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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.
| 251. |
A particle of mass 100 g moving at an initial speed u collides with another particle of same mass kept initially at rest. If the total kinetic energy becomes 0.2 J after the collision what could be minimum and the maximum value of u. |
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Answer» Correct Answer - A::B In elastic collision, velocities are interchanged. So, `v` is minimum `:. 1/2xx0.1xxv^2=0.2` `:. V=2m//s=`minimum value In perfectly inelastic collision, speed of combined mass will remain half. `:. 1/2xx0.1xx(v/2)^2=0.2` `:. v=2sqrt2m//s`=maximum value |
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| 252. |
A particle of mass `2kg` moving with a velocity `5hatim//s` collides head-on with another particle of mass `3kg` moving with a velocity `-2hatim//s`. After the collision the first particle has speed of `1.6m//s` in negative x-direction, Find (a) velocity of the centre of mass after the collision, (b) velocity of the second particle after the collision. (c) coefficient of restitution. |
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Answer» Correct Answer - A::B::C::D (a) `v_c=(m_1u_1+m_2u_2)/(m_1+m_2)=0.8hatim//s` (b) `v_1=-1.6hatim//s` From COM, `m_1u_1+m_2u_2=m_1v_1_m_2v_2` `implies v_2=2.4hatim//s` (c) `e=(v_2-v_1)/(u_1-u_2)=4/7` |
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| 253. |
A bullet mass 10gm is fired from a gun of mass 1 kg . If the recoil velocity is 5 `m//s`, the velocity of the muzzle is |
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Answer» Conservation of linear momentum gives `m_(1)v_(1) + m_(2)v_(2) = 0 rArr m_(1)v_(1) = -m_(2)v_(2)` `rArr v-(1) = -(m_(2)v_(2))/(m_(1) = 500 m//s` Given, `m_(1) = 10 g = (10/1000) kg` `m_(2) = 1kg` and `v_(2) = -5 m//s` `therefore` Velocity of muzzle, `v-(1) = (+1 xx 5)/(10//1000)= 500 m//s` |
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| 254. |
A bag of mass `M` hangs by a long massless rope. A bullet of mass in, moving horizontally with velocity `u`, is caught in the bag. Then for the combined (bag `+` bullet) system, just after collisionA. momentum is `"mu"M//(M+m)`B. kinetic energy is `"mu"^2//2`C. momentum is `"mu"(M+m)//M`D. kinetic energy is `m^(2)u^(2)//2(M+m)` |
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Answer» Correct Answer - D Let combined velocilty of the system be `v` `(m+MV)v="mu"impliesv=("mu")/(m+M)` `KE=1/2(m+M)v^(2)=(m^(2)u^(2))/(2(M+m))` |
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| 255. |
Assertion: Center of mass and center of gravity of a body will coincide on moon. Reason: There is no gravity on moon.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 Atmosphere is absent on the surface of moon. Gravity is present. |
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| 256. |
A small ball of mass `1 kg` is kept in circular path of radius `1 m` Inside a concentric smooth horizontal fixed casing of radius `R`. Angular speed of the ball in the circular motion is `1 rads^(-1)`. At a certain moment the string, which kept the ball in the circular path breaks and the ball goes off tangentially to the wall of rigid casing and bounces off elastically and again hits the casing and bounces off. This way, the ball traces a regular hexagon. Consider all the collisions to be elastic. Following quantities of the ball will remain a constant before and after any collisionA. linear momentumB. kinetic energy, angular momentum about the centre of the circleC. velocity, angular momentum about the centre of the circle, kinetic energyD. none of these |
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Answer» Correct Answer - B Since an impulse acts on the ball during each collision, so linear momentum does not remain constant. The impulsive force is directed towards the centre of hexagon, therefore the lever arm of this imupulsive force is zero. As are result, it does not produce any torque and angular momentum remains costant. Further, kinetic energy also remains same for elastic collision. |
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| 257. |
A ball of mass `m` is thrown at an angle of `45^(@)` to the horizontal,, from the top of a `65 m` high tower `AB` as shown in Fig. at `t = 0`. Another identical ball is thrown with velocity `20 m//s` horizontally towards `AB` from the top of a `30 m` high tower `CD 1 s` after the projection of the first ball. Both the balls move the same vertical plane. If they collide in mid-air, . the velocity of combined ball just after they struck together isA. `-5i-20jm//s`B. `10i-20jm//s`C. `-10i-20jm//s`D. `-5i+10jm//s` |
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Answer» Correct Answer - A When two balls collide, their height from ground is same. Let it be `h`. Then vertical displacement of the first and second balls is `(65 - h) m` and `(30 - h) m`, respectively. Let the time of flight of the first ball up to the instant be `n` seconds. Then the time of flight that of second ball is `(n - 1)` seconds. Considering vertically downward component of motion of first ball, `u=10sqrt(2)sin45^(@)=10m//s` `a=g=10ms^(-1), s=y_(1)=(65-h)t=ns` `s=ut+1/2at^(2)` `(65-h)=10n+5n^(2)`………i From eqn i and ii `n=2s, h=25m` Displacement `AC=` Horizontal distance moved by first ball up to the instant of collision `+` that moved by the second ball up to the same instant `:.AC = (10sqrt(2)cos45^(@))n+20(n-1)=40m` Just before collision, vertically downward component of velocity of first ball is `v_(y1)=(10sqrt(2)sin45^(@))+gn=30m//s` and that of second ball is `v_(y2)=g(n-1)=10 m//s` According to the law of conservation of momentum, vertically downward component `v_(y)`, of velocity of combined body is given by `mv_(y1)+mv_(y2)=(m+m)v_(y)` or `v_(y)=20 ms^(-1)` Similarly, horizontally leftward component `y`, of combined body (just after collision) is given by `mxx20-mxx10sqrt(2)cos45^(@)=(m+m)v_(x)` `implies v_(x)=5m//s` |
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| 258. |
A ball of mass `m` is thrown at an angle of `45^(@)` to the horizontal,, from the top of a `65 m` high tower `AB` as shown in Fig. at `t = 0`. Another identical ball is thrown with velocity `20 m//s` horizontally towards `AB` from the top of a `30 m` high tower `CD 1 s` after the projection of the first ball. Both the balls move the same vertical plane. If they collide in mid-air, the distance `AC` isA. `20m`B. `30m`C. `40m`D. `20m` |
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Answer» Correct Answer - C When two balls collide, their height from ground is same. Let it be `h`. Then vertical displacement of the first and second balls is `(65 - h) m` and `(30 - h) m`, respectively. Let the time of flight of the first ball up to the instant be `n` seconds. Then the time of flight that of second ball is `(n - 1)` seconds. Considering vertically downward component of motion of first ball, `u=10sqrt(2)sin45^(@)=10m//s` `a=g=10ms^(-1), s=y_(1)=(65-h)t=ns` `s=ut+1/2at^(2)` `(65-h)=10n+5n^(2)`………i From eqn i and ii `n=2s, h=25m` Displacement `AC=` Horizontal distance moved by first ball up to the instant of collision `+` that moved by the second ball up to the same instant `:.AC = (10sqrt(2)cos45^(@))n+20(n-1)=40m` Just before collision, vertically downward component of velocity of first ball is `v_(y1)=(10sqrt(2)sin45^(@))+gn=30m//s` and that of second ball is `v_(y2)=g(n-1)=10 m//s` According to the law of conservation of momentum, vertically downward component `v_(y)`, of velocity of combined body is given by `mv_(y1)+mv_(y2)=(m+m)v_(y)` or `v_(y)=20 ms^(-1)` Similarly, horizontally leftward component `y`, of combined body (just after collision) is given by `mxx20-mxx10sqrt(2)cos45^(@)=(m+m)v_(x)` `implies v_(x)=5m//s` |
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| 259. |
A wooden block of mass `10 g` is dropped from the top of a tower `100 m` high. Simultaneously, a bullet of mass `10 g` is fired from the foot of the tower vertically upwards with a velocity of `100 m//s`. If the bullet is embedded in it, how high will the block rise above the top of tower before it starts falling? `(g=10m//s^2)` A. `75m`B. `85m`C. `80m`D. `10m` |
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Answer» Correct Answer - A The bullet and block will meet after tme `t=h/(u_(rel))=100/100=1` During this time, distance travelled by the block, `s_(1)=1/2"gt"^(2)=1/2xx10xx1^(2)=5m` Distance travelled by the bullet, `s_(2)=100-s_(1)=96m` Velocity of the bullet before collision `u_(2)=u-"gt"=100-10xx1=90m//s` Velocity of the block before collision. `u_(1)"gt"=10m//s` Let `V` be the combined velocity after collision. According to the law of conservation of momentum. `m_(1)u_(1)+m_(2)u_(2)=(m_(1)+m_(2))V` (Velocity in upward direction in considered positive) Solving we get `V=40m//s` Maximum height risen by the block `=(V^(2))/(2g)=80m` Height reached above the top of the tower is `80-s_(1)=80-5=75m` |
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| 260. |
Two particles of mass `m_(1)` and `m_(2)` are projected from the top of a tower. The particle `m_(1)` is projected vertically downward with speed u and `m_(2)` is projected horizontally with same speed. Find acceleration of CM of system of particles by neglecting the effect of air resistance. |
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Answer» As effect of air is neglected, therefore the only force acting on the particles is the gravitational force is downward direction. Let the point of projection is taken as origin and downward direction as negative Y-axis, then acceleration of 1st point mass `a_(1) = -gj` acceleration of 2nd point mass `a_(2)` = `-gj` `a_(CM)` = `(m_(1)a_(1) `+ `m_(2)a_(2))/(m_(1) + m_(2)` = `(m_(1)(-gj) + m_(2)(-gj))/(m_(1) + m_(2)` `rArr a_(CM) = -gj` i.e., acceleration of CM is equal to acceleration due to gravity and is in downward direction. Note. If large number of particles are projected under the effect of gravity only indifferent direction, then acceleration of CM is equal to the acceleration due to gravity irrespective of direction of projection of particles. |
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| 261. |
A ball of mass `m` is thrown at an angle of `45^(@)` to the horizontal,, from the top of a `65 m` high tower `AB` as shown in Fig. at `t = 0`. Another identical ball is thrown with velocity `20 m//s` horizontally towards `AB` from the top of a `30 m` high tower `CD 1 s` after the projection of the first ball. Both the balls move the same vertical plane. If they collide in mid-air, the height from the ground where the two balls will collide isA. `15m`B. `25m`C. `10m`D. `20m` |
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Answer» Correct Answer - B When two balls collide, their height from ground is same. Let it be `h`. Then vertical displacement of the first and second balls is `(65 - h) m` and `(30 - h) m`, respectively. Let the time of flight of the first ball up to the instant be `n` seconds. Then the time of flight that of second ball is `(n - 1)` seconds. Considering vertically downward component of motion of first ball, `u=10sqrt(2)sin45^(@)=10m//s` `a=g=10ms^(-1), s=y_(1)=(65-h)t=ns` `s=ut+1/2at^(2)` `(65-h)=10n+5n^(2)`………i From eqn i and ii `n=2s, h=25m` Displacement `AC=` Horizontal distance moved by first ball up to the instant of collision `+` that moved by the second ball up to the same instant `:.AC = (10sqrt(2)cos45^(@))n+20(n-1)=40m` Just before collision, vertically downward component of velocity of first ball is `v_(y1)=(10sqrt(2)sin45^(@))+gn=30m//s` and that of second ball is `v_(y2)=g(n-1)=10 m//s` According to the law of conservation of momentum, vertically downward component `v_(y)`, of velocity of combined body is given by `mv_(y1)+mv_(y2)=(m+m)v_(y)` or `v_(y)=20 ms^(-1)` Similarly, horizontally leftward component `y`, of combined body (just after collision) is given by `mxx20-mxx10sqrt(2)cos45^(@)=(m+m)v_(x)` `implies v_(x)=5m//s` |
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| 262. |
A ball of mass `m` is thrown at an angle of `45^(@)` to the horizontal,, from the top of a `65 m` high tower `AB` as shown in Fig. at `t = 0`. Another identical ball is thrown with velocity `20 m//s` horizontally towards `AB` from the top of a `30 m` high tower `CD 1 s` after the projection of the first ball. Both the balls move the same vertical plane. If they collide in mid-air, the two balls will collide at time `t =`A. `2s`B. `5s`C. `10s`D. `3s` |
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Answer» Correct Answer - A When two balls collide, their height from ground is same. Let it be `h`. Then vertical displacement of the first and second balls is `(65 - h) m` and `(30 - h) m`, respectively. Let the time of flight of the first ball up to the instant be `n` seconds. Then the time of flight that of second ball is `(n - 1)` seconds. Considering vertically downward component of motion of first ball, `u=10sqrt(2)sin45^(@)=10m//s` `a=g=10ms^(-1), s=y_(1)=(65-h)t=ns` `s=ut+1/2at^(2)` `(65-h)=10n+5n^(2)`………i From eqn i and ii `n=2s, h=25m` Displacement `AC=` Horizontal distance moved by first ball up to the instant of collision `+` that moved by the second ball up to the same instant `:.AC = (10sqrt(2)cos45^(@))n+20(n-1)=40m` Just before collision, vertically downward component of velocity of first ball is `v_(y1)=(10sqrt(2)sin45^(@))+gn=30m//s` and that of second ball is `v_(y2)=g(n-1)=10 m//s` According to the law of conservation of momentum, vertically downward component `v_(y)`, of velocity of combined body is given by `mv_(y1)+mv_(y2)=(m+m)v_(y)` or `v_(y)=20 ms^(-1)` Similarly, horizontally leftward component `y`, of combined body (just after collision) is given by `mxx20-mxx10sqrt(2)cos45^(@)=(m+m)v_(x)` `implies v_(x)=5m//s` |
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| 263. |
A system of men and trolley is shown in Fig. To the left, end of the string, a trolley of mass `M` is connected on which a man of mass `m` is standing. To the right end of the string another trolley of mass `m` is connected on which a man of mass `M` is standing. Initially, the system is at rest. All of a sudden both the men leap upwards simultaneously with the same velocity `u` w.r.t. ground. Find the impulse generated in the string connecting the trolleys during this process.A. `(M"mu")/(m+M)`B. `((M^(2)+m^(2))m)/(m+M)`C. `(m^(2)u)/(m+M)`D. `(M^(2)u)/(m+M)` |
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Answer» Correct Answer - B `I+I_(1)-Mu_(0)="mu"-Mu_(0)="mu"-(M(m-M))/(m+m)u=((m^(2)+M^(2))u)/(M+m)` |
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| 264. |
An object of mass `10 kg` is launched from the ground at `t = 0`, at an angle of `37^(@)` above the horizontal with a speed of `30 m//s`. At some time after its launch, an explosion splits the projectile into two pieces. One piece of mass `4 kg` is observed at (`105 m, 43 m`) at `t =2 s`. Find the location of second piece at `t = 2 s`? A. `(10,2)`B. `(48,16)`C. `(10,-2)`D. information insufficient |
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Answer» Correct Answer - D As only the gravity force is acting on the system, the centre of mass of the system follows a parabolic path. At `t=2s`, `x_(CM)=30cos37^(@)xx2=48m` `y_(CM)=30sin37^(@)xx2-1/2xx10xx2^(2)=16m` Let coordinates of the second piece, i.. `6kg` piece be `(x,y)` Then `x_(CM)=48=(6x+4xx105)10impliesx=10m` `y_(CM)=16=(6y+4xx43)/10impliesy=-2m` Negative value of `y` shows that the second piece collides the ground before `t=2s`. |
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| 265. |
Look at the drawing given in the figure which has been drawn with ink of uniform line-thickness. The mass of ink used to draw each of the two inner circles, and each of the two line segments is m. The mass of the ink used to draw the outer circle is 6 m. The coordinates of the centres of the different parts are: outer cicle (0,0), left circle (-a, a), right inner circle (a,a), vertical line (0,0) and horizontal line (0 ,-a). The y-coordinate of the centre of mass of the ink in this drawing is A. `(a)/(10)`B. `(a)/(8)`C. `(a)/(12)`D. `(a)/(3)` |
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Answer» Correct Answer - A `Y_(cm) = (m_(1)y_(1) + m_(2)y_(2) + m_(3)y_(3))/(M_(1) + M_(2) + M_(3))` `Y_(cm) = (6m(o) + m( + a) + m(a) + m(-a) + m(0))/(10m), Y_(cm) = (a)/(10)` |
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| 266. |
The quantities remaining constant in colision areA. momentum, kinetic energy and temperatureB. momentum and kinetic energy but not temperatureC. momentum and temperature but not kinetic energyD. momentum but neither kinetic energy nor temperature |
| Answer» Correct Answer - D | |
| 267. |
A heavy ring fo mass m is clamped on the periphery of a light circular disc.A small particle having equal mass is clamped at the centre of the disc. The system is rotated in such a way that the centre of mass moves in a circle of radius r with a uniform speed v. We conclude that an external forec :A. `(mv^2)/r` must be acting on the central particleB. `(2mv^2)/r` must be acting on the central particleC. `(2mv^2)/r` must be acting on the systemD. `(2mv^2)/r` must be acting on the ring |
| Answer» Correct Answer - C | |
| 268. |
A particle of mass `2 kg` is moving with velocity `vec(v)_(0) = (2hat(i)-3hat(j))m//s` in free space. Find its velocity `3s` after a constant force `vec(F)= (3hat(i) + 4hat(j))N` starts acting on it. |
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Answer» `vecp_f=vecp_i+vecI_(mp)rarr" "mvecv_f=mvecv_0+vecFDeltat` Substituting given values, we have `2vecV_f=2(2hati-3hatj)+(3hati+4hatj)xx3=13hati+6hatj` `vecv_f=(6.5hati+3hatj)m//s` |
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| 269. |
A small block of mass `M` moves on a frictionless surface of an inclined plane, as shown in the figure. The angle of the incline suddenly changes from `60^(@)` to `30^(@)` at point `B`. The block is many at rest at `A`. Assume that collisions between the block id the incline are totally inelastic. The speed of the block at point `C`, immediately before it leaves the second incline A. `sqrt(120) m//s`B. `sqrt(105) m//s`C. `sqrt(90)m//s`D. `sqrt(75) m//s` |
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Answer» Correct Answer - D Height fallen by the block from `B` to `C` `h_(2)=3sqrt(3) tan30^(@)=3m` Let `v_(3)` be the speed of the block, at point `C` just before it leaves the second incline, then `v_(3)=sqrt(v_(2)^(2)+2gh_(2))=sqrt(45+2xx10xx3)=sqrt(105) m//s` |
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| 270. |
Each of the blocks shown in figure has mass 1 kg. The rear block moves with a speed of 2 m/s towards the front block kept at rest. The spring attached to the front block islight and has a spring constant 50 N/m. find the maximum compression of the spring. |
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Answer» Maximum compression will take place when the blocks move with equal velocity. As no net exterN/Al force acts on the system of the two blocks, the totla linear momentum will remain constant. If V is the common speed at maximum compression we hae , `(1kg)(2m/s)=(1kg)V+(1kg)V` or, `V=1m/s` INitial kinetic energy `=1/2 (kg)(2m/s)^2=2J` FiN/Al kinetic energy `=1/2(1kg)(1m/s)^2+1/2(1kg)(1m/s)^2` `=1j` `The kinetic energh lost is stored as the elastic energy in spring. Hence, `1/2(50N/m)x^2=2J-1=1J` `or x=0.2m` |
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| 271. |
A ball kept in a close box moves in the box making collisions with the walls. The box is kept on a smooth surface. The velocity of the centre of massA. of the box remains constantB. of the box plus the ball sytem remains constantC. of the ball remains constantD. of the ball relative to the box remains constant. |
| Answer» Correct Answer - B | |
| 272. |
A body moving towards a finite body at rest collides with it. It is possible thatA. both the bodies come to restB. both the bodies move after collisionC. the moving body comes to rest and the stationary body starts movingD. the stationary body remains stationary, the moving body changes its velocity |
| Answer» Correct Answer - B::C | |
| 273. |
A uniform sphere is place on a smooth horizontal surface and as horizontal force F is appied on it at a distance h above the surface. The acceleratioin of the centreA. is maximum when h=0B. is maximum when h=RC. is maximum when h=2RD. is independent of h |
| Answer» Correct Answer - D | |
| 274. |
A ball hits a floor and rebounds after an inelastic collision. In this caseA. the momentum of the ball just after the collision is same as that just before the collisionB. the mechanical energy of the ball remains the same during the collisionC. the total momentum of the ball and the earth is conservedD. the total energy of theball and the earth remains the same |
| Answer» Correct Answer - C::D | |
| 275. |
A cart A of mass 50 g moving at a speed of 20 km/h hits a lighter cart B of mass 20 kg moving towardsd it at a speed of 10 km/h. The two carts cling to each other. Find the speed of the combined mass after the collision. |
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Answer» This is an eample of inelastic collision. As the carts move towards each other, their momenta have opposite sign. If the common speed after the collision is v , momentum conservation gives `(50kg)(20km/h)-(20kg)(10 km/h)=(70kg)V` or `V=80/7km/h`. |
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| 276. |
A ball of mas m moving t a speed v makes a head on collisiobn with a identicalbl at rest. The kinetic energy of the blls after the collision is three fourths of the original. Find the coefficient of retitutioin. |
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Answer» Mas of 1st ball =m and speed =v Mass of 2nd ball =m Let fiN/Al velocities of 1st and 2nd ball are `v_1 and v_2` respectively Using law of conservation of momentum `m(v_1+v_2)=mv`……i `rarr v_1+v_2=v`…………ii `also v_1-v_2=ev` Given that fiN/Al K.E.`=3/4 initial K.E. `rarr 1/2 mv_1^2+1/2mv_2^2=3/4xx1/2mv^2` `rarr v_1^2+v_2^2=3/4v^2` `rarr v_1^2+v_2^2=3/4v^2` `rarr ((v_1+v_2)^2+(v_1-v_2)^2)/2=3/4v^2` `rarr ((1+e^2)v^2)/2=3/4v^2` rarr 1+e^2=(3/2)` `rarr e^2=1/2` `-e=1/sqrt2` |
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| 277. |
A bullet hits a lock kept at rest on a smooth horizontal surface andgets embedded into it. Which of the following does not change?A. a linear momentum of the blockB. kinetic energy of the blockC. gravitational potential energy of the blockD. temperature of the block |
| Answer» Correct Answer - C | |
| 278. |
Calculate the velocity of the centre of mss of the system of particles shown in figure. |
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Answer» `m_1=1.0kg` `V_1=(-15 cos 37^0veci-1.5sin37^0vecj)` `m_2=1.2kg` `V_2=0.4vecj` `m_3=1.5kg` `V_3=-1.0cos37^0veci+1.0sin37^0vecj`)` `m_4=0.50kg` `V_4=3.0veci` `m_5=1.0kg` `V_5=2.0cos37^0veci-2.0sin37^0vecj` `V_(cm)=(m_1 v_1+m_2v_2+m_2v_3+m_4v_4+m_5v_5)/(m_1+m_2+m_3+mK_4+m_5)` `1/(1.0+1.2+1.5+1.0+0.50)^([1.0(-1.5xx4/5veci-1.5xx3/5vecj)+...-2.0xx3/5vecj])` `=0.20m/s at 45^0` below the direction towards right, |
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| 279. |
A bullet of mas m moving at a speed v hits as ball of mass M kept at rest. A small prt hving mass m breaks from the ball and sticks to the bullet. The remaining ball is found to move at a speed `v_1` in the direction of the bullet. Find the velocity of the ulet after the collision. |
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Answer» Mass of the bullet =m and speed =v Mass of the ball =M `m^1`= fractioN/Al mass from the ball Using law of conservation of momentum `mv+0=(m^1+m)v^1+(M-m^)v_1` Where `v^1=fiN/Al velocity of the bullet + fractioN/Al mass` `rarr v^1=(mv-(M-m)v_1)/(m+m^1)` |
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| 280. |
A bomb of mass 9kg explodes into two pieces of masses 3 kg and 6kg. The velocity of 3kg mass is `16ms^(-1)`. The velocity of 6kg mass isA. `4ms^(-1)`B. `8ms^(-1)`C. `16ms^(-1)`D. `32ms^(-1)` |
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Answer» Correct Answer - B `p_(3) = p_(6)` or `3 xx 16=6 xx v` or `v=8 ms^(-1)`. Therefore, the velocity of 6 kg mass is `8ms^(-1)`. |
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| 281. |
A bomb travelling in a parabolic path under the effect of gravity, explodes in mid air. The centre of mass of fragments will :A. Move vertically upwards and then downwardsB. Move vertically downwardsC. Move in irregular pathD. Move in the parabolic path which the unexploded bomb would have travelled. |
| Answer» Correct Answer - D | |
| 282. |
Abullet of mss 20 g moving horizontally at speed of 300 m/s is fired into wooden block of mas 500 suspended by a long string. The bullet criosses the block and emerges on the other side. If the centre of mass of the block rises through a height of 20.0 cm, find teh speed of tbe bullet as it emerges from the block. |
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Answer» Mass of bullet `m=20 m =0.02kg ` Mass of wooden block `M=500 gm=0.5kg` Velocity of the bullet with which it strikes `u=300m/sec` Let tehh bullet emerges out with velociyt V and the velocity of block `v^1` As per law of conservation of momentum `mu=Mv^1+m_1v`.......i Again applying work energy principle for the block after the collision `0-(1/2)Mx(v^10^2=-M.g.h` (where h=0.2m) `rarr (v^2)^2=2gh` `=sqrt(20xx2.0)=2m/sec` Substituting the value `v^1` in equation i we get `0.02xx300=2+0.2xxv` `rarr v=(6-1)/0.02` `=5/0.02=250m/sec` |
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| 283. |
A bomb of `1kg` is thrown vertically up with speed `100m//s` After 5 seconds it explodes into two parts. One part of mass `400gm` goes down with speed `25m//s` What will happen to the other part just after explosion .A. `100 ms^(-1)` upwardB. `600 ms^(-1)` upwardC. `100 ms^(-1)` downwardD. `300 ms^(-1)` upward |
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Answer» Correct Answer - A Velocity of particle after 5s `v = u-gt=100-10xx5` `=100-50=50ms^(-1)` (upwards) Conservation of linear momentum gives `Mv = m_(1)v_(1) + m_(2)v_(2)`................(i) Taking upward direction positive, `v_(1) = -25 ms^(-1), v=50ms^(-1)` `M=1kg, m_(1) = 400g = 0.4Kg` `m_(2) = M-m_(1) = 1-0.4=0.6kg` From eq (i), `1 xx 50=0.4xx(-25)+0.6v_(2)` or `v_(2) = 100 ms^(-1)` (upwards)` |
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| 284. |
A bullet of mass `20 g` and moving with `600 m/s` collides with a block of mass `4 kg` hanging with the string. What is the velocity of bullet when it comes out of block, if block rises to height `0.2 m` after collision?A. `200 ms^(-1)`B. `150 ms^(-1)`C. `400 ms^(-1)`D. `300 ms^(-1)` |
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Answer» Correct Answer - A Velocity of block just after collisioin = sqrt(2gh)` `=sqrt(2 xx 10xx 0.2)` `2ms^(-1)` Now applying conservation of linear momentum just before and just after collision. `0.02 xx 600 = 4 xx2+0.02xxv` `v=200 ms^(-1)` |
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| 285. |
A man standing on the edge of the terrace of a high rise building throws a stone, vertically up with at speed of `20 m//s`. Two seconds later, an identical stone is thrown vertically downwards with the same speed of `20 m`,. ThenA. the relative velocity between the two stones remain, constant till one hits the groundB. both will have the same kinetic energy, when they hit the groundC. the time interval between their hitting the ground it `2s`D. if the collision on the ground is perfectly elastic, both will rise to the same height above the ground |
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Answer» Correct Answer - A::B::C::D a. `v_(1)=20-"gt"` `v_(2)=-20-"gt"(t-2)` `v_(1)-v_(2)=[20-"gt"]+[20+g(t-2)]=20m//s` Clearly it is time independent i.e., relative velocity between the two stones remains constant. b. in both the cases, the total energy remains the same. c. and d. these are simple concepts. |
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| 286. |
Two blocks of masses `m_(1)` and `m_(2)`, connected by a weightless spring of stiffness `k` rest on a smooth horizontal plane as shown in Fig. Block 2 is shifted a small distance `x` to the left and then released. Find the velocity of centre of mass of the system after block 1 breaks off the wall. |
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Answer» If `m_(2)` is shifted by a distance x and released, the mass `m_(1)` will break off from the wall when the spring restores its natural length and `m_(2)` will start going towards right. At the time of breaking `m_(1), m_(2)` will be going towards right with a velocity `v`, which is given as `1/2kx^(2)=1/2m_(2)v^(2)impliesv=(sqrt(k/m_(2)))x` and the velocity of the centre of mass at this instant is `v_(CM)=(m_(1)xx0+m_(2)xxv)/(m_(1)+m_(2))=((sqrt(m_(2)k)x))/(m_(1)+m_(2))` |
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| 287. |
The bob A of a simple pendulum released from `30^(@)` to the vertical hits another bobo B of the same mass at rest on a table as shown in figure. How high does the bob A rise after the collision ? Neglect the size of the bobs and assume the collision to be elastic. |
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Answer» Correct Answer - bob A does not rise In this elastic collision velocity of masses are exchaged. So `v_(A) = 0 rArr A` does not rise |
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| 288. |
Two identical ball bearings in contact with each other and resting on a frictionless table are hit heat-on by another ball bearing of the same mass moving initially with a speed `V` as shown in figure. If the collision is elastic, which of the following (figure) is a possible result after collision ?A. B. C. D. |
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Answer» Correct Answer - B When two bodies of equal masses collides elastically, their velocities are interchanged. When ball 1 collides with ball-2, then velocity of ball-1, `v_(1)` becomes zero and velocity of ball-2, `v_(2)` becomes v, i.e., similarly. `v_(1) = 0 rArr v_(2) = v` When ball 2 collides will ball 3 `v_(2) = 0, v_(3) = v` |
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| 289. |
A shell is fired from a cannon with a speed of `100 m//s` at an angle `30^(@)`with the vertical (`y`-direction). At the highest point of its trajectory, the shell explodes into two fragments of masses in the ratio `1:2`. The lighter fragment moves vertically upwards with an initial speed of `200 m//s`. What is the speed of the heavier fragment at the time of explosion? |
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Answer» The velocity of shell at the highest point is `v=u sin theta= 100 xx sin 30^(@) = 50 m//s`. Let in be the mass of the shell. Then the mass of the lighter fragment is `m//3` and that of heavier fragment is `2m//3`. Initial momentum of the shell before explosion is `mv=50m` As no external force is acting on the shell, we can conserve momentum of the shell before and after its explosion. In `x`-direction `mv=(2m)/3v_(2)costhetaimpliesv_(2)costheta=3/2v`..............i in `y`- direction `0=m/3v_(1)-(2m)/3v_(2)sinthetaimpliesv_(2)sintheta=(v_(1))/2`........ii ltbr `Squaring and adding eqn i and ii we get `v_(2)^(2)=9/4v^(2)+1/4v_(1)^(2)` `implies v_(2)=1/2sqrt(9v^(2)+v_(1)^(2))=1/2sqrt(9xx2500+40000)=125m//s` |
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| 290. |
A shell is fired from a cannon with a speed of `100 m//s` at an angle `60^(@)` with the horizontal (positive `x`-direction). At the highest point of its trajectory, the shell explodes in to two equal fragments. One of the fragments moves is the speed of the other fragment at the time of explosion. |
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Answer» The velocity of the shell at the highest point of tarajector is `v_(M)=ucostheta=100 cos 60^(@)=50 m//s` Let `v_(1)` be the speed of the fragment which moves long the negative `x`-direction and the other fragment has speed `v_(2)`, which must along the positive `x`-direction. now from, momentum conservation, we have `mv=(-m)/2 v_(1)=m/2v_(2)` or `2v=v_(2)-v_(1)` or `v_(2)=2v+v_(1)=(2xx50)+50=150m//s` |
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| 291. |
Two `20 kg` cannon balls are chained together and fired horizontally with a velocity of `200 m//s` from the top of a `30 m` wall. The chain breaks during the flight of the cannon balls and one of them strikes the ground at `t=2s`, at a distance of `250 m` from the foot fo the wall, and `5 m` to the right of line of fire determine the position of the other cannon ball at that instant Neglect the resistance of air. |
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Answer» As no external force acts in `z` direction, hence `z` - coordinate of the centre of mass of he ball should be zero. To make `z`-coordinate zero other ball should fall symmetricaly with respect to `z` axis. Hence `z`-coordinate of other ball `=-5m`. The balls do not have any exeternal force in `x` direction. Hence in `x` direction the centre of mass should move with constant velocity. `x` coordinate of centre of mass at `t=1.5 s=200xx2=400m` Hence `x_(CM)=(m_(1)x_(1)+m_(2)x_(2))/(m_(1)+m_(2))` `400=(20xx250+20x_(2))/(20+20)` ` x_(2)=800-250=550m` Height fallen by centre of mass at `t=2s`, `h=1/2xx10xx(2)^(2)=20m` Hence `y` coordinate of centre of mass `=30-20=10n` Hence `y_(CM)=(m_(1)y_(1)+m_(2)y_(2))/(m_(1)+m_(2))` `10=(20xx0+20xxy_(2))/(20+20)` `implies y_(2)=20m` |
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| 292. |
A cannon and a supply of cannon balls are inside a sealed rail road car. The cannon fires to the right, the car recoils to the left. The canon balls remain in the car after hitting the far wall. Show that no matter how the cannon balls are fired, the rail road car cannot travel more than `L`. assuming it starts from rest. |
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Answer» Initially, the whole system is at rest, so `v_(CM)=0`. As there is no external force acting on the system`vecv_(CM)=`constant `= 0`. So position of centre of mass of system remains fixed. `x_(CM)=(mx_(1)+Mx_(2))/(m+M)`…………i where in is mass of the cannon balls and `M` that of the (car `+` cannon) system. `As /_x_(CM)=0` therefore `m/_x_(1)+m/_x_(2)=0`.....ii As cannon balls cannot leave the car, so maximum displacement of the balls relative to the car is `L` and in doing so the car will shift a distance `/_x_(2)=D`(say) relative to the ground, opposite to the displacement of the balls, then the displacement of balls relative to ground will be `/_x_(1)=L-D`.........iii Substituting the value of `/_x_(1)` from eqn iii in eqn ii we get `m(L-D)-MD=0` `impliesD=(mL)/(M+m)=L/(1+M/m)` `implies DltL` i.e. rail road car cannot travel more than `L`, |
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| 293. |
A railway flat car has an artillery gun installed on it. The combined system has a mass `M` and moves with a velocity `V`. The barrel of the gun makes an angle a with the horizontal. A shell of mass `m` leaves the barrel at a speed `v` relative to the barrel. The speed of the flat car so that it may stop after the firing isA. `(mv)/(M+m)`B. `((Mv)/(M+m))cosalpha`C. `(mv)/(M+m))cosalpha`D. `(M+m))cosalpha` |
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Answer» Correct Answer - C As the car stops so velocity of bullet after firing is same w.r.t ground and w.r.t car. Conservng momentum in horizontal direction. `(M+m)v=mvcosalphaimpliesV=(mvcosalpha)/(M+M)` |
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| 294. |
A ball of mass `m` is projected with a speed `v` into the barrel of a spring gun of mass `M` initially at rest lying on a frictionless surface. The mass sticks in the barrel at the point of maximum compression in the spring. The fraction of kinetic energy of the ball stored in the spring isA. `m/M`B. `M/(m+M)`C. `m/(M+M)`D. none of these |
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Answer» Correct Answer - B At maximum compression when `m` sticks to `M`, `mv=(M+m)V` ` V=(mv)/(M+m)` Fraction of `KE` stored in spring `f=(K_(i)-K_(f))/K_(i)` `:. F=1-(1/2(M+m)V^(2))/(1/2mv^(2))impliesf=1(m/(M+m))=M/(M+m)` |
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| 295. |
Statement -1 : When a girl jumbs from a boat, the boat slightly moves away from the shore. and Statement-2 : The total linear momentum of an isolated system remain conserved.A. Statement -1 is True, Statement -2 is True , Statement -2 is a correct explanation for Statement -1.B. Statement -1 is True, Statement -2 is 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 | |
| 296. |
A shell of mass m is moving horizontally with velocity `v_(0)` and collides with the wedge of mass M just above points A, as shown in the figure. As a consequences, wedge starts to move towards left and the shell returns with a velocity in xy-plance. The principle of conservation of momentum can be applied for A. system (m + M) along any directionB. system (m+M) along verticalC. system (m + M) horizontallyD. None of the above |
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Answer» Correct Answer - C In horizontal direction, net force on the system is zero. Therefore, principle of conservation of momentum can be applied for system (m+M) horizontally. |
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| 297. |
Assertion: In the figure shown, linear momentum of system (of blocks A and B) moves towards right during motion of block A over the block B. 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. |
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Answer» Correct Answer - A Froce of friction will act as B towards right (from ground). |
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| 298. |
Two blocks of mass m and 2m are kept on a smooth horizontal surface. Theya are connected by an ideal spring of force constant k. Initially the spring is unstretched. A constant force is applied to the heavier block in the direction shown in figure. Suppose at time t displacement of smaller block is `x_(1)` then displacement of the heavier block at this moment would be A. `x/2`B. `(Ft^(2))/(6m) + x/3`C. `x/3`D. `(Ft^(2))/(4m) - x/2` |
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Answer» Correct Answer - D `X_CM = (mx + 2mx_(2m))/(m + 2m)` `therefore 1/2a_(CM)^(t)^2 = (x+2x_(2m))/3` `therefore 3/2(F/(3m)t^(2) = x+2x_(2m)` `therefore X_(2m) = (Ft^(2))/(4m)) - x/2` |
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| 299. |
A light spring of constant `k` is kept compressed between two blocks of masses `m` and `M` on a smooth horizontal surface. When released, the block acquire velocities in opposite directions. The spring loses contact with the blocks when it acquires natural length. if the spring was initially compressed through a distance `x`, find the final speeds of the two blocks. |
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Answer» Consider the two blocks plus the spring to the the system, no external force acts on this suystem in horizontal direction. Hence the linear momentum will remain constant. Suppose, the block of mass `M` moves with a speed `V` and the other block with a speed `v` after losing contact with spring. From conservation of linear momentum in horizontal direction, we have `MV-mv=0` or `V=(mv)/M`........i Initially, the mechanical energy of the system `=1/2MV^(2)` finally the mechanical energy of the system `=1/2mv^(2)+1/2MV^(2)` As there is no friction, mechaical energy will remain conserved, therefore, `1/2m^(2)+1/2MV^(2)=1/2k^(2)`.........ii Solving Eqn i and ii, we get `v=[(kM)/(m(M+m))]^(1/2)x` and `V=[(kM)/(M(M+m))]^(1/2)x` |
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| 300. |
Assertion: A constant force F is applied on two blocks and one spring system as shown in figure. Velocity of centre of mass increases linearly with time. Reason: Acceleration of centre of mass is constant.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. |
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Answer» Correct Answer - A `a_(CM)=(F)/(3m)=cosnt ant` `:. `v_(CM)=a_(CM)t=(F)/(3m)t` or `v_(CM)prop t` |
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