Category: Machine Kinematics Objective Questions

250+ TOP MCQs on Types of Links and Answers

Machine Kinematics Multiple Choice Questions on “Types of Links”.

1. When the nature of contact between the element of a pair is such that it can only slide relative to the other, the pair is known as a
a) screw pair
b) spherical pair
c) turning pair
d) sliding pair
Answer: d
Clarification: When the nature of contact between the element of a pair is such that one element can turn abut the other by screw threads, the pair is known as a screw pair. When the nature of contact between the element of a pair is such that it can only slide relative to the other, the pair is known as a sliding pair.

2. When the nature of contact between the element of a pair is such that it can turn or revolve about a fixed axis, the pair is known as a rolling pair.
a) True
b) False
Answer: b
Clarification: When the nature of contact between the element of a pair is such that it can turn or revolve about a fixed axis, the pair is known as a turning pair.

3. When the nature of contact between the element of a pair is such that one element can turn abut the other by screw threads, the pair is known as a
a) screw pair
b) spherical pair
c) turning pair
d) sliding pair
Answer: a
Clarification: When the nature of contact between the element of a pair is such that one element can turn abut the other by screw threads, the pair is known as a screw pair.

4. A sliding pair has a completely constrained motion.
a) True
b) False
Answer: a
Clarification: When the nature of contact between the element of a pair is such that it can only slide relative to the other, the pair is known as a sliding pair and it has a completely constrained motion.

5. Which of the following is an example of sliding pair?
a) Piston and cylinder of a reciprocating steam engine
b) Shaft with collars at both ends fitted into a circular hole
c) Lead screw of a lathe with nut
d) Ball and a socket joint
Answer: a
Clarification: When the nature of contact between the element of a pair is such that it can only slide relative to the other, the pair is known as a sliding pair and can be found in piston and cylinder of a reciprocating steam engine.

6. The ball and socket joint is an example of screw pair.
a) True
b) False
Answer: b
Clarification: The ball and socket joint is an example of spherical pair. When the two elements of a pair are connected in such a way that one element turns or swivels about the other fixed element, the pair formed is called a spherical pair.

7. Which of the following is an open pair?
a) Ball and socket joint
b) Journal bearing
c) Lead screw and nut
d) Cam and follower
Answer: a
Clarification: None

8. The lead screw of a lathe with nut forms a
a) rolling pair
b) sliding pair
c) screw pair
d) turning pair
Answer: c
Clarification: When the nature of contact between the element of a pair is such that one element can turn abut the other by screw threads, the pair is known as a screw pair. The lead screw of a lathe with nut is an example of screw pair.

9. Which of the following is a turning pair?
a) Piston and cylinder of a reciprocating steam engine
b) Shaft with collars at both ends fitted into a circular hole
c) Lead screw of a lathe with nut
d) Ball and a socket joint
Answer: b
Clarification: When the two elements of a pair are constrained in such a way that one can only turn or revolve about a fixed axis of another link, the pair is known as turning pair.

10. When the two elements of a pair have a surface contact when relative motion takes place and the surface of one element slides over the surface of the other, the pair formed is known as a
a) lower pair
b) higher pair
c) self-closed pair
d) force-closed pair
Answer: a
Clarification: When the two elements of a pair have surface contact when relative motion, takes place and the surface of one element slides over the surface of the other, the pair formed is known as lower pair. It will be seen that sliding pairs, turning pairs and screw pairs form lower pairs.

250+ TOP MCQs on Velocity of a point Link on a Link By Instanteneous Centre Method and Answers

Machine Kinematics online test on “Velocity of a point Link on a Link By Instanteneous Centre Method”.

1. There are two points P and Q on a planar rigid body. The relative velocity between the two points
a) should always be along PQ
b) can be oriented along any direction
c) should always be perpendicular to PQ
d) should be along QP when the body undergoes pure translation
Answer: c
Clarification: Velocity of any point on a link with respect to another point (relative velocity) on the same link is always perpendicular to the line joining these points on the configuration (or space) diagram.
vQP = Relative velocity between P & Q
= vP − vQ always perpendicular to PQ.

2. For a four-bar linkage in toggle position, the value of mechanical advantage is
a) 0.0
b) 0.5
c) 1.0
d) ∞
Answer: d
Clarification: When the mechanism is toggle,then β = 00 and 1800.
So M.A = ∞.

3. The number of inversion for a slider crank mechanism is
a) 6
b) 5
c) 4
d) 3
Answer: c
Clarification: For a 4 bar slider crank mechanism, there are the number of links or inversions are 4. These different inversions are obtained by fixing different links once at a time for one inversion. Hence, the number of inversions for a slider crank mechanism is 4.

4. Match the item in columns I and II
Column I Column II
P. Addendum 1. Cam
Q. Instantaneous centre of velocity 2. Beam
R. Section modulus 3. Linkage
S. Prime circle 4. Gear
a) P-4, Q-2, R-3, S-1
b) P-4, Q-3, R-2, S-1
c) P-3, Q-2, R-1, S-4
d) P-3, Q-4, R-1, S-2
Answer: b
Clarification: Column I Column II
P. Addendum 4. Gear
Q. Instantaneous centre of velocity 3. Linkage
R. Section modulus 2. Beam
S. Prime circle 1. Cam
So correct pairs are, P-4, Q-3, R-2, S-1.

5. Match the items in columns I and II
Column I Column II
P. Higher Kinematic Pair 1. Grubler’s Equation
Q. Lower Kinemation Pair 2. Line contact
R. Quick Return Mechanism 3. Euler’s Equation
S. Mobility of a Linkage 4. Planar
5. Shaper
6. Surface contact
a) P-2, Q-6, R-4, S-3
b) P-6, Q-2, R-4, S-1
c) P-6, Q-2, R-5, S-3
d) P-2, Q-6, R-5, S-1
Answer: d
Clarification: In this question pair or mechanism is related to contact & machine related to it.
Column I Column II
P. Higher Kinematic Pair 2. Line Contact
Q. Lower Kinematic Pair 6. Surface Contact
R. Quick Return Mechanism 5. Shaper
S. Mobility of a Linkage 1. Grubler’s Equation

So correct pairs are, P-2, Q-6, R-5, S-1.

6. In a four-bar linkage, S denotes the shortest link length, L is the longest link length, P and Q are the lengths of other two links. At least one of the three moving links will rotate by 3600 if
a) S + L < P + Q
b) S + L > P + Q
c) S + P < L + Q
d) S + P > L + Q
Answer: a
Clarification: Here P,Q,R, & S are the lengths of the links.
According to Grashof’s law : “For a four bar mechanism, the sum of the shortest and longest link lengths should not be greater than the sum of remaining two link lengths, if there is to be continuous relative motion between the two links
S + L < P + Q.

7. The number of degrees of freedom of a planar linkage with 8 links and 9 simple revolute joints is
a) 1
b) 2
c) 3
d) 4
Answer: c
Clarification: Given l= 8, j= 9
We know that, Degree of freedom,
n =3(l − 1)−2j = 3(8 − 1)−2 x 9 = 3.

8. The lengths of the links of a 4-bar linkage with revolute pairs are p,q,r, and s units. given that pa) link of length p
b) link of length q
c) link of length r
d) link of length s
Answer: a
Clarification: Given p“Double crank” mechanism occurs, when the shortest link is fixed. From the given pairs p is the shortest link. So, link of length p should be fixed.

9. When a cylinder is located in a Vee-block, the number of degrees of freedom which are arrested is
a) 2
b) 4
c) 7
d) 8
Answer: b
Clarification: Number of degrees of freedom = 2 & movability includes the six degrees of freedom of the device as a whole, as the ground link were not fixed. So, 4 degrees of freedom are constrained or arrested.

10. The minimum number of links in a single degree-of-freedom planar mechanism with both higher and lower kinematic pairs is
a) 2
b) 3
c) 4
d) 5
Answer: c
Clarification: From the Kutzbach criterion the degree of freedom,
n = 3(l − 1) − 2j − h
For single degree of Freedom (n = 1),
1 = 3(l − 1) − 2j − h
3l − 2j − 4 − h = 0 …(i)
The simplest possible mechanisms of single degree of freedom is four-bar mechanism. For this mechanism j = 4, h = 0
From equation (i), we have
3l − 2 x 4 − 4 − 0 = 0
or, l = 4.

11. The total number of instantaneous centres for a mechanism consisting of n links are
a) n/2
b) n
c) n – 1/2
d) n(n – 1)/2
Answer: d
Clarification: The number of instantaneous centres in a constrained kinematic chain is equal to the number of possible combinations of two links. The number of pairs of links or the number of instantaneous centres is the number of combinations of n links taken two at a time. Mathematically, number of instantaneous centres, N = n(n – 1)/2 where n = Number of links.

12. According to Aronhold Kennedy’s theorem, if three bodies move relatively to each other, their instantaneous centres will lie on a
a) straight line
b) parabolic curve
c) ellipse
d) none of the mentioned
Answer: a
Clarification: The Aronhold Kennedy’s theorem states that if three bodies move relatively to each other, they have three instantaneous centres and lie on a straight line.

13. In a mechanism, the fixed instantaneous centres are those centres which
a) remain in the same place for all configurations of the mechanism
b) vary with the configuration of the mechanism
c) moves as the mechanism moves, but joints are of permanent nature
d) none of the mentioned
Answer: a
Clarification: Fixed instantaneous centres remain in the same place for all configurations of the mechanism. The permanent instantaneous centres move when the mechanism moves, but the joints are of permanent nature.

14. The instantaneous centres which vary with the configuration of the mechanism, are called
a) permanent instantaneous centres
b) fixed instantaneous centres
c) neither fixed nor permanent instantaneous centres
d) none of the mentioned
Answer: c
Clarification: Fixed instantaneous centres remain in the same place for all configurations of the mechanism. The permanent instantaneous centres move when the mechanism moves, but the joints are of permanent nature. Neither fixed nor permanent instantaneous centres vary with the configuration of the mechanism.

15. When a slider moves on a fixed link having curved surface, their instantaneous centre lies
a) on their point of contact
b) at the centre of curvature
c) at the centre of circle
d) at the pin joint
Answer: b
Clarification: When the slider link moves on fixed link having constant radius of curvature, the instantaneous centre lies at the centre of curvature i.e. the centre of the circle, for all configuration of the links.

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250+ TOP MCQs on Steering Gear Mechanism and Answers

Machine Kinematics Multiple Choice Questions on “Steering Gear Mechanism”.

1. The tilting of the front wheels away from the vertical is called
a) caster
b) camber
c) toe-in
d) toe-out
Answer: b
Clarification: The angle between the vertical line from the centre point of the tyre and the central line of the tyre is known as the camber angle.

2. In the steering gear, a gear sector or toothed roller is meshed with a
a) ball bearing
b) roller bearing
c) worm
d) steering wheel
Answer: c
Clarification: A rotary valve power steering gear for the integral system uses recirculating ball-type worm and wheel steering gear.

3. The only service that a steering linkage normally requires is
a) tie-rod adjustment
b) lubrication
c) ball-joint replacement
d) none of the mentioned
Answer: a
Clarification: In the linkage-type power steering system, the power cylinder is not a part of the steering system. Instead, the power cylinder is fitted into the steering linkage.

4. Caster action on the front wheels of a vehicle will
a) make it easier for the driver to take corners
b) help reduce the load on the king-pins
c) automatically achieve the straight wheel position
d) none of the mentioned
Answer: c
Clarification: Negative caster produces directional stability to the vehicle keeping the wheel position straight.

5. Too much toe-in will be noticed by
a) excessive tyre wear because of taking corners
b) steering wander
c) feathering of tyres
d) light steering
Answer: a
Clarification: The toe-in ensures parallel rolling of the front wheels. It stabalises steering and prevents side slipping and excessive wear of the tyres.

6. Hard steering is a result of
a) very loose steering linkage
b) worn out steering linkage
c) too loose front wheel bearings
d) incorrect lubricant
Answer: d
Clarification: Hard steering is caused because of the following reasons:
a) incorrect lubricant
b) broken or bent steering arms or knuckles
c) too tight spherical ball joints
d) insufficient lubricant
e) low or uneven pressure.

7. Excessive play or looseness in the steering system is caused by
a) worn out front wheel bearings
b) broken or bent steering arms or knuckles
c) too tight spherical ball joints
d) insufficient lubricant
Answer: a
Clarification: Excessive play or looseness in the steering system is the result of
a) very loose steering linkage
b) worn out steering linkage
c) too loose front wheel bearings
d) worn out front wheel bearings
e) loose steering gear flexible coupling
f) worn out steering gear flexible coupling.

8. Erratic steering is caused due to
a) worn out brake lining
b) broken or bent steering arms or knuckles
c) too tight spherical ball joints
d) insufficient lubricant
Answer: a
Clarification: It is caused due to following reasons
a) worn out brake lining
b) brake lining choked with oil
c) brake lining choked with brake fluid.

9. Wheel wobbles occur due to
a) inoperative stabilizer
b) wheel out of balance
c) bent kin-pin
d) bent steering knuckle
Answer: b
Clarification: It is caused due to following reasons
a) wheel out of balance
b) worn joints in assembly
c) loose wheel bearing.

10. Wheel wobbling can be fixed by
a) adjusting and repairing it
b) repairing the stabilizer
c) replacing kin-pin
d) replacing brake lining
Answer: a
Clarification: It can be fixed by the following remedies
a) adjusting and repairing it
b) balancing wheels
c) adjusting bearings.

11. Hard steering can be fixed by
a) replacing the bent or broken parts
b) replacing worn out parts
c) tightening the loose bearings
d) none of the mentioned
Answer: a
Clarification: It can be fixed by the following remedies
a) inflate the tyres to correct pressure
b) replace all tight ball joints
c) replacing the bent or broken parts.

12. Erratic steering can be adjusted by
a) replacing worn out parts
b) replacing the brake lining
c) tightening the loose bearings
d) none of the mentioned
Answer: b
Clarification: It can be fixed by the following remedies
a) replacing the brake lining
b) locating and removing the cause for choke
c) removing the cause for choke due to the brake fluid.

13. Excessive play can be fixed by
a) replacing the bent or broken parts
b) replacing the brake lining
c) tightening the loose bearings
d) none of the mentioned
Answer: c
Clarification: It can be fixed by the following remedies
a) replacing worn out parts
b) replacing worn out couplings
c) tightening the loose bearings.

250+ TOP MCQs on Belt, Rope and Chain Drives and Answers

Machine Kinematics Multiple Choice Questions on “Belt, Rope and Chain Drives”.

1. The velocity ratio of two pulleys connected by an open belt or crossed belt is
a) directly proportional to their diameters
b) inversely proportional to their diameters
c) directly proportional to the square of their diameters
d) inversely proportional to the square of their diameters

Answer: b
Clarification: It is the ratio between the velocities of the driver and the follower or driven.
Let d1 = Diameter of the driver,
d2 = Diameter of the follower,
N1 = Speed of the driver in r.p.m., and
N2 = Speed of the follower in r.p.m.
∴ Length of the belt that passes over the driver, in one minute
= π d1.N1
Similarly, length of the belt that passes over the follower, in one minute
= π d2 . N2
Since the length of belt that passes over the driver in one minute is equal to the length of belt that passes over the follower in one minute, therefore
π d1.N1 = π d2 . N2
∴ Velocity ratio, N2/N1 = d1/d2.

2. Two pulleys of diameters d1 and d2 and at distance x apart are connected by means of an open belt drive. The length of the belt is
a) π /2 (d1 + d2) 2x + (d1 + d2)2/4x
b) π /2 (d1 – d2) 2x + (d1 – d2)2/4x
c) π /2 (d1 + d2) 2x + (d1 – d2)2/4x
d) π /2 (d1 – d2) 2x + (d1 + d2)2/4x

Answer: c
Clarification: None.

3. In a cone pulley, if the sum of radii of the pulleys on the driving and driven shafts is constant, then
a) open belt drive is recommended
b) cross belt drive is recommended
c) both open belt drive and cross belt drive are recommended
d) the drive is recommended depending upon the torque transmitted

Answer: b
Clarification: In a cross belt drive, both the pulleys rotate in opposite directions. If sum of the radii of the two pulleys be constant, then length of the belt required will also remain constant, provided the distance between centres of the pulleys remain unchanged.

4. Due to slip of the belt, the velocity ratio of the belt drive
a) decreases
b) increases
c) does not change
d) none of the mentioned

Answer: a
Clarification: The result of the belt slipping is to reduce the velocity ratio of the system. As the slipping of the belt is a common phenomenon, thus the belt should never be used where a definite velocity ratio is of importance.

5. When two pulleys of different diameters are connected by means of an open belt drive, then the angle of contact taken into consideration should be of the
a) larger pulley
b) smaller pulley
c) average of two pulleys
d) none of the mentioned

Answer: b
Clarification: None.

6. The power transmitted by a belt is maximum when the maximum tension in the belt (T) is equal to
a) TC
b) 2TC
c) 3TC
d) 4TC

Answer: c
Clarification: When the power transmitted is maximum, 1/3rd of the maximum tension is absorbed as centrifugal tension.
T = 3TC
where TC = Centrifugal tension.

7. The velocity of the belt for maximum power is
a) √T/3m
b) √T/4m
c) √T/5m
d) √T/6m

Answer: a
Clarification: We know that T1 = T– TC and for maximum power TC = T/3
T1 = T – T/3 = 2T/3

the velocity of the belt for the maximum power, v = √T/3m
where m = Mass of the belt in kg per metre length.

8. The centrifugal tension in belts
a) increases power transmitted
b) decreases power transmitted
c) have no effect on the power transmitted
d) increases power transmitted upto a certain speed and then decreases

Answer: c
Clarification: None.

9. When the belt is stationary, it is subjected to some tension, known as initial tension. The value of this tension is equal to the
a) tension in the tight side of the belt
b) tension in the slack side of the belt
c) sum of the tensions in the tight side and slack side of the belt
d) average tension of the tight side and slack side of the belt

Answer: d
Clarification: When the driver starts rotating, it pulls the belt from one side (increasing tension in the belt on this side) and delivers it to the other side (decreasing the tension in the belt on that side). The increased tension in one side of the belt is called tension in tight side and the decreased tension in the other side of the belt is called tension in the slack side.

10. The relation between the pitch of the chain ( p) and pitch circle diameter of the sprocket (d) is given by
a) p = d sin (600/T)
b) p = d sin (900/T)
c) p = d sin (1200/T)
d) p = d sin (1800/T)

Answer: d
Clarification: It is given by p = d sin (1800/T).

250+ TOP MCQs on Comparison Between Involute and Cycloidal Gears and Answers

Machine Kinematics Multiple Choice Questions on “Comparison Between Involute and Cycloidal Gears”.

1. The velocity of sliding _____________ the distance of the point of contact from the pitch point.
a) is directly proportional to
b) is inversaly proportional to
c) is equal to cosɸ multiplied by
d) does not depend upon
Answer: a
Clarification: The velocity of sliding is the velocity of one tooth relative to its mating tooth along the common tangent at the point of contact.

2. In involute gears, the pressure angle is
a) dependent on the size of teeth
b) dependent on the size of gears
c) always constant
d) always variable
Answer: c
Clarification: None

3. In full depth 140 involute system, the smallest number of teeth in a pinion which meshes with rack without interference is
a) 12
b) 16
c) 25
d) 32
Answer: d
Clarification: The minimum number of teeth on the pinion in order to avoid interference for 14.50 full depth involute are 32 and for 200 full depth involute teeth are 18.

4. The pressure angle for involute gears depends upon the size of teeth.
a) True
b) False
Answer: b
Clarification: In a gear drive, the shape of the tooth depends upon the pressure angle.

5. The contact ratio is given by
a) Length of the path of approach/Circular pitch
b) Length of the path of recess/Circular pitch
c) Length of the arc of contact/Circular pitch
d) Length of the arc of approach/cosɸ
Answer: c
Clarification: None

6. For an involute gear, the ratio of base circle radius and pitch circle radius is equal to
a) sinɸ
b) cosɸ
c) secɸ
d) cosecɸ
Answer: b
Clarification: None

7. Which of the following statement is correct for gears?
a) The addendum is less than the dedendum
b) The pitch circle diameter is the product of module and number of teeth
c) The contact ratio means the number of pairs of teeth in contact
d) All of the mentioned
Answer: d
Clarification: None

8. In a gear having involute teeth, the normal to the involute is a tangent to the
a) pitch circle
b) base circle
c) addendum circle
d) dedendum circle
Answer: b
Clarification: Addendum circle is the circle drawn through the top of the teeth and is concentric with the pitch circle.
Dedendum circle is the circle drawn through the bottom of the teeth. It is also called root circle.
Pitch circle is an imaginary circle which by pure rolling action, would give the same motion as the actual gear.

9. The centre distance between two meshing involute gears is equal to
a) sum of base circle radii/cosɸ
b) difference of base circle radii/cosɸ
c) sum of pitch circle radii/cosɸ
d) difference of pitch circle radii/cosɸ
Answer: a
Clarification: None

10. When the tip of a tooth undercuts the root on its mating gear, it is known as interference.
a) True
b) False
Answer: a
Clarification: None

250+ TOP MCQs on Kinetics of Motion and Answers

Machine Kinematics Multiple Choice Questions on “Kinetics of Motion”.

1. The force which acts along the radius of a circle and directed ____________ the centre of the circle is known as centripetal force.
a) away from
b) towards
c) at the
d) none of the mentioned

Answer: b
Clarification: Centripetal force acts radially inwards and is essential for circular motion.

2. The unit of mass moment of inertia in S.I. units is
a) m4
b) kgf-m-s2
c) kg-m2
d) N-m

Answer: c
Clarification: Moment of inertia is the distance, from a give reference, where the whole mass of body is assumed to be concentrated to give the same value of I. The unit of mass moment of inertia in S.I. units is kg-m2.

3. Joule is a unit of
a) force
b) work
c) power
d) none of the mentioned

Answer: b
Clarification: In S.I. system of units, the practical unit of work is N-m. It is the work done by a force of 1 newton, when it displaces a body through 1 metre. The work of 1 N-m is known as joule (briefly written as J ) such that 1 N-m = 1 J.

4. The energy possessed by a body, for doing work by virtue of its position, is called
a) potential energy
b) kinetic energy
c) electrical energy
d) chemical energy

Answer: a
Clarification: Potential energy is the energy possessed by a body for doing work, by virtue of its position.
Kinetic energy is the energy possessed by a body, for doing work, by virtue of its mass and velocity of motion.

5. When a body of mass moment of inertia I (about a given axis) is rotated about that axis with an angular velocity, then the kinetic energy of rotation is
a) 0.5 I.ω
b) I.ω
c) 0.5 I.ω2
d) I.ω2

Answer: c
Clarification: When a body of mass moment of inertia I (about a given axis) is rotated about that axis, with an angular velocity ω, then it possesses some kinetic energy. In this case,
Kinetic energy of rotation = 1/ 2I.ω2

When a body has both linear and angular motions e.g. in the locomotive driving wheels and wheels of a moving car, then the total kinetic energy of the body is equal to the sum of kinetic energies of translation and rotation.
∴ Total kinetic energy = 1/ 2mv2 +1/ 2I.ω2

6. The wheels of a moving car possess
a) potential energy only
b) kinetic energy of translation only
c) kinetic energy of rotation only
d) kinetic energy of translation and rotation both.

Answer: d
Clarification: in the locomotive driving wheels and wheels of a moving car, then the total kinetic energy of the body is equal to the sum of kinetic energies of translation and rotation.

7. The bodies which rebound after impact are called
a) inelastic bodies
b) elastic bodies
c) solid bodies
d) none of the mentioned

Answer: b
Clarification: The bodies, which rebound after impact are called elastic bodies and the bodies which does not rebound at all after its impact are called inelastic bodies.

8. The coefficient of restitution for inelastic bodies is
a) zero
b) between zero and one
c) one
d) more than one

Answer: a
Clarification: The process of regaining the original shape is called restitution. Inelastic bodies can not regain their original shapes. Therefore their coefficient of restitution is zero.

9. Which of the following statement is correct ?
a) The kinetic energy of a body during impact remains constant.
b) The kinetic energy of a body before impact is equal to the kinetic energy of a body after impact.
c) The kinetic energy of a body before impact is less than the kinetic energy of a body after impact.
d) The kinetic energy of a body before impact is more than the kinetic energy of a body after impact.

Answer: d
Clarification: Total kinetic energy of the system before impact,
E1 = 1/2 m1 (u1)2 + 1/2 m2 (u2)2

When the two bodies move with the same velocity v after impact, then
Kinetic energy of the system after impact,

E2= 1/2( m1 + m2) v2

∴ Loss of kinetic energy during impact,
EL = E1 – E2

10. A body of mass m moving with a constant velocity v strikes another body of same mass m moving with same velocity but in opposite direction. The common velocity of both the bodies after collision is
a) v
b) 2 v
c) 4 v
d) 8 v

Answer: b
Clarification: If the body will move in opposite direction a negative sign would be there.
We know that Common velocity = V1 – 2
Here both the velocities are same.
Therefore Common velocity = V – (-V)
= V + V = 2V