1496 + Interview Questions in KINEMATICS IN GENERAL AWARENESS Page 23 InterviewSolution

1101.	On the basis of frame of reference in how many types is the motion divided? Write the names.
Answer» On the basis of frame of reference, the motion is divided into three types which are given below : One dimensional motion. Two dimensional motion. Three dimensional motion.

1101.

On the basis of frame of reference in how many types is the motion divided? Write the names.

Answer»

On the basis of frame of reference, the motion is divided into three types which are given below :

One dimensional motion.
Two dimensional motion.
Three dimensional motion.

Discussion

1102.	Differentiate between distance and displacement.
Answer» Distance and displacement are two quantities that may seem to mean the same thing yet have distinctly different definitions and meanings. (i) Distance : Distance is a scalar quantity that refers to the length of the path covered by the object regardless of its starting or ending position. In other words, distance refers to the length of the entire path travelled by the object. Unit of distance = meter in M.K.S. system = centimeter in C.G.S. system Unit of displacement = meter in M.K.S. system = centimeter in C.G.S. system. (ii) Displacement : Displacement is a vector quantity that refers to the shortest distance between the two positions of the object i.e, the difference between the final and initial positions of the object, in a given time. Its direction is from initial to final position of the object. It is represented by the vector drawn from the initial position to its final position.

1102.

Differentiate between distance and displacement.

Answer»

Distance and displacement are two quantities that may seem to mean the same thing yet have distinctly different definitions and meanings.

(i) Distance : Distance is a scalar quantity that refers to the length of the path covered by the object regardless of its starting or ending position. In other words, distance refers to the length of the entire path travelled by the object.
Unit of distance = meter in M.K.S. system
= centimeter in C.G.S. system
Unit of displacement = meter in M.K.S. system
= centimeter in C.G.S. system.

(ii) Displacement : Displacement is a vector quantity that refers to the shortest distance between the two positions of the object i.e, the difference between the final and initial positions of the object, in a given time. Its direction is from initial to final position of the object. It is represented by the vector drawn from the initial position to its final position.

Discussion

1103.	Time taken from Jaipur to Ajmer by car is 1.5 h. If the average velocity of car is 80 km/h then how much is the distance between Jaipur and Ajmer?
Answer» Given; time t = 1.5 hr.; \(\overline{v}\) = 80km/hr s = ? ∵ \(\overline{v}=\frac{s}{t} \Rightarrow s=\overline{v} \times t\) ∴ s = 80 × 1.5 = 120 km

Discussion

1104.	How much is the displacement in one cycle of a circular motion?
Answer» Zero, because initial and final points are same.

Discussion

1105.	Explain translatory motion with examples.
Answer» Translatory motion : When an object moves from one place to other place with respect to the frame of reference, then this motion is said to be translatory motion. Example : Motion of a car or bus on a straight road.

Discussion

1106.	Explain the differences between average and instantaneous velocity with examples.
Answer» Average velocity : Average velocity of a body is defined as the change in position or displacement (Δx) divided by time interval (Δt) in which that displacement occurs. \(\overrightarrow{v_{a v}}=\frac{\Delta \vec{x}}{\Delta t}=\frac{\overrightarrow{x_{2}}-\overrightarrow{x_{1}}}{t_{2}-t_{1}}\) Instantaneous velocity : The instantaneous velocity of a body is the velocity of the body at any instant of time or at any point of its path. \(\vec { v } =\lim _{ \Delta t\rightarrow 0 } \frac { \Delta \vec { x } }{ \Delta t } =\frac { d\vec { x } }{ dt } \) \(\vec{v}=\frac{d \vec{x}}{d t}\) Velocity can be positive, negative or zero. By studying speed and velocity we come to the result that at any time interval average speed of an object is equal or more than the average velocity but instantaneous speed is equal to instantaneous velocity.

1106.

Explain the differences between average and instantaneous velocity with examples.

Answer»

Average velocity : Average velocity of a body is defined as the change in position or displacement (Δx) divided by time interval (Δt) in which that displacement occurs.
\(\overrightarrow{v_{a v}}=\frac{\Delta \vec{x}}{\Delta t}=\frac{\overrightarrow{x_{2}}-\overrightarrow{x_{1}}}{t_{2}-t_{1}}\)

Instantaneous velocity : The instantaneous velocity of a body is the velocity of the body at any instant of time or at any point of its path.

\(\vec { v } =\lim _{ \Delta t\rightarrow 0 } \frac { \Delta \vec { x } }{ \Delta t } =\frac { d\vec { x } }{ dt } \)
\(\vec{v}=\frac{d \vec{x}}{d t}\)
Velocity can be positive, negative or zero.
By studying speed and velocity we come to the result that at any time interval average speed of an object is equal or more than the average velocity but instantaneous speed is equal to instantaneous velocity.

Discussion

1107.	What is negative acceleration called?
Answer» Negative acceleration called Retardation.

Discussion

1108.	Write the formula of speed.
Answer» Speed = Distance/Time

Discussion

1109.	`{:("Column A", "Column B"),((A)"Rest and motion",(a)" car taking a turn"),((B)"Oscillatory and periodic",(b)"Relative"),((C)"Translatory and rotatory",(c)"motion of the bob of a simple pendulum"),((D)"Variable velocity",(d)"Motion of the wheels of a bicycle in motion"):}`A. ` A to b, B to c, C to d, D to a`B. ` A to b, B to c, C to d, D to d `C. `A to c, B to b, C to d, D to a`D. `A to a, B to c, C to b, D to d`
Answer» Correct Answer - A A -B Rest and motion are relative. B-c Motion of the bob of a simple pendulum is oscillatory as well as periodic. C-d The wheels of a bicyle in motion perform both rotatory and translatory motions. D-a A car taking a turn is example of varable velocity.

Discussion

1110.	How much would be the displacement if a person walks 4 m towards east then 3m towards the south and then 4m towards west.
Answer» 3m towards south.

Discussion

1111.	An object moves the first half of the total distance with a speed of 2 m `s^(-1)` . If the average of the body is ` 3m s ^(-1)` , the speed of the body when it travelles the remaining distance is _______ ` m s^(-1)`A. 3B. 6C. 4D. 2
Answer» Let the total distance be 2 d Then, `(d)/(t_(1))=2" "(1)` `(d)/(t_(2))=x,` `(2d)/(t_(1)+t_(2))=3rArrd=(3(t_(1)+t_(2)))/(2)=rArrt4t_(1)=3t_(1)+3t_(2)` `t_(1)=3t_(2):. (d)/(t_(1//3))=xrArr(3d)/(t_(1))=x` From (1), `3(2)=x` `x=6 m s^(-1)`

Discussion

1112.	The ratiro of unit of accceleration and velocity gives unit of the physical quantity _______A. timeB. frequencyC. amplitudeD. speed
Answer» Correct Answer - B Unit of acceleration m `s^(-2)` Unit of velocity = ` m s^(-1)` The ratio of unit of acceleration to the velocity `= ( m s^(-2))/(m s^(-1)) =s ^(-1)` It is the unit of frequency.

Discussion

1113.	An electric fan rotates 100 times in 50s. If the of its wing from its axis of rotation is 0.5 m, then the speed of particle at the edge of the wing is ______ m `s^(-1)`A. `2pi`B. `0.5 pi`C. `pi`D. 2
Answer» Correct Answer - A Radius of the circular path coverd by the fan = 0.5m. The number of rotations per second= 100 Speed = `("distance")/("time")= (100 xx (2pir))/50` `(100 xx 2pi xx 0.5)/50 = 2pi m s^(-1)` `(4m + 3m + 5m)/(20s +10s +30s)= (12 m)/(60 s) = 0.2 m s^(-2)`

Discussion

1114.	An electric fnas rotates 100 times in 50s . If the blength of its wing from its axis of rotation is `0.5` m, then the speed of particle at the edge of the wing is _________ `m s^(-1)`A. `2pi`B. `0.5 pi`C. `pi`D. 2
Answer» Radius of the circle path covered by the fan=0.5 m the number of rotation of rotations per socond =100 `"Speed"=("distance")/("time")=(100xx(2pir))/(50)` `=(100xx2pixx0.5)/(50)=2pi ms^(-1)`

Discussion

1115.	An electric fan rotates 100 times in one second. If the length of its wing from its axis of rotation is 0.5 m . Then the speed of the fan is ______ m `s^(-1)`A. `100 pi`B. `50 pi`C. `200 pi`D. 100
Answer» Correct Answer - A Radius of the circular path traced by fan = 0.5 m the number of rotation per second = 100 Speed - `("distance")/("time")= (100 xx (2pir))/ (1 "sencond")` `100 xx 2 pi xx 0.5 ` `50 xx 2pi= 100 pi m s^(-1)`

Discussion

1116.	A canon can fire shells at speed u. Inclination of its barrel to the horizontal can be changed in steps of `Delta theta = 1^(@)` ranging from `theta_(1) = 15^(@)` to `theta_(2) = 85^(@)`. Let `R_(n)` be the horizontal range for projection angle `theta = n^(@)`. `Delta R_(n) = \|R_(n) - R_(n+1)\|` For what value of n the value of `DeltaR_(n)` is maximum? Neglect air resistance. (ii) A small water sprinkler is in the shape of a hemisphere with large number of uniformly spread holes on its surface. It is placed on ground and water comes out of each hole with speed u. Assume that we mentally divide the ground into many small identical patches – each having area `DeltaS`. What is the distance of a patch from the sprinkler which receives maximum amount of water ?
Answer» Correct Answer - (i) `n = 84^(@)` (ii) `(u^(2))/(g)`

Discussion

1117.	A jugglar maintains four balls in motion, vaking each in turn rise to a height of `20 .0 m` from his hand .Find the velocity with which the jugglar project these balls and the position of other three balls at the instant when the foruth ball is just leaving the hand of jugglar. Take `g= 10 m//s^2.
Answer» Taking vertical upward motion of a ball, we have ` S= 20 m , a=- 10 m//s^2 , ltbRgt ` u?, v=0, t=?` As ` v^2 =u^2 + 2 aS` so, ` 0=u^2 =2 ( -10) xx 20` or ` u = 20 ms^(-1)` ltbRgt Also ` v=u + at ` or ` t=(v-u) /a = )0-20)/(-10) =2s` so each ball will return to the hand of jugglar after ` 4 s`. To maintain proper distance, the balls must be projected upwards with an interval of time ` = 4/4 = 1 s`. When the fourth ball is in hand, the third ball has travelled for ` 1 s`, second ball fro ` 2` s and dirst ball for 3 s`. (i) Fro Third ball, ltbRgt ` S= ut + 1/2 at^2 = 20 xx 1 + 1/2 (-10) 1^2 = 15 m` Thus third ball will be ` 15 m` ablove the ground going upwards. (ii) For second ball, ` S =20 xx 2 + 1/2 ( -10) xx 2^2 = 20 m` The second ball will be ` 20 m` abover the ground just at rest. (iii) For first ball, ` S= 20 xx 3 ss 1/2 + ( -1 0) xx 3^2 = 15 m` The first ball will be ` 15 m above the ground moving dounwards.

Discussion

1118.	A juggler maintains four balls in motion, making each to them to rise a height of `20 m` from his hand. What time interval should he maintain, for the alphaer distance between them.A. `3s`B. `3/2s`C. `1s`D. `2s`
Answer» Correct Answer - C

Discussion

1119.	The range of a projectile fired at an angle of 15° is 50 m. If it is fired with the same speed at an angle of 45°, its range will be(a) 25 m (b) 37 m (c) 50 m (d) 100
Answer» (d) 100 m. Explanation: In the first case Range = 50 = (u² /g) Sin(2x15°) = (u² /2g) => (u² /g)=100 In the second case range = (u² /g) Sin(2x45°) = (u² /g) Sin90° =100 m

1119.

The range of a projectile fired at an angle of 15° is 50 m. If it is fired with the same speed at an angle of 45°, its range will be(a) 25 m (b) 37 m (c) 50 m (d) 100

Answer»

(d) 100 m.

Explanation:

In the first case Range = 50 = (u² /g) Sin(2x15°) = (u² /2g) => (u² /g)=100 In the second case range = (u² /g) Sin(2x45°) = (u² /g) Sin90° =100 m

Discussion

1120.	The figure-1.125 shows the acceleration versus time graph of a train. If it starts from rest, the distance it travels before it comes to rest is : A. 30 mB. 26 mC. 13 mD. 40 m
Answer» Correct Answer - B

Discussion

1121.	The component of a vector `r` along X-axis will have maximum value ifA. (a) ` vec r` is along positve Y-axisB. (b) ` vec r` is along positve X-axisC. (c ) `vec r` makes an angle of ` 45^@` with the X-axisD. (d) ` vec r` is along negative Y-axis
Answer» Correct Answer - B If ` vec r` makes an angle ` theta` with x-axis, theta component of ` vec r` is along x-axis = ` r cos theta`. It will be maximum if ` cos = max =m 1` or ` theta0^2` i.e. vec r` along positve x-axis.

Discussion

1122.	A car is moving along X- axis. As shown in figure it moves from O to P in 18 seconds and return from P to Q in 6 second. What are the average velocity and average speed of the car in going fromO to P From O to P and hack to Q
Answer» O to P, Average velocity =20 ms O to P and back to Q Average velocity = 10 ms^-1 Average speed = 20 ms^-1

1122.

A car is moving along X- axis. As shown in figure it moves from O to P in 18 seconds and return from P to Q in 6 second. What are the average velocity and average speed of the car in going fromO to P From O to P and hack to Q

Answer»

O to P, Average velocity =20 ms
O to P and back to Q

Average velocity = 10 ms^-1

Average speed = 20 ms^-1

Discussion

1123.	What is the vector sum of n coplanar forces, each of magnitude F, if each force makes an angle of \(\frac{2π}{n}\) with the preceding force?
Answer» Resultant force is zero.

Discussion

1124.	What is the vector sum of n coplanar forces, each of magnitude F, if each force makes an angle of 2π/n with the preceding force ?
Answer» Resultant force is zero.

Discussion

1125.	Find the vector sum of `N` coplanar forces, each of the magnitude `F`,when each force makes an angle of `2pi//N` with that preceding it.
Answer» Let (A) be the magnitude of each vector and the first vector is along x-axis. Then the various vectors will be making angles with the x-axis ltbRgt as ` 0, ( 2 pi) /N, 9 4 pi)/N, ( 6 pi) /N ,……( (N- a)/N 2 pi`. Let ` R_x ` and ` R_y` be the (X) and Y-component of resultant vector ` vec T`. The ` R-x = A cos 0^@ + A cos ( 2 pi) /N + A cos ( 4 pi) N + .....` ` + A cos ( N- 10 ( 2 pi)/N = A sum _ ( i=0)^(( N-1)) cos ( 2 pi i)/N` ltbRgt and ` R_y = A sun_ (i=0)^(( N-1)) sin ( 2 pi i) /N` But ` sun_(i=0)^(( N-1)) cos ( 2 pi i)/N = sum_(i=0)^((N-1)) sin ( 2pi i) /N =0` :. ` R_x = R_y =0`. Hence, ` R= sqrt (R_x^2 + R_y^2) =0` Note. the total angle subtende by (N) vectors with the inital cector `= ( 2 pi // N) xx N= 2 pi `. It means these (N) vectors can be represented by the (N) sides of a closed polygon taken in one order .Hence their vector sum is zero.

Discussion

1126.	The speed -time graph of a particle moving along a fixed direction id shown ifn Fig. 2 (b) . 31. Fid (i) distance travelled by the particle between ` 0 sec` to ` 10 sec` (ii) average speed between thid interval (iii) the time when the speed was minimum (iv) the time when the speed was maximum. .
Answer» (a) Distanc travelled by the particle between ` 0` to ` 10` s will be =area of ` Delta OAB`, whose base of ` 10 s` and height is `12 ms^(-1)` ` =1/2 xx 10 xx 12 = 60 m` Average speed ` = (60)/(10) = 6 ms^(-1)` (b) Let `S_1` and `S_2` be the distances coverd by the particle in the time in terval ` t_1 =2 s to ` 5 s` and ` t_2 = 5 s` to 6 s, then total distance covered in time inerval `t=2 to ` 6 s` will be ` S= S_1` ....(i) To finds ` S_1`, let us consider ` u_1` is the velocity of particle agter ` 2 secpmds` and ` a_1` is the accelration of the particle during the time interval zero to ` 5 seocond` . ltbRgt The ` u= 0, = 12 m//s, a= a_1 ` and ` t= 5 s`. ltbRgt We have, `a_1 = (v-u)/t = ( 12-0)/5 = (120/5 = 2.4 ms^(-20` :. ` u-1 = u + a_1 t= 0 + 2.4 xx 2 = 4.8 ms^(-1)` Thus for tje distanc etravelled by particle in `3 secons` (i.e. time inrerval ` 2 s` to 5 s) we have ltbRgt ` u_1 = 4.8 ms^(-1), t-1 =3 s, a_1 = 2.4 ms^(-20, S_1 = ?` As ` S_1 = u-1 t_1 = 1/2 a_1 r_1^2` :. ` S-1 = 4.8 xx 3 + 1/2 xx 2. 4 xx 3^2 = 25 .2 m` To find ` S_2`. Let ` a-2` be the the accelration of the particle during the motion, ` t= 5 to ` t= 10 s`. We have ` a_2 = ( 0-12)/( 10- 5) =- 2. 4 ms^(2)` Taking motion of the particle in time interval ` t= 5 s tp ` t=6 s`, we have ltbRgt ` u_2 = 12 ms^(-1), a-2 =- 2.4 ms^(-2) , t_2 = 1 s, S_2 = ?` As ` S-2 = u-2 t_2 + 1/2 a_2 t_2^2` :. ` S_2 = 12 xx 1 = 1/2 (- 2. 4) 1^2 = 10.8 m` :. Total distance travelled, `S= 25. + 10 .8 = 36 m` Acerage speed = 936)/( 6- 2) = 9360/4 = 9 ms^(-1)`.

Discussion

1127.	The speed -time graph of a particle moving along a fixed direction id shown ifn Fig. 2 (b) . 31. Fid (i) distance travelled by the particle between ` 0 sec` to ` 10 sec` (ii) average speed between thid interval (iii) the time when the speed was minimum (iv) the time when the speed was maximum. .
Answer» (i) Distance trvallled by particle in time interval `0 ` to ` 10` seconds is S=area under velocity-time graph ltBrgt ` 1/2 xx ( 10 - 0) xx ( 12 -0) = 60 m` (ii) Average speed , ` v_(av) = ( Total distance travelled) / ( total time taken) = ( 60)/( 10) = 6 ms^(-1)` ltbRgt (ii) Speed of particle is minimum at ` t= 0 s` and ` t= 10 s` (iv) Speed of particle is maximum at ` t= 5s `.

Discussion

1128.	A point moves with uniform acceleration and `v_(1), v_(2)`, and `v_(3)` denote the average velocities in the three successive intervals of time `t_(1).t_(2)`, and `t_(3)` Which of the following Relations is correct?.A. `(v_(1)-v_(2)):(v_(2)-v_(3)=(t_(1)-t_(2):(t_(2)+t_(3)`.B. `(v_(1)-v_(2)):(v_(2)-v_(3)=(t_(2)-t_(2):(t_(2)+t_(3)`C. `(v_(1)-v_(2)):(v_(2)-v_(3)=(t_(1)-t_(2):(t_(2)+t_(3)`D. `(v_(1)-v_(2)):(v_(2)-v_(3)=(t_(1)-t_(2):(t_(2)+t_(3)`
Answer» Correct Answer - B Suppose `u` be the initial velocity. Velocity after time `t_(1)`: `v_(11) =u+at_(1)` Velocity after time `t_(1) +t_(2)`: `v_(22) =u +a (t_(1) +t_(2)` Velocity after time `t_(1) +t_(2) +t_(3)`: `v 3 =u+a (t_(1) +t_(2) +t_(3))` Now `v_(1) =(u+v_(11))/(2) =(u+u+at_(1))/(2) =u+(1)/(2) at_(1)` `v_(2) =(v_(11) +v_(22))/(2) =u+at_(1) +(1)/(2_(2))` `v_(3) =(v_(33)+ v_(33))/(2) =u+at_(2) + (1)/(2) at_(3)` So `v_(1)-v_(2) =- (1)/(2) a(t_(1) +t_(2))` `v_(2)-v_(3) =-(1)/(2) a(t_(2)-t_(3))` `(v_(1)-v_(2))`: `(v_(2)-v_(3) =(t_(1) +t_(2))` : `(t_(2) +t_(3))`.

Discussion

1129.	Velocity of a particle is `v=(2hat i + 3hat j - 4hat k) m//s` and its acceleration is zero. State whether it is 1-D, 2-D or 3-D motion?
Answer» Correct Answer - A Since, acceleration of the particle is zero. Therefore, it is uniform motion or motion in a straight line. So, it is one dimensional motion.

Discussion

1130.	Two particles moving initially in the same direction undergo a one-dimensional, elastic collision. Their relative velocities before and after the collision are Vector v1 and Vector v2. Which of the following is not correct?(a) \| Vector v1 \| = \| Vector v2 \|(b) Vector v1 = - Vector v2 only if the two are of equal mass.(c) Vector v1 . Vector v2 = - \| Vector v1 \|2(d) \| Vector v2 . Vector v1 \| = Vectoe v2\|2
Answer» Correct Answer is: (b) Vector v₁ = - Vector v₂ only if the two are of equal mass. For a head-on elastic collision, relative velocity after collision = - relative velocity before collision. This is independent of the masses of the particles.

1130.

Two particles moving initially in the same direction undergo a one-dimensional, elastic collision. Their relative velocities before and after the collision are Vector v1 and Vector v2. Which of the following is not correct?(a) | Vector v1 | = | Vector v2 |(b) Vector v1 = - Vector v2 only if the two are of equal mass.(c) Vector v1 . Vector v2 = - | Vector v1 |2(d) | Vector v2 . Vector v1 | = Vectoe v2|2

Answer»

Correct Answer is: (b) Vector v₁ = - Vector v₂ only if the two are of equal mass.

For a head-on elastic collision,

relative velocity after collision

= - relative velocity before collision.

This is independent of the masses of the particles.

Discussion

1131.	A force Vector F = - K(yi + xj), where k is a positive constant, acts on a particle moving in the xy plane. Starting from the origin, the particle is taken along the positive x-axis to the point (a, 0), and then parallel to the y-axis to the point (a, a). The total work done by the force on the particle is(a) -2ka2 (b) 2ka2 (c) -ka2 (d) ka2
Answer» Correct answer is: (c) -ka² For the first displacement, y = 0. Hence F_x = 0 and no work is done. For the second displacement, F_y = - ka and Δy = a. Work = F_y Δ_y = - ka² .

1131.

A force Vector F = - K(yi + xj), where k is a positive constant, acts on a particle moving in the xy plane. Starting from the origin, the particle is taken along the positive x-axis to the point (a, 0), and then parallel to the y-axis to the point (a, a). The total work done by the force on the particle is(a) -2ka2 (b) 2ka2 (c) -ka2 (d) ka2

Answer»

Correct answer is: (c) -ka²

For the first displacement, y = 0. Hence F_x = 0 and no work is done. For the second displacement, F_y = - ka and Δy = a.

Work = F_y Δ_y = - ka² .

Discussion

1132.	Two particles moving initially in the same direction undergo a one- dimensional , elastic collision . Their relative velocities before and after the collision are `vec(v_(1))` and `vec(v_(2))`. Which of the following is not correct?A. `\|vec(v_(1))\|=\| vec(v_(2))\|`B. `vecv_(1)=-vecv_(2)` only if the two are of equal mass.C. `vecv_(1).vecv_(2)=-\|vecv_(1)\|^(2)`D. `\|vec(v_(2)).vec(v_(1))\|=\|vec(v_(2))\|^(2)`
Answer» Correct Answer - B

Discussion

1133.	In a one-dimensional collision between two particles, their relative velocity is `vec(v)_(1)` before the collision and `vec(v)_(2)` after the collision.A. `vec(v)_(1) = vec(v)_(2)` if the collision is elasticB. `vec(v)_(1) =- vec(v)_(2)` if the collision is elasticC. `\|vec(v)_(2)\| = \|vec(v)_(1)\|` in all casesD. `vec(v)_(1) =- k vec(v)_(2)` in all cases, where `k ge 1`
Answer» Correct Answer - B::C::D

Discussion

1134.	In a one-dimensional collision between two particles, their relative velocity is Vector v1 before the collision and Vector v2 after the collision.(a) Vector v1 = Vector v2 if the collision is elastic.(b) Vector v1 = Vector v2 if the collision is elastic.(c) \|Vector v1\| = \|Vector v2\| in all cases.(d) Vector v1 = - k Vector v2 in all cases, where k ≥ 1.
Answer» Correct Answer is: (b, c, & d)

Discussion

1135.	A long block A is at rest on a smooth horizontal surface. A small block B, whose mass is half of A, is placed on A at one end and projected along A with some velocity u. The coefficient of friction between the blocks is μ.(a) The blocks will reach a final common velocity μ/3.(b) The work done against friction is two-thirds of the initial kinetic energy of B.(c) Before the blocks reach a common velocity, the acceleration of A relative to B is 2/3 μg.(d) Before the blocks reach a common velocity the acceleration of A relative to B is 3/2 μg.
Answer» Correct Answer is: (a, b, & d) As there are no external forces acting on the ‘A + B’ system, its total momentum is conserved. If the masses of A and B are 2m and m respectively, and v is the final common velocity, mu = (m + 2m)v or v = u/3. Work done against friction = loss in KE = 1/2 mu² - 1/2 (3m)v² = 1/2 mu² - 1/2 (3m) u²/9 = 1/2 mu² [ 1 - 1/3 ] = 2/3 x 1/2 mu² . The force of friction between the blocks is μmg. Acceleration of A (to the right) = a₁ = μmg/2m = μg/2. Acceleration of B (to the left) = a₂ = μmg/m = μg. Acceleration of A relative to B = a₁ = (-a₂) = 3/2 μg.

1135.

A long block A is at rest on a smooth horizontal surface. A small block B, whose mass is half of A, is placed on A at one end and projected along A with some velocity u. The coefficient of friction between the blocks is μ.(a) The blocks will reach a final common velocity μ/3.(b) The work done against friction is two-thirds of the initial kinetic energy of B.(c) Before the blocks reach a common velocity, the acceleration of A relative to B is 2/3 μg.(d) Before the blocks reach a common velocity the acceleration of A relative to B is 3/2 μg.

Answer»

Correct Answer is: (a, b, & d)

As there are no external forces acting on the ‘A + B’ system, its total momentum is conserved. If the masses of A and B are 2m and m respectively, and v is the final common velocity, mu = (m + 2m)v or v = u/3.

Work done against friction = loss in KE = 1/2 mu² - 1/2 (3m)v²

= 1/2 mu² - 1/2 (3m) u²/9 = 1/2 mu² [ 1 - 1/3 ] = 2/3 x 1/2 mu² .

The force of friction between the blocks is μmg.

Acceleration of A (to the right) = a₁ = μmg/2m = μg/2.

Acceleration of B (to the left) = a₂ = μmg/m = μg.

Acceleration of A relative to B = a₁ = (-a₂) = 3/2 μg.

Discussion

1136.	A hollow vertical cylinder of radius r and height h has a smooth internal surface. A small particle is placed in contact with the inner side of the upper rim, at point A, and given a horizontal speed u, tangentical t the rim. it leaves the lower rim at point B, vertically below A. If n is an integer thenA. `(u)/(2pir) sqrt(2h//g) = n`B. `(h)/(2pir) = n`C. `(2pir)/(h) = n`D. `(u)/(sqrt(2gh)) = n`
Answer» Correct Answer - A

Discussion

1137.	A block of mass m is pushed towards a movable wedge of mass ηm and height h, with a velocity u. All surfaces are smooth. The minimum value of u for which the block will reach the top of the wedge is(a) √(2gh)(b) η √(2gh)(c) √(2gh (1 + 1/η))(d) √(2gh ( 1 - 1/η))
Answer» Correct answer is: (c) √(2gh (1 + 1/η)) The center of mass of the ‘block plus wedge’ must move with speed mu / m + ηm = u/1+η = v_CM. ∴ 1/25 mu² - mgh = 1/2 ( m + ηm) v²_CM.

1137.

A block of mass m is pushed towards a movable wedge of mass ηm and height h, with a velocity u. All surfaces are smooth. The minimum value of u for which the block will reach the top of the wedge is(a) √(2gh)(b) η √(2gh)(c) √(2gh (1 + 1/η))(d) √(2gh ( 1 - 1/η))

Answer»

Correct answer is: (c) √(2gh (1 + 1/η))

The center of mass of the ‘block plus wedge’ must move with speed

mu / m + ηm = u/1+η = v_CM.

∴ 1/25 mu² - mgh = 1/2 ( m + ηm) v²_CM.

Discussion

1138.	A man who can swim at a speed v relative to the water wants to cross a river of width d, flowing with a speed u. The point opposite him across the river is P. (a) The minimum time in which he can cross the river is d / v .(b) He can reach the point P in time d / v .(c) He can reach the point P in time d /√( v2 - u2)(d) He cannot reach P if u > v.
Answer» Correct Answer is: (a, c, & d)

Discussion

1139.	A strip of wood of length l is placed on a smooth horizontal surface. An insect starts from one end of the strip, walks with constant velocity and reaches the other end in time t1. It then flies off vertically. The strip moves a further distance l in time t2.(a) t2 = t1 (b) t2 < t1(c) t2 > t1(d) Either (b) or (c) depending on the masses of the insect and the strip
Answer» Correct answer is: (c) t₂ > t₁ The strip and insect will move in opposite directions such that their centre of mass is stationary. Their relative velocity is l/t₁ and each will have speed < l/t₁. When the insect flies off, the strip will continue to move at its previous speed and hence will cover a further distance l in time t₂ > t₁.

1139.

A strip of wood of length l is placed on a smooth horizontal surface. An insect starts from one end of the strip, walks with constant velocity and reaches the other end in time t1. It then flies off vertically. The strip moves a further distance l in time t2.(a) t2 = t1 (b) t2 < t1(c) t2 > t1(d) Either (b) or (c) depending on the masses of the insect and the strip

Answer»

Correct answer is: (c) t₂ > t₁

The strip and insect will move in opposite directions such that their centre of mass is stationary. Their relative velocity is l/t₁ and each will have speed < l/t₁. When the insect flies off, the strip will continue to move at its previous speed and hence will cover a further distance l in time t₂ > t₁.

Discussion

1140.	A long block A is at rest on a smooth horizontal surface. A small block B, whose mass is half of A, is placed on A at one end and projected along A with some velocity u. The coefficient of friction between the blocks is `mu`: A. The blocks will a final common velocity `(u)/(3)`.B. The work done against friction is two-thirds of the initial kinetic energy of B.C. Before the blocks reach a common velocity, the acceleration of A relative to B is `(2)/(3)mu g`.D. Brfore the blocks reach a common velocity the acceleration of A relative to B is `(3)/(2)mu g`.
Answer» Correct Answer - A::B::D

Discussion

1141.	In the figure, the block B of mass m starts from rest at the top of a wedge W of mass M. All surfaces are without friction. W can slide on the ground. B slides down onto the ground, moves along it with a speed v, has an elastic collision with the wall, and climbs back onto W.(a) B will reach the top of W again. (b) From the beginning, till the collision with the wall, the centre of mass of ‘B plus W’ does not move horizontally. (c) After the collision, the centre of mass of ‘B plus W’ moves with the velocity 2mv / m + M.(d) When B reaches its highest position on W, the speed of W is 2mv / m + M .
Answer» Correct Answer is: (a, c, & d) The horizontal momentum of the system is conserved (= 0) till the collision, as there are no horizontal forces acting on the system. At the collision, an additional impulse 2mv is given by the wall to the system.

1141.

In the figure, the block B of mass m starts from rest at the top of a wedge W of mass M. All surfaces are without friction. W can slide on the ground. B slides down onto the ground, moves along it with a speed v, has an elastic collision with the wall, and climbs back onto W.(a) B will reach the top of W again. (b) From the beginning, till the collision with the wall, the centre of mass of ‘B plus W’ does not move horizontally. (c) After the collision, the centre of mass of ‘B plus W’ moves with the velocity 2mv / m + M.(d) When B reaches its highest position on W, the speed of W is 2mv / m + M .

Answer»

Correct Answer is: (a, c, & d)

The horizontal momentum of the system is conserved (= 0) till the collision, as there are no horizontal forces acting on the system. At the collision, an additional impulse 2mv is given by the wall to the system.

Discussion

1142.	The displacement (x) of a particle depends on time (t) as x = αt2 - βt 3. (a) The particle will return to its starting point after time α/β.(b) The particle will come to rest after time 2α/3β. (c) The initial velocity of the particle was zero but its initial acceleration was not zero. (d) No net force will act on the particle at t = α/3β
Answer» Correct Answer is: ( a, b, c, & d)

Discussion

1143.	A man of mass m stands on a long flat car of mass M, moving with velocity V. If he now begins to run with velocity u, with respect to the car, in the same direction as V, the velocity of the car will be(a) V - mu/M (b) V - mu/(m + M) (c) V + mu/(m + M) (d) V - u(M - m)/(M + m)
Answer» Correct Answer is: (b) V - mu/(m + M) Momentum of the ‘man + car’ system = (m + M)V = constant. When the man begins to run, let v_M = velocity of man and v_C = velocity of car. ∴ velocity of man with respect to the car = u = v_M - v_C. By conservation of momentum, (m + M) V = mv_M + Mv_C. Rearranging, v_C = V - mu/m + M.

1143.

A man of mass m stands on a long flat car of mass M, moving with velocity V. If he now begins to run with velocity u, with respect to the car, in the same direction as V, the velocity of the car will be(a) V - mu/M (b) V - mu/(m + M) (c) V + mu/(m + M) (d) V - u(M - m)/(M + m)

Answer»

Correct Answer is: (b) V - mu/(m + M)

Momentum of the ‘man + car’ system = (m + M)V = constant. When the man begins to run, let v_M = velocity of man and v_C = velocity of car.

∴ velocity of man with respect to the car = u = v_M - v_C.

By conservation of momentum, (m + M) V = mv_M + Mv_C.

Rearranging, v_C = V - mu/m + M.

Discussion

1144.	The two blocks A and B of equal mass are initially in contact when released from rest on the inclined plane. The coefficients of friction between the inclined plane and A and B are μ1 and μ2 respectively.(a) If μ1 > μ2, the blocks will always remain in contact. (b) If μ1 < μ2, the blocks will slide down with different accelerations. (c) If μ1 > μ2, the blocks will have a common acceleration 1/2 (μ1 + μ2)gsin θ.(d) If μ1 < μ2, the blocks will have a common acceleration μ1μ2 / μ1 + μ2 sin θ.
Answer» Correct Answer is: (a, b) The block with lower value of μ will tend to have greater acceleration down the slope.

1144.

The two blocks A and B of equal mass are initially in contact when released from rest on the inclined plane. The coefficients of friction between the inclined plane and A and B are μ1 and μ2 respectively.(a) If μ1 > μ2, the blocks will always remain in contact. (b) If μ1 < μ2, the blocks will slide down with different accelerations. (c) If μ1 > μ2, the blocks will have a common acceleration 1/2 (μ1 + μ2)gsin θ.(d) If μ1 < μ2, the blocks will have a common acceleration μ1μ2 / μ1 + μ2 sin θ.

Answer»

Correct Answer is: (a, b)

The block with lower value of μ will tend to have greater acceleration down the slope.

Discussion

1145.	The ring R in the arrangement shown can slide along a smooth, fixed, horizontal rod XY. It is attached to the block B by a light string. The block is released from rest, with the string horizontal.(a) One point in the string will have only vertical motion. (b) R and B will always have momenta of the same magnitude. (c) When the string becomes vertical, the speeds of R and B will be inversely proportional to their masses. (d) R will lose contact with the rod at some point.
Answer» Correct Answer is: (a, c) There are no horizontal forces acting on the ‘R plus B’ system. Hence, its centre of mass will move down vertically, and horizontal momentum will be conserved.

1145.

The ring R in the arrangement shown can slide along a smooth, fixed, horizontal rod XY. It is attached to the block B by a light string. The block is released from rest, with the string horizontal.(a) One point in the string will have only vertical motion. (b) R and B will always have momenta of the same magnitude. (c) When the string becomes vertical, the speeds of R and B will be inversely proportional to their masses. (d) R will lose contact with the rod at some point.

Answer»

Correct Answer is: (a, c)

There are no horizontal forces acting on the ‘R plus B’ system. Hence, its centre of mass will move down vertically, and horizontal momentum will be conserved.

Discussion

1146.	A particle moves along the x-axis as follows: it starts from rest at t = 0 from a point x = 0 and comes to rest at t = 1 at a point x = 1. No other information is available about its motion for the intermediate time (0 < t < 1). If α denotes the instantaneous acceleration of the particle then(a) α cannot remain positive for all t in the interval 0 ≤ t ≤ 1 (b) \|α\| cannot exceed 2 at any point in its path (c) \|α\| must be ≥ 4 at some point or points in its path (d) α must change sign during the motion, but no other assertion can be made with the information given
Answer» Correct Answer is: (a ,c)

Discussion

1147.	A point moving along the x-direction starts from rest at x=0 and comes to rest at x=1 after 1 s Its accelertion at any point is denoted by `alpha` Which of following is not correct ?A. `alpha` must change sign during the motion .B. `\|alpha\|`ge 4 units st some or all points during the motion.C. It is not possible to specify an uppwer limit for `\|alpha\|` from the given data.D. `\|alpha\|` connot be less that 1/2 during the motion.
Answer» Correct Answer - D

Discussion

1148.	A point moving along the x-direction starts from rest at x = 0 and comes to rest at x = 1 after 1 s. Its acceleration at any point is denoted by α. Which of the following is not correct?(a) α must change sign during the motion. (b) \| α \| ≥ 4 units at some or all points during the motion. (c) It is not possible to specify an upper limit for \| α \| from the given data. (d) \| α \| cannot be less that 1/2 during the motion.
Answer» Correct Answer is: (d) \| α \| cannot be less that 1/2 during the motion. The simplest solution to the motion is shown by the plot OAB. The area under the v~t plot must be equal to 1 (displacement). For this, V/2 = 1 or V = 2. The acceleration V/1/2 = 4 units. ∴ slope of OA = 4. Taking OAB as reference, the condition of the motion can be satisfied by any other curve, e.g., OCB, as long as the area under it is equal to 1. Also, the slope of the curve at any point gives the acceleration α. Points D and E have α < 4. Points below D have α > 4 and above D have α > 4. As the shape of the curve OCB is arbitrary, it is not possible to set upper or lower limits on the acceleration.

1148.

A point moving along the x-direction starts from rest at x = 0 and comes to rest at x = 1 after 1 s. Its acceleration at any point is denoted by α. Which of the following is not correct?(a) α must change sign during the motion. (b) | α | ≥ 4 units at some or all points during the motion. (c) It is not possible to specify an upper limit for | α | from the given data. (d) | α | cannot be less that 1/2 during the motion.

Answer»

Correct Answer is: (d) | α | cannot be less that 1/2 during the motion.

The simplest solution to the motion is shown by the plot OAB. The area under the v~t plot must be equal to 1 (displacement).

For this, V/2 = 1 or V = 2.

The acceleration V/1/2 = 4 units.

∴ slope of OA = 4.

Taking OAB as reference, the condition of the motion can be satisfied by any other curve, e.g., OCB, as long as the area under it is equal to 1. Also, the slope of the curve at any point gives the acceleration α. Points D and E have α < 4. Points below D have α > 4 and above D have α > 4.

As the shape of the curve OCB is arbitrary, it is not possible to set upper or lower limits on the acceleration.

Discussion

1149.	Two blocks are in contect moving on an inclined plane of inclination `alpha` with the horizontal. The masses of the blocks are equal to `m_(1)` and `m_(2)` and the coefficients of friction between the inclined plane and the blocks are equal to `mu_(1)` and `mu_(2) (mu_(1) gt mu_(2))` respectively. Find (a) common acceleration of the blocks and (b) the contact force between the blocks. A. If `mu_(1)gt mu_(2)`, the blocks will always remain in contact.B. If `mu_(1)lt mu_(2)`, the blocks will slide down with different accelerations.C. If `mu_(1)gt mu_(2)`, the blocks will have a common acceleration `(1)/(2)(mu_(1)_mu_(2))g sin theta`.D. If `mu_(1)lt mu_(2)`, the blocks will have a common acceleration `(mu_(1)mu_(2)g)/(mu_(1)+mu_(2))sin theta`.
Answer» Correct Answer - A::B

Discussion

1150.	The ring R in the arrangement shown can slide along a smooth, fixed, horizontal rod XY. If is attached to the block B by a light sring. The blocks is released from rest, with the string horizontal.A. One point in the string will have only vertical motionB. R and B will always have momenta of the same magnitudeC. When the string becomes vertical, the speeds of R and B will be inversely proportional to their massesD. R will lose contact with the rod at some point
Answer» Correct Answer - A::C

Discussion

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