the speed of the bullet. Take `g=10 m/s^2` |
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Answer» Let the speed of the bullet b v. Let the common velocity of the bullet and the bob, after the bullet is embedded ito the bob, is V. By the principle of conservation of linear momentum, `V=((0.05kg)v)/(0.45kg+0.05kg)=v/10` The string becomes loose and the bob will go up with a deceleration of `g=10m/s^2`. As it comes to rest at a height of 1.8 m, usn the equation `v^2=u^2+2ax, `1.8m=((v/10)^2)/(2xx10m/s^2)` or `v=60 m/s |
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| 143. |
Assertion: A body is thrown with a velocity u inclined to the horizontal at some angle. It moves along a parabolic path and falls to the ground, Linear momentum of the body, during its motion, will remain conserved. Reason: Throughout the motion of the body, a constant force acts on it.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. |
| Answer» Correct Answer - D | |
| 144. |
A particle of mass 1 kg is projected upwards with velocity `60ms^(-1)`. Another particle of mass 2kg is just dropped from a certain height. After 2s, match the following. [Take, g `=10 ms^(-2)]``{:("column1","column2"),("Accelertion of CM","P Zero"),("B Velocity of CM","Q 10 SI unit"),("C Displacement of CM","20 SI unit"),("-","S None"):}` |
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Answer» Correct Answer - A `rightarrow` q, B`rightarrow`p, C `rightarrow` r From `a_(CM)= ((m_(1)a_(1) + m_(2)a_(2))/(m_(1) + m_(2))` `=((1)(-10)+2(-10))/(3) = -10 ms^(-2)` `u_(CM) = ((m_(1)u_(1) + m_(2)u_(2))/(m_(1) + m_(2)`, after 2s `=((1)(40)+(2)(20))/(3) = (40+40)/3 = 80/3` |
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| 145. |
A projectile is fired at a spedd of 100 m/s at an angel of `37^0` above the horizontal. At the highest point, the projectile breaks into two parts of mass ratio 1:3 the smaller coming to rest. Findthe distance from the launching point to the where the heavier piece lands. |
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Answer» See ure. At the highest point, the projectrile has horizontal velocity. The lighter part comes to rest. Hence the heavier part willl move with inceased horizontal velocity. In vertical direction both parts have zero velocity and undergo same acceleration, hence they willl cover equal vertical displacements in a given time. Thus both will hit the ground together As interN/Al forces do not affect the motion of teh centre of mass, the centre of mass hits the ground at teh positin where the origiN/Al projectile would have landed. Te range of the origiN/Al projectile is `x_(CM)=(2u^2sinthetacostheta)/g=(2xx10^4xx3/5xx4/5)/10m` =960m. The centre of mass will hit the grouond at this position. As the smaller block comes to rest after breaking. It falls down vertically and hits the ground at half othe range i.e., at `x=480m`. If the heavier block hits the ground at `x_2` then `x_(CM)=(m_1x_1+m_2x_2)/(m_2+m_2)` `960m=(M/4xx480m+(3M)/4xxx_2)` or, `x_2=1120m` |
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| 146. |
Two particles of masses 2 kg and 4 kg are approaching each other with acceleration `1 ms^(-2)` and `2 ms^(-2)`, respectively, on a smooth horizontal surface. Find the acceleration of center of mass of the system. |
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Answer» The acceleration of center of mass of the system `a_(CM) = |m_(1)a_(2) + m_(2)a_(2)|/(m_(1) + m_(2))` `rArr a_(CM) = |m_(1)a_(2) + m_(2)a_(2)|/(m_(1) + m_(2))` `=(2xx 1-4xx2)/(2+4) = -1 ms^(2)` [-ve sign because direction of 4 kg is opposite to that of 2kg] Since `m_(2)a_(2) gt m_(1)a_(1)` so the direction of acceleration of center of mass will be directed towards `a_(1)` |
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| 147. |
Two particles of mass 1 kg and 0.5 kg are moving in the same direction with speed of `2 m//s` and 6 m/s respectively on a smooth horizontal surface. The speed of centre of mass of the system isA. `(10)/(3) m//s`B. `(10)/(7) m//s`C. `(11)/(2) m//s`D. `(12)/(3) m//s` |
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Answer» Correct Answer - A `V_(cm)=(1xx2+1/2xx6)/(1+1//2)=10/3 m//sec` |
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| 148. |
A particle of mass 4 m which is at rest explodes into three fragments. Two of the fragments each of mass m are found to move with a speed v each in mutually perpendicular directions. The total energy released in the process of explosion is ............A. `(2//3)mv^(2)`B. `(3//2) mv^(2)`C. `(4//3) mv^(2)`D. `(3//4) mv^(2)` |
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Answer» Correct Answer - B `mv_(1)+mvj+2mv_(3)=0` `vec(v)_(3)=-((vi+vj))/2=-v/2(i+j)=-v/(sqrt(2))` `k_(f)=1/2 mv^(2)+1/2 mv^(2)+1/2 2m(v^(2))/2` `k_(f)=(3mv^(2))/2` |
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| 149. |
A ball of mass 50 gm is dropped from a height h=10m. It rebounds losing 75 percent of its kinetic energy. If it remains in contanct with the ground for `Deltat=0.01` sec. the impulse of the impact force is (take `g=10 m//s^(2)`)A. 1.3 N-sB. 1.06 N-sC. 1300 N-sD. 1.05 |
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Answer» Correct Answer - B `v_(1)=sqrt(2gh)=sqrt(2xx10xx10)=10sqrt(2)` `k_(2)=1/4k_(1)rArr v_(2)^(2)=1/4v_(1)^(2)` `:. v_(2)=(v_(1))/2=5sqrt(2)` `|DeltaP|=|-mv_(2)-(mv_(1))|=m|-v_(2)-v_(1)|` `|DeltaP|=50xx10^(-3)xx3/2xx10sqrt(2)=(15xx10^(-1))/(sqrt(2))` `J=DeltaP=1.05 N-s` |
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| 150. |
An ideal spring is permanently connected between two blocks of masses `M` and `m`. The blocks-spring system can move over a smooth horizontal table along a straight line along the length of the spring as shown in Fig.The blocks are brought nearer to compress the spring and then released. In the subsequent motion, A. initially they move in opposite directions with velocities inversely proportional to their massesB. the ratio of their velocities remains constantC. linear momentum and energy of the system remain conservedD. the two blocks will Oscillate about their centre of mass which remains stationary |
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Answer» Correct Answer - A::B::C::D Since the system of two blocks and a spring is placed on as smooth horizontal floor, therefore no external horizontal force acts on the system. Since the system is initially at rest, therefore centre of mass of the system will remain stationary and the block will oscillate with the same frequency. it means option d is correct. Since the system is initially ast rest and centre of mass stationary, therefore centre of mas cannot move or the momentum of the system remains equal to zero. it means, moment of two blocks are numerically equal but they are in opposite directions. therefore, ratio of magnitude of velocity of the two blocks will be in inverse ratio of their masses. Hence, option a is correct. Since ratio of mass is constant. therefore the ratio of their velocity remain constant. it means that option b is also correct. Since the system is on smooth horizontal floor, therefore no energy loss takes place. hence, linear momentum and energy of the system remain conserved. therefore, option c is also correct. |
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