InterviewSolution
Saved Bookmarks
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. |
Two blocks `A` and `B` are joined by means of a slacked string passing over a massless pulley as shown in Figure. The system is released from rest and it becomes taut when `B` falls a distance `0.5 m`. a. Find the common velocity of the two blocks just after the string becomes taut. b. Find the magnitude of impulse on the pulley by the clamp during the small interval while string becomes taut. |
|
Answer» Correct Answer - `sqrt((10))/3m//s, 4/5sqrt(5)Ns` Velocity of `B` just before the string is taut `v_(B)=sqrt(2g/_h)=sqrt10m//s` a. common velocity =v `:. (m_(A)+m_(B))v=m_(B)v_(B)impliesv=v_(B)/3=(sqrt(10))/3m//s` b. magnitude of impulse on `A =` magitude of impulse on `B=1(sqrt(10)-(sqrt(10))/3)=2/3sqrt(10)m//s` Therefore, impulse on pulley `=sqrt(2)` impulse on `A=4/3sqrt(5)Ns` |
|
| 2. |
A body of mass `1 kg` moving with velocity `1 m//s` makes an elastic one dimesional collision with an identical stationary body. They are in contact for brief time `1 sec`. Their force of interaction increases form zero to `F_(0)` linearly in time `0.5s` and decreases linearly to zero in further time `0.5 sec` as shown in figure. Find the magnitude of force `F_(0)` in newton. |
|
Answer» Correct Answer - 2 In the one dimensional elastic collision with one body at rest, the body moving initially comes to rest & the on which was at rest earlier starts moving the velocity that first body had before collision. So, if `m` & `V_(0)` be the mass & velocity of body, the change in momentum `= mV_(0) rArr int Fdt = mV_(0) rArr int Fdt = mV_(0) rArr F = (2mV_(0))/(Deltat) = 2 N` |
|
| 3. |
Consider a two particle system with particles having masses `m_1 and m_2` if the first particle is pushed towards the centre of mass through a distance d, by what distance should the second particle is moved, so as to keep the center of mass at the same position?A. dB. `(m_(2))/(m_(1))d`C. `(m_(1))/(m_(1)+m_(2))d`D. `(m_(1))/(m_(2))d` |
|
Answer» Correct Answer - D `m_(1)Deltax_(1)=m_(2)Delta_(2)` `m_(1)d=m_(2)Deltax_(2)` `Deltax_(2)=(m_(1)d)/(m_(2))` |
|
| 4. |
Two paricle A and B initially at rest, move towards each other under mutual force of attraction. At the instant when the speed of A is V and the speed of B is 2V, the speed of the centre of mass of the system isA. `3`B. `v`C. `1.5v`D. zero |
|
Answer» Correct Answer - D Net external force on system is zero. Therefore centre of mass always remains at rest. |
|
| 5. |
A body of mass `m = 3.513 kg` is moving along the x-axis with a speed of `5.00ms^(-1)`. The magnetude of its momentum is recorded asA. `17.565 kg ms^(-1)`B. `17.56 kgms^(-1)`C. `17.57 kg ms^(-1)`D. `17.6 kg ms^(-1)` |
| Answer» Correct Answer - D | |
| 6. |
Two paricle A and B initially at rest, move towards each other under mutual force of attraction. At the instant when the speed of A is V and the speed of B is 2V, the speed of the centre of mass of the system isA. velocity remains constantB. zeroC. 2vD. 3v/2 |
| Answer» Correct Answer - B | |
| 7. |
Two paricle A and B initially at rest, move towards each other under mutual force of attraction. At the instant when the speed of A is V and the speed of B is 2V, the speed of the centre of mass of the system isA. zeroB. vC. 1.5vD. 3v |
|
Answer» Correct Answer - A `P_(i)=0` `therefore` `P_(f) = 0` or center of mass should be at rest at all instants. |
|
| 8. |
When two blocks connected by a spring move towards each other under mutual interactionA. their velocities are equal and opposite.B. their accelerations are unequal and oppositeC. the force acting on them are equal and oppositeD. their momentum are equal and opposite |
|
Answer» Correct Answer - C::D The spring pulls both the blocks with same force. Hence force on both blocks is equal and opposite since no net external force acts on system and its initial moment is zero, therefore net momentum of system is always zero, hence momentum of blocks are equal and opposite. |
|
| 9. |
A rocket, with an initial mass of `1000kg`, is launched vertically upwards from rest under gravity. The rocket burns fuel at the rate of `10kg` per second. The burnt matter is ejected vertically downwards with a speed of `2000ms^-1` relative to the rocket. If burning ceases after one minute, find the maximum velocity of the rocket. (Take g as constant at `10ms^-2`) |
|
Answer» Correct Answer - A::B::C After `1`min total `600kg` gases will be burnt. `v=u-"gt"+V_r1n(m_0/m)` `=0-10xx60+(2000)1n(1000/400)` `=1232.6m//s` |
|
| 10. |
An initially stationary box on a frectionless floor explodes into two places, places A with mass `m_(A)` and piece B with mass `m_(B)`. Two pieces then move across the floor along x-axis. Graph of position versus time for the two pieces are given. If all the graphs are possible then, in which of the following cases external force must be acting on the box :-A. IIB. VC. VID. I |
|
Answer» Correct Answer - C::D `v_(cm)` is not zero in graph (I), (III) and (VI) |
|
| 11. |
Comprehension # 5 One particle of mass `1 kg` is moving along positive x-axis with velocity `3 m//s`. Another particle of mass `2 kg` is moving along y-axis with `6 m//s`. At time `t = 0, 1 kg` mass is at `(3m, 0)` and `2 kg` at `(0, 9m), x - y` plane is the horizontal plane. (Surface is smooth for question 1 and rough for question 2 and 3) If both the particles have the same value of coefficient of friction `mu = 0.2`. The centre of mass will stop at time `t =` ..........s :-A. `1.5`B. `4.5`C. `3.0`D. `2.0` |
|
Answer» Correct Answer - C `vec(a)_(cm) = (m_(1)vec(a)_(1) + m_(2)vec(a)_(2))/(m_(1) + m_(2)) = (1 (-2hat(i)) + 2 (-2 hat(j)))/(1 + 2)` `= -(2)/(3)(hat(i) + 2hat(j))m//s^(2)` By using `vec(v)_(cm) = vec(u)_(cm) + vec(a)_(cm)t`, we get `t = 3s` |
|
| 12. |
Comprehension # 5 One particle of mass `1 kg` is moving along positive x-axis with velocity `3 m//s`. Another particle of mass `2 kg` is moving along y-axis with `6 m//s`. At time `t = 0, 1 kg` mass is at `(3m, 0)` and `2 kg` at `(0, 9m), x - y` plane is the horizontal plane. (Surface is smooth for question 1 and rough for question 2 and 3) The centre of mass of the two particles is moving in a straight line which equation is :-A. `y = x + 2`B. `y = 4x + 2`C. `y = 2x - 4`D. `y = 2x + 4` |
|
Answer» Correct Answer - B `vec(r )_(cm) = (m_(1)vec(r )_(10) + m_(2) vec(r )_(2))/(m_(1) + m_(2)) = (1 (3 hat(i)) + 2 (9 hat(j)))/(1 + 2) = (hat(i) + 6hat(j))m` `vec(v)_(cm) = (m_(1) vec(v)_(1) + m_(2)vec(v)_(2))/(m_(1) + m_(2)) = (1(3hat(i)) + 2(6hat(j)))/(1 + 2) (hat(i) + 4hat(j))m//s` Now `Deltavec(r )_(cm) = vec(v)_(cm) t rArr vec(r )_(cm) - vec(r)_(cm_(0)) = vec(v)_(cm) t` `rArr (x - 1)hat(i) + (y - 6) hat(j) = (hat(i) + 4hat(j)) t` `rArr x = 1 + t` and `y = 6 + 4t rArr y = 4x + 2` |
|
| 13. |
Comprehension # 8 An initially stationary box on a frectionless floor explodes into two places, places A with mass `m_(A)` and piece B with mass `m_(B)`. Two pieces then move across the floor along x-axis. Graph of position versus time for the two pieces are given. Which graphs pertain to physically possible explosions ?A. II, VB. VIC. I, IIID. IV |
|
Answer» Correct Answer - A::D By COLM `0 = m_(A) v_(A) + m_(B) v_(B), vec(v)_(B) = -(m_(A))/(m_(B)) vec(v)_(A)` Both velocity are opposite in direction `:.` II, IV, V |
|
| 14. |
Comprehension # 5 One particle of mass `1 kg` is moving along positive x-axis with velocity `3 m//s`. Another particle of mass `2 kg` is moving along y-axis with `6 m//s`. At time `t = 0, 1 kg` mass is at `(3m, 0)` and `2 kg` at `(0, 9m), x - y` plane is the horizontal plane. (Surface is smooth for question 1 and rough for question 2 and 3) Co-ordinates of centre of mass where it will stop finally are :-A. `(2.0 m, 14.25 m)`B. `(2.25 m, 10 m)`C. `(3.75 m, 9 m)`D. `(1.75 m, 12 m)` |
|
Answer» Correct Answer - D By using `s = ut + (1)/(2)at^(2)` for individual partiles For `1^(st)` particle : `s_(x) = (3) (1.5) + (1)/(2) (2) (1.5)^(2) = 2.25 m` For `II^(nd)` particle `= s_(y) = (6) (3) + (1)/(2)(-2)(3)^(2) = 9m` Therefore `X_(cm) = (2.25)/(3) + 1 = 1.75 m` and `y_(cm) = (2(9))/(3) + 6 = 12m` |
|
| 15. |
Comprehension # 6 A `1 kg` block is given a velocity of `15 m//s` towards right over a very long rough plank of mass `2 kg` as shown in figure. Momentum of both the blocks are equal at time `t =` ……. Second :-A. `1.75`B. `1.875`C. `2.5`D. `1.25` |
|
Answer» Correct Answer - C `v = u + at rArr 5 = 15 - (4) t rArr t = 2.5 sec` |
|
| 16. |
The figure shown the velocity as a function of the time for an object with mass `10kg` being pushed along a frictioniess horizontal surface sby external horizontal force. At `t=3s`, the force stops pushing and the object moves freely. It then collides head on and sticks to another object of mass `25g` A. External force acting on the system is `50 N`B. Velocity of the `2^(nd)` particle just before the collision is `1 m//s`C. Before collision both bodies are moving in the same direcitonD. Before collision, bodies are moving in opposite direction with respect to each other |
|
Answer» Correct Answer - A::B::C `F_(ext) = (Delta(mv))/(Deltat) = (10 xx (15 - 0))/(3) = 50N` `COLM : m_(1)u_(1)+ m_(2)u_(2) = (m_(1) + m_(2))v` `rArr 10 xx 15 + 25 xx u_(2) = (10 + 25)5 rArr u_(2) = 1 m//s` |
|
| 17. |
A block of mass `m` is pushed with a velocity `v_(0)` along the surface of a trolley car of mass `M`. If the horizontal ground is smooth and the coefficient of kinetic friction between the block of plank is `mu`. Find the minimum distance of relative sliding between the block and plank. |
|
Answer» Method 1: Using ground frame method decide system: block and plank Observation: No external force acting on system nn horizontal direction Conclusion linear momentum of the system will be conserved. As friction in kinetic nature and system. when relative sliding between block and troley stops, velocity `V`. Problem solving: Using conservation of linear momentum in horizontal direction. ` mv_(0)+0=(m+M)VimpliesV=(mv_(0))/((m+M)`.........i now using WE theorem from ground frame. `W_(ext)+w_(int)=/_K` `W_(ext)=0` `W_(int)=W_("friction")=-fx=-mumgx` `0+(-mumgx)=(1/2(m+M)V^(2)-1/2mv_(0)^(2))`..........ii From i and ii relative distance of sliding `x=v_(0)^(2)/(2(1+m/M)mug)` Method 2: work energy thoerem from centre of mass frame Considering the system from `CM` frame `W_(ext)+W_("int") =(/_K)_(cm)=(k_(f)-k_(i))_(cm)` Initial kinetic energy of system w.r.t `CM` `(K_(i))_(cm)=1/2(mM)/((m+M) v_(0)^(2)` Final kinetic energy of sytem w.r.t `CM` `(K_(f))_(cm)=0` or `|(V_("rel"))f|=0` hence `0+(-mumgx)=0-1/2(mM)/((m+M))v_(0)^(2)` `impliesx=v_(0)^(2)/(2(1+m/M)mug` |
|
| 18. |
Considering a system having two masses `m_(1)` and `m_(2)` in which first mass is pushed towards centre of mass by a distance a, the distance required to be moved for second mass to keep centre of mass at same position is :- A. `(m_(1))/(m_(2))a`B. `(m_(1)m_(2))/(a)`C. `(m_(2))/(m_(1))a`D. `((m_(2)m_(1))/(m_(1) + m_(2)))a` |
|
Answer» Correct Answer - A `Deltabar(x) = (m_(1).Deltax_(1) + m_(2).Deltax_(2))/(m_(1) + m_(2))` `rArr 0 = (m_(1)a + m_(2)Deltax_(2))/(m_(1) + m_(2))` `rArr Deltax_(2) = -(m_(1)a)/(m_(2))` |
|
| 19. |
Three identical particles with velocities `v_0hati`, `-3v_0hatj` and `5v_0hatk` collide successively with each other in such a way that they form a single particle. The velocity vector of resultant particle isA. (a) `v_0/3(hati+hatj+hatk)`B. (b) `v_0/3(hati-hatj+hatk)`C. (c) `v_0/3(hati-3hatj+hatk)`D. (d) `v_0/3(hati-3hatj+5hatk)` |
|
Answer» Correct Answer - D `p_1=p_f` `:.m(v_0hati)+m(-3v_0hatj)+m(5v_0hatk)` `=3mv` `:. v=(v_0)/(3)(hati-3hatj+5hatk)` |
|
| 20. |
You are supplied with three identical rods of same length and mass. If the length of each rod is `2pi`. Two of them are converted into rings and then placed over the third rod as shown in figure. If point A is considered as origin of the coordinate system the coordinate of the center of mass will be (you may assume AB as x-axis of the coordinate system). A. `(pi/2,1/3)`B. `(pi/2,2/3)`C. `(pi,1/3)`D. `(pi,2/3)` |
|
Answer» Correct Answer - D `2piR = 2pi therefore R=1` `y_(CM) = (m xx 0 + m xx 1 + mxx1)/(m+m+m) = 2/3` `x_9CM) = (m(pi)+ m(0) + m(2pi))/(m + m + m) = pi` |
|
| 21. |
Three rods of the same mass are placed as shown in the figure. Calculate the coordinates of the centre of mass of the system. |
|
Answer» Correct Answer - `((a)/(3),(a)/(3))` `x_(cm) = ((M xx (a)/(2)) + (M xx 0) + (M xx (a)/(2)))/(M + M + M) = (a)/(3)` `y_(cm) = ((M xx 0) + (M xx (a)/(2)) + (M xx (a)/(2)))/(M + M + M) = (a)/(3)` |
|
| 22. |
Centre of mass of two unifrom rods of same length but made up of different materials & kept as shown, if the meeting point is the origin of co-ordinates :- A. `(L//2, L//2)`B. `(2L//3, L//2)`C. `(L//3, L//3)`D. `(L//3, L//6)` |
|
Answer» Correct Answer - D Equation of line joining the CM of two rods `(x)/(L//2) + (y)/(L//2) = 1` coordinate `((L)/(3),(L)/(6))` satisfies this equation. |
|
| 23. |
Two point masses `m_1` and `m_2` are connected by a spring of natural length `l_0`. The spring is compressed such that the two point masses touch each other and then they are fastened by a string. Then the system is moved with a velocity `v_0` along positive x-axis. When the system reached the origin, the string breaks `(t=0)`. The position of the point mass `m_1` is given by `x_1=v_0t-A(1-cos omegat)` where A and `omega` are constants. Find the position of the second block as a function of time. Also, find the relation between A and `l_0`. |
|
Answer» Correct Answer - `x_(2) = v_(0)t + (m_(1))/(m_(2)) A (1 - cos omegat), l_(0) = ((m_(1))/(m_(2)) + 1) A` The string snaps and the spring force comes into play. The spring force beings an internal force for the two mass-spring system will not be able to change the velocity of centre of mass. This means the location of centre of mass at time t will be `v_(0)t` `x_(cm) = (m_(1)x_(1) + m_(2)x_(2))/(m_(1) + m_(2)) = v_(0)t` `rArr m_(1)[v_(0)t - A(1 - cos omegat)] + m_(2)x_(2) = v_(0)tm_(1) + v_(0)tm_(2)` `rArr m_(2)x_(2) = v_(0)tm_(1) + v_(0)tm_(2) - v_(0)tm_(1) + m_(1)A (1 - cos omegat)` `rArr m_(2)x_(2) = v_(0)tm_(2) + m_(1)A(1 - cos omegat)` `rArr x_(2) = v_(0)t + (m_(1))/(m_(2)) A(1 - cos omegat)` (b) Given that `x_(1) = v_(0)t - A(1 - cos omegat)` `:. (dx_(1))/(dt) = v_(0) - Aomega sin omegat` `:. (d^(2)x_(1))/(dt^(2)) = -Aomega^(2) cos omegat` ....(i) This is the acceleration of mass `m_(1)`. When the spring comes to its natural length instantaneously then `(d^(2)x_(1))/(dt^(2)) = 0` and `x_(2) - x_(1) = l_(0)` `:. [v_(0)t + (m_(1))/(m_(2)) A(1 - cos omegat)] - [v_(0)t - A (1 - cos omegat)] = l_(0)` `((m_(1))/(m_(2)) + ) A ( 1 - cos omega t ) = l_(0)` Also when `(d^(2)x_(1))/(dt^(2)) = 0, cos omegat = 0` from (1) `:. l_(0) = ((m_(1))/(m_(2)) + 1) A` |
|
| 24. |
`N` beads identical beads are resting on a smooth horizontal wire which is circular at the end with radius `r = 0.5 m` as shown in the figure. Find the minimum velocity which should be imparted to the first bead such that `nth` bead will fall in the tank after completing full circle in vertical plane as shown in the figure. Take all the collisions between the beads elastic `(e = 1)`. |
|
Answer» Correct Answer - 5 Speed of the `nth` ball `=sqrt((5gr))` Since `e=1` speed of 1st ball `=` speed of `nth` ball `=sqrt((5gr))=5ms^(-1)` |
|
| 25. |
Two beads `A` and `B` of masses `m_(1)` and `m_(2)` respectively, are threaded on a smooth circular wire of radius a fixed in a vertical plane. `B` is stationary at the lowest point when `A` is gently dislodged from rest at the highest point. A collided with `B` at the lowest point. The impulse given to `B` due to collision is just great enough to carry it to the level of the centre of the circle while `A` is immediately brought to rest by the impact. What is the coefficient of restituting between the beads?A. `1`B. `1/2`C. `1/3`D. `1/sqrt(2)` |
|
Answer» Correct Answer - D `e=(v_(2)-v_(1))/(u_(1)-u_(2))=(sqrt(2ga)-0)/(2sqrt(ga)-0)=1/sqrt2` |
|
| 26. |
The Fig. showns a string of equally placed beads of mass m, separated by distance. The beads are free to slide without friction on a thin wire. A constant force F act on the first bead initially at rest till it makes collision with the second bead. The second bead then collides with the third and so on. Supposed that all collisions are elastic, A. Speed of the first immediately before and immediately after its collision with the second bead is `sqrt((2Fd)/(m))` and zero respectivelyB. Speed of the first bead immediately before and immediately after its collision with the second bead is `sqrt((2Fd)/(m))` and `(1)/(2) sqrt((2Fd)/(m))` respectivelyC. Speed of the second bead immediately after collision with third bead is zeroD. The average speed of the first bead is `(1)/(2)sqrt(2Fd)/(m)` |
|
Answer» Correct Answer - A::C In elastic head on collision if the masses of the collision bodies are eaqual, the velocities after for `Ist` bead, `Fd = (1)/(2) m u^(2) rArr u = sqrt((2Fd)/(m))` |
|
| 27. |
A block of mass `m` at rest is acted upon by a force `F` for a time t. The kinetic energy of block after time t isA. `(F^(2)t^(2))/(m)`B. `(F^(2)t^(2))/(2m)`C. `(F^(2)t^(2)/(3m)`D. `(Ft)/(2m)` |
|
Answer» Correct Answer - B Kinetic energy `E = p^(2)/(2m) = (Ft)^(2)/(2m) = (F^(2)t^(2))/(2m)` |
|
| 28. |
A particle of mass 2kg is moving in free space with velocity `vecv_0=(2hati-3hatj+hatk)m//s` is acted upon by force `vecF=(2hati+hatj-2hatk)N`. Find velocity vector of the particle 3s after the force starts acting. |
|
Answer» `vecp_f=vecp_i+vecI_mp rarr mvecv_f=mvec v_0+vecFDeltat` Substituting given value, we have `2vecv_f=2(2hat i-3hatj+hatk)+(2hati+hatj-2hatk)xx3=10hati-3hatj-4hatk` `vecv_f=(5hati-1.5hatj-2hatk)m/s` |
|
| 29. |
A box of mass `m=2 kg` resting on a frictionless horizontal ground is acted upon by a horizontal force F, which varies as shown. Find speed of the particle when force ceases to act. |
|
Answer» `vec(p)_(f)= vec(p)_(i) + vec(I)_(m)p rarr m vec(v)_(f) = m vec(v)_(i) + int_(t)^(f) vec(F)dt` `2v=2 xx 0 + 1/2 xx 20 xx 4` `v=20 m//s` |
|
| 30. |
As shown, a big box of mass M is resting on a horziontal smooth florr. On the bottom of the box there is a small block of mass m. The block is given an intial speed `v_(0)` relative to the floor, and starts to bounce back and forth between the two walls of the box. Find the final speed of the box when the block has finally come to rest in the box :- |
|
Answer» Correct Answer - D `F_(ext) =` zero, `vec(P) =` constant `P_(i) = P_(f)` Finally moving together with velocity v `mv_(0) = (m + M)v` `v = (mv_(0))/((m + M))` |
|
| 31. |
The elastic collision between two bodies, A and B, can be cosidered using the following model. A and B are free to move along a common line without friction. When their distance is greater than `d = 1 m`, the interacting force is zero , when their distance is less d, a constant repulsive force `F = 6 N` is present. The mass of body A is `m_(A) = 1 kg` and it is initially at rest, the mass of body B is `m_(B) = 3 kg` and it is approaching body A head-on with a speed `v_(0) = 2 m//s`. Find th eminimum distance between A and B :- A. `0.25 m`B. `0.50 m`C. `0.75 m`D. `1 m` |
|
Answer» Correct Answer - C `F_(ext) = 0` on system containing A & B `vec(P)` is conserved At minimum seperation both will move with same velocity `P_(1) = P_(f)` `2 xx 3 = 3 xx v` `v = (3)/(2)` `W_(all) = DeltaKE` `W_(F) = DeltaKE` `-6x = (1)/(2) xx 3 xx 2^(2) - (1)/(2) xx 4 xx ((3)/(2))^(2)` `x = 0.25` Minimum separation `= d - x = 1 - 0.25 = 0.75` |
|
| 32. |
Two blocks of masses 10 kg and 4 kg are connected by a spring of negligible mass and placed on a frictionless horizontal surface. An impulse gives a velocity of `14 m//s` to the heavier block in the direction of the lighter block. The velocity of the centre of mass isA. `30m//s`B. `20 m//s`C. `10 m//s`D. `5 m//s` |
|
Answer» Correct Answer - C Just after giving the impulse: `V_(c)=(10xx14+4xx10)/(10+4)=10m/s` Since spring force is internal force it cannot change the linear momentum of the (two mass `=` spring) system. Therefore `V_(c)` remains the same |
|
| 33. |
Two blocks of masses 10 kg and 4 kg are connected by a spring of negligible mass and placed on a frictionless horizontal surface. An impulse gives a velocity of `14 m//s` to the heavier block in the direction of the lighter block. The velocity of the centre of mass isA. `30 m//s`B. `20 m//s`C. `10 m//s`D. `5 m//s` |
|
Answer» Correct Answer - C Just after collision `v_(c ) = (10 xx 14 + 4 xx 0)/(10 + 4) = 10 m//s` since spring force is internal force, it cannot change the linear momentum of the (two mass + spring) system. Therefore `v_(c )` remains the same. |
|
| 34. |
Two pendulums each of length `l` are initially situated as shown in Fig. The first pendulum is released and strikes the second. Assume that the collision is completely inelastic and neglect the mass of the string and any frictional effects. How high does the centre of mass rise after the collision? A. `d[(m_(1))/((m_(1)+m_(2)))]^2`B. `d[(m_(1))/((m_(1)+m_(2)))]`C. `d[((m_(1)+m_(2)))/(m_(2))]^(2)`D. `[(m_(2))/((m_(1)+m_(2)))]^(2)` |
|
Answer» Correct Answer - A Applying the law of conservatioin of momentum `m_(1) v_(1)=(m_(1)+m_(2))V` …………i where `v_(1)=sqrt(2gd)` is the velocity with which `m_(1)` collides with `m_(2)` therefore `V=m_(1)/((m_(1)+m_(2)))sqrt(2gd)` Now, let the centre of mass rise through a height `h` after collision. In this case, the kinetic enerty of `m_(1)+m_(2)` system is converted into potential energy at maximum height `h` `implies 1/2(m_(1)+m_(2))V^(2)=(m_(1)+m_(2))gh` `implies1/2(m_(1)+m_(2)){(m_(1))/(m_(1)+m_(2))}^(2) 2gh=(m_(1)+m_(2))gh` `implies h=d{m_(1)/(m_(1)+m_(2))}^(2)` |
|
| 35. |
Assertion: In inelastic collision, linear momentum of system does not remain constant during collision. But before collision and after collision, it is constant. Reason: In elastic collision, momentum remains constant during collision also.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 - B Momentum is always constant during any type of collision |
|
| 36. |
Assertion: Two blocks of masses `m_(A)` and `m_(B)` `(gtm_(A)` are thrown towards each other with same speed over a rough ground. The coefficient of friction of both the blocks with ground is same. Initial velocity of CM is towards left. Reason: Initial acceleration of center of mass is towards right.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 - B `v_(CM) = (m_(A)(+v) + m_(B)(-v))/(m_(A) + m_(B)) = -ve` Both A and B will have same acceleration (ug) on A towsards left and on B towards right. Since, B have more mass so acceleration of CM will be towards right. |
|
| 37. |
Two blocks of masses 10 kg and 4 kg are connected by a spring of negligible mass and placed on a frictionless horizontal surface. An impulse gives a velocity of `14 m//s` to the heavier block in the direction of the lighter block. The velocity of the centre of mass isA. `30ms^(-1)`B. `20ms^(-1)`C. `10ms^(-1)`D. `5ms^(-1)` |
|
Answer» Correct Answer - C Velocity of center of mass, `v_(CM) = (10(14)+4(0))/(10 +4) = 10 ms^(-1)` |
|
| 38. |
Two blocks, with masses `m_(1) 2.5 kg` and `m_(2) = 14 kg`, approach each other along a horizontal, frictionless track, The initial velocities of the blocks are `v_(1) = 12.0 m//s` to the right an d`v_(2) = 3.4 m//s` to the left. The two blocks then collide and stick together. Which of the graph could represent the force of block `1` on block `2` during the collision ?A. B. C. D. |
| Answer» Correct Answer - A | |
| 39. |
Two blocks of masses `5 kg and 2 kg` are placed on a frictionless surface and connected by a spring. An external kick gives a velocity of `14 m//s` to the heavier block in the direction of lighter one. The magnitudes of velocities of two blocks in the centre of mass frame after the kick are respectively : |
|
Answer» Given, `m_(1) = 5` kg, `m_(2) = 2 kg`, `v_(1) = 14ms^(-1)`, `v_(2) = 0`, `v_(cm) = ?` `therefore v_(CM) = (m_(1)v_(1) + m_(2)v_(2))/(m_(1) + m_(2))` = (5 xx 14 +2 xx 0)/(5 +2)` `=70/7 = 10 ms^(-1)` in the direction of lighter one. |
|
| 40. |
`{:(,"Column I",,"Column II",),((A),"Elastic collision",(p),KE " is conserved",),((B),"Inelastic collision",(q),KE "after collision = KE before collision" ,),((C ),"Perfectly inelastic collision",(r ),"KE after collision" != "KE before collision",),(,,(s),"particles stick after collision",),(,,(t),"Linear momentum is conserved",),(,,(u),"Relative velocity of separation after is zero",):}` |
| Answer» Correct Answer - (A) `rarr` q,t; (B) `rarr` r,t; (C ) `rarr` r,s,t,u | |
| 41. |
Two blocks of masses `m_(1)= 2 kg` and `m_(2) = 4 kg` are moving in the same direction with speeds `v_(1) = 6 m//s` and `v_(2) = 3 m//s`, respectively on a frictionless surface as shown in the figure. An ideal spring with spring constant `k = 30000 N//m` is attached to the back side of `m_(2)`. Then the maximum compression of the spring after collision will be A. `0.06m`B. `0.04m`C. `0.02m`D. none of these |
|
Answer» Correct Answer - C At the time of maximum compression, the speeds of blocks will be the same. Let the speed be `v` and maximum compression be `x`. Applying conservation of momentum, `(m_(1)+m_(2))v=m_(1)v_(1)+m_(2)v_(2)` `impliesV=4m//s` Applying conservation of mechanical energy, `1/2kx^(2)+1/2(m_(1)+m_(2))v^(2)=1/2m_(1)v_(2)^(2)+1/2m_(2)v_(2)^(2)` solving we get `x=0.02m` |
|
| 42. |
Two blocks of masses `m_(1) = 2 kg` and `m_(2) = 5` kg are moving in the same direction along a frictionless surface with speeds `10 m//s` and `3 m//s`, respectively, `m_(2)` being ahead of `m_(1)`. An ideal spring with `k = 1120 N//m` is attached to the back side of `m_(2)`. Find the maximum compression of the spring when the blocks collide. What are the final velocities of the blocks when they separate? |
|
Answer» Form conservation of momentum `v=(m_(1)v_(1)+m_(2)v_(2))/(m_(1)+m_(2))=5m//s` conservationof energy `1/2m_(1)v_(1)^(2)+1/2m_(2)v_(2)^(2)=1/2(m_(1)+m_(2))v^(2)+1/2kx^(2)` `implies kx^(2)=(m_(1)v_(1)^(2)+m_(2)v_(2)^(2)-(m_(1)-m_(2))v^(2))/k` `implies x=sqrt(40/1120)=0.25m` when the blocks have equa speeds, spring has maximum commpression. after his instasnt,the spring againexpands, and after sometimes `m_(1)` loses contact with the spring. Let `v_(1)` and `v_(2)` be the velocities of the blocks after they lose contact. `m_(1)u_(1)+m_(2)u_(2)=m_(1)v_(1)+m_(2)u_(2)` and `v_(1)-v_(2)=-1(u_(1)-u_(2))` `[e=1,` because there is no loss in `KE`] Solving for `v_(1)` and `v_(2)` we get `v_(1)=0m//s` and `v_(2)=7m//s` |
|
| 43. |
Assertion : A moving ball having an inelastic collison withh another moving ball can have larger kinetic energy after collision. Reason: During a collision between two bodies, transfer of energy mayh take place from one body to another.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 | |
| 44. |
Three point masses `m_(1), m_(2) and m_(3)` are placed at the corners of a thin massless rectangular sheet `1.2m xx 1m)` as shown. Center of mass will be located at the point. A. (0.8,0.6)mB. (0.6, 0.8)mC. (0.4, 0.4)mD. (0.5, 0.6)m |
|
Answer» Correct Answer - C `m_(1) = 1.6 kg, (x_(1),y_(1)) = (0,0)` `m_(2) = 2kg, (x_(2),y_(2)) = (1.2,0)` `m-93) = 2.4 kg, (x_(3),y_(3)) = (0,1)` `therefore` Coordinates of center of mass will be `x_(CM) = (m_(1)x_(1) +m_(2)x_(2) + m_(3)x_(3))/(m_(1)+m_(2)+m_(3))` `((1.6)(0) + (2)(1.2)+(2.4)(0))/(1.6 + 2 + 2.4)` `x_(CM) = 0.4m` `y_(CM) = ((m_(1)y_(1) + m_(2)y_(2)+m_(3)y_(3))/(m_(1)+m_(2)+m_(3))` `=((1.6)(0)+(2)(0) + (2.4)(1))/(1.6 + 2+2.4)` `y_(CM) = 0.4`m `therefore Coordiantes of center of mass = (0.4,0.4)m |
|
| 45. |
Two blocks of msses 10 kg and 30 kg are placed along a vertical line. The first block is raised through a height of 7 cm. By what distance should the second mass be moved to raise the centre of mass by 1 cm? |
|
Answer» Two masses `m_1 and m_2` ar kept in a vertical line `m-1=10kg, m_2=30kg` The first block is raised through a height of 7 cm. The centre of mass raised by 1 cm. `:.1=(m_1y_1+m_2y_2)/(m_1+m_2)` `=(10xx7+30xxy_2)/40` `=70+30xxy_2)/40` `rarr 70+30y_2=40` `rarr 30y_2=-30` `rarr y_2=-1` the 30 kg body should be displaced 1 cm downward in order to raise the centre of mass through 1 cm. |
|
| 46. |
A circular plate of diameter d is kept in contact with a square plate of edge d as shown in figure. The densit of the material and the thickness are same . Everywhere. The centre of mass of the composite system will beA. inside the circular plateB. inside the squre plateeC. at the point of contactD. outside the system |
| Answer» Correct Answer - B | |
| 47. |
A man is sitting in a boat floating in water of a pond. There are heavy stones placed in the boat.A. When the man throws the stomes in water from the boat, the level of boat goes down.B. When the man throws the stomes in water from the boat, the level of boat rise upC. When the man drinks some water from the pond, the level of boat goes down.D. When the man drinks some water from the pond, the level of boat remains unchanged. |
|
Answer» Correct Answer - (B, D) For (A/B) : Force of buouanvy increases. Therefore level of boat rise up. for (C/D) : When man drinks some water, the level of boat remains unchanged. |
|
| 48. |
A man of mass M stands at one end of a stationary plank of length L, lying on a smooth surface. The man walks to the other end of the plank. If the mass of the plank is `M//3`, the distance that the man moves relative to the ground isA. `(3L)/4`B. `L/4`C. `(4L)/5`D. `L/3` |
|
Answer» Correct Answer - B Let plank moves x distance in opposite direction. Then, displacement of man relative to ground will be, (L-x), Applying `m_(B)x_(B) = m_(L)x_(L)` or `M(L-x) =M/3x` Solving the equation we get, `x = (3L)4` `therefore` Displacement of man relative to ground = L - (3L)/4 = L/4` |
|
| 49. |
Two blocks of equal mass `m` are connected by an unstretched spring and the system is kept at rest on a frictionless horizontal surface. A constant force `F` is applied on the first block pulling away from the other as shown in Fig. If the extension of the spring is `x_(0)` at time `t`, then the displacement of the second block at this instant isA. `((Ft^(2))/(2m) - x_(0))`B. `(1)/(2) ((Ft^(2))/(2m) + x_(0))`C. `(1)/(2)((2Ft^(2))/(m) - x_(0))`D. `(1)/(2)((Ft^(2))/(2m) - x_(0))` |
| Answer» Correct Answer - D | |
| 50. |
Two blocks of equal mass `m` are connected by an unstretched spring and the system is kept at rest on a frictionless horizontal surface. A constant force `F` is applied on the first block pulling away from the other as shown in Fig. If the extension of the spring is `x_(0)` at time `t`, then the displacement of the second block at this instant isA. `(Ft^(2))/(2m)-x_(0)`B. `1/2((Ft^(2))/(2m)+x_(0))`C. `1/2((2F^(2))/m-x_(0))`D. `1/2((Ft^(2))/(2m)-x_(0))` |
|
Answer» Correct Answer - D Suppose the displacement of the first block is `x_(1)` and that of the second is `x_(2)` then `x=(mx_(1)+mx_(2))/(2m), (Ft^(2))/(4m)=(x_(1)+x_(2))/2` or `x_(1)+x_(2)=(Ft^(2))/(2m)`……….i Further the extension of the spring is `x_(1)-x_(2)`. Therefore `x_(1)-x_(2)=x_(0)`.....ii From eqn i and ii `x_(1)=1/2((Ft^(2))/(2m)+x_(0))impliesx_(2)=1/2((Ft^(2))/(2m)-x_(0))` |
|