**Question 1** => Four particles A, B, C, D having masses m, 2m, 3m, 4m respectively are placed in order at the corners of a square of side a. Locate where would be centre of mass of this four particle system. (See the below diagram)

**Solution** => If we consider this particle system with respect to usual Coordinate System. Then

Particle | mass | x-coordinate | y-coordinate |
---|---|---|---|

A | m | 0 | 0 |

B | 2m | a | 0 |

C | 3m | a | a |

D | 4m | 0 | a |

So X-Coordinate of Centre of Mass of this system will be

= (m X 0 + 2ma + 3ma + 4m X 0) / (m + 2m + 3m + 4m) = a/2**X-Coordinate of Centre of Mass = a/2**

Y-Coordinate of Centre of Mass = (m X 0 + 2m X 0 + 3ma + 4ma) / (m + 2m + 3m + 4m) = 7a/10**Y-Coordinate of Centre of Mass = 7a/10 **

So Position of **Centre of Mass will be (a/2, 7a/10)** with respect to origin of Coordinate System.

**Question 2** => Two identical uniform rods AB and CD, each of length L jointed to form a Shape. Locate the centre of mass of the Shape. The centre of mass of a uniform rod is at the middle point. Find Centre of Mass of Shape.

**Solution** => Let the mass of each rod be m. Take the centre C of the rod AB as the origin and CD as the Y-axis. The rod AB has mass m and its centre of mass is at C. For the calculation of the centre of mass of the combined system, AB may be replaced by a point particle of mass m placed at the point C. Similarly the rod CD may be replaced by a point particle of mass m placed at the centre E of the rod CD. Thus, the frame is equivalent to a system of two particles of equal masses m each, placed at C and E. The centre of mass of this pair of particles will be at the middle point F of CE. The centre of mass of the frame is, therefore, on the rod CD at a distance L/4 from C.

So **Centre of mass of this figure will be at distance L/4 from point C on rod CD**.

**Question 3** => Each of block shown in below picture has mass of 1 kg, the left block moves with a speed of 2 m/s towards right block (this block is kept at rest position). The spring attached to the front block is light and has a spring constant of 50 N/m. Find what would be maximum compression of the spring when left block hits it.

**Solution** => Maximum compression of spring will take place when blocks move with equal velocity. As there is no external force acting on this system of blocks. So total linear momentum of two blocks system should remain constant.

Suppose **V** be **common speed of blocks** at maximum compression of spring.

As Per Momentum Conversation Law momentum would remain same.**Before compression total momentum = After compression total momentum** (*Equation 1*)

Before compression total momentum = 1 X 2 = 2 kg ms^{-1}

After compression total momentum = 1 X V + 1 X V = 2V kg ms^{-1}

Putting these two equations together in *Equation 1* => 2V = 2 so **V = 1 ms ^{-1}**

So when there is maximum compression blocks would be moving with velocity of 1 ms

^{-1}.

But we need to

**find out maximum compression of spring/Maximum compression length of spring**.

Also as there is no external force being applied to system so Kinetic Energy of left block motion would become elastic potential energy of spring.

**Elastic Potential Energy of Spring = Before compression kinetic energy – After compression kinetic energy**

Before compression kinetic energy = 1 X 2

^{2}/2 = 2 Joule (Only One Block is Moving)

After compression kinetic energy = 1 X 1

^{2}/2 + 1 X 1

^{2}/2 = 1 Joule (Both blocks are moving)

So Potential Energy of Spring = 2 – 1 = 1 Joule

By formula Elastic Potential Energy of Spring = kx

^{2}/ 2 ( Where K is Spring Constant)

Putting values together we get 1 = 50 X x

^{2}/2

After solving this we get

**x = 0.2 m**

So

**Maximum Compression of spring in this particular scenario will be 0.2 metres**.

**Question 4** => A cart A of mass 50 kg moving at a speed of 20 km/h hits a lighter cart B of mass 20 kg moving towards 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.

**Solution** => This is an example 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

(50 kg) (20 km/h) – (20 kg) (10 km/h) = (70 kg) V

Solving the above equation gives **V = 80/7 km/h**

Thus **Speed of Combined Mass after collision will be 80/7 km/h**

**Question 4** => A block of mass m moving at speed **v** collides with another block of mass **2m** at rest. The lighter block comes to rest after the collision. Find the coefficient of restitution.

**Solution** => Suppose the second block moves at speed **v’** towards right after the collision. From the principle of conservation of momentum, mv = 2mv’ or v’ =v/2.

So **v’ =v/2**

So Velocity of Separation = v / 2 and Velocity of approach = v.

By formula Coefficient of Restitution = Velocity of Separation / Velocity of approach = (v/2)/v = 1/2 = 0.5**Coefficient of Restitution = 0.5**

**Question 5** => Three particles of masses 0.50 kg, 1.0 kg and 1.5 kg are placed at the three corners of a right angled triangle of sides 3.0 cm, 4.0 cm and 5.0 cm as shown in below picture. Locate where would the centre of mass of this system will be.

**Solution** => Let’s assume this triangle with respect to a Coordinate System. So line 0.50 kg, 1.0 kg, 4 cm will be X-axis and line 0.50 kg, 1.5 kg, 3 cm will be Y-axis.

So now let’s see what will be positions of the particles 0.50 kg, 1.0 kg, 1.5 kg with respect to this Coordinate System.

Particle mass | X-Coordinate | Y-Coordinate |
---|---|---|

0.50 kg | 0 | 0 |

1.0 kg | 4.0 cm | 0 |

1.5 kg | 0 | 3.0 cm |

X-Coordinate of Centre of Mass of this System = (m_{1}x_{1} + m_{2}x_{2} + m_{3}x_{3}) / (m_{1} + m_{2} + m_{3})

=> (0.50 X 0 + 1 X 4 + 1.5 X 0) / (0.50 + 1 + 1.5) = 4 / 3 = 1.3 cm**X-Coordinate of Centre of Mass of this System = 1.3 cm**

Y-Coordinate of Centre of Mass of this System = (m_{1}y_{1} + m_{2}y_{2} + m_{3}y_{3}) / (m_{1} + m_{2} + m_{3})

=> (0.50 X 0 + 1 X 0 + 1.5 X 3) / (0.50 + 1.0 + 1.5) = 1.5 cm**Y-Coordinate of Centre of Mass of this System = 1.5cm**

So Centre of Mass of this System will be at Position (1.3, 1.5) with respect to Coordinate System. Meaning it will be 1.3 cm to right of 0.50 kg and will be 1.5 cm above line 4 cm.

**Question 6** => The density of a linear rod of length L varies as p = A + Bx where x is the distance between from left end of rod. Find out where the centre of mass of this rod be.

**Solution** => For solving this problem let’s suppose some parameters about Rod.

Let cross sectional area of rod = **a**

So Mass of a little component of rod of lenght **dx** = (A+Bx)adx

So Centre of Mass of Rod = (Integration of x with respect to mass) / (Integration of Mass change rate across rod)

Centre of Mass of Rod = ∫x dm / ∫ dm = ∫ x(A + Bx)adx / ∫ (A + Bx)adx = (AL^{2}/2 + BL^{3}/3) / (AL + BL^{2}/2)**Centre of Mass of Rod = (3AL + 2BL ^{2}) / 3(2A + BL)**

**Question 7** => Think about a two particle system with masses m_{1} and m_{2}, if in case first particle m_{1} is moved towards centre of mass by distance d, then by what distance should second particle be moved so as to keep centre of mass at the same position?

**Solution** => Suppose distance of m_{1} from centre of mass C is x_{1}

Suppose distance of m_{2} from centre of mass C is x_{2}

Suppose mass m_{2} is moved through a distance d^{‘} towards C so as to keep centre of mass stay at posititon C.

That’s why **m _{1}x_{1} = m_{2}x_{2}** (

*Equation 1*)

**m**(

_{1}(x_{1}– d) = m_{2}(x_{2}– d^{‘})*Equation 2*)

Subtracting

*Equation 2*from

*Equation 1*we get

**d**

^{‘}= m_{1}d / m_{2}**Question 8** => A Military Personal fires 50g bullets from a machine gun each of bullet fires at speed of 1.0 km/s. If personal fired 20 bullets in 4 seconds. Then calculate what average force does personal exert against the machine gun during firing of these bullets?

**Solution** => Momentum of Each Bullet Fired = 0.050 kg X 1000 ms^{-1} = 50 kgms^{-1}

Rate of change of Momentum when bullets are fired= (50 kgms^{-1} X 20)/4 = 250N

So In order to fire 20 bullets in 4 seconds with speed of 1.0km/s, Military Personal have to exert a force of 250N against the run to hold it properly.

**Question 9** => A Space Shuttle travelling at speed of 4000 km/h with respect to earth disconnects and ejects some mass backward. This ejected mass weighs one fifth of the mass of Space Shuttle left. If shuttle ejects this module at a speed of 100 km/h with respect to state of shuttle before ejection. Then find out what would be the final velocity of shuttle.

**Solution** => Let’s suppose that Mass of shuttle including ejected mass is M

So as per question, Mass of ejected mass will be M/6**Linear momentum of Shuttle before ejection = M X 4000 km/h**

Velocity of Ejected Module with respect to earth = Velocity of Ejected Module with respect to shuttle + Velocity of Shuttle with respect to Earth

Velocity of Ejected Module with respect to earth = -100 km/h + 4000 km/h = 3900km/h**Velocity of Ejected Module with respect to earth = 3900 km/h**

Let final velocity of shuttle after ejection of mass = V

Then Total Final Momentum of System(Includes momentum of both Shuttle after mass ejection and Momentum of ejected mass) = 5MV/6 + 3600M/6**Final Momentum of System = 5MV/6 + 3600M/6**

As Per law of Conversation of Momentum, **Linear momentum of Shuttle before ejection** = **Final Momentum of System**

So **M X 4000** = **5MV/6 + 3600M/6** => Upon Solving this we get **V = 4020 km/h**

So **final velocity of Shuttle after ejection of mass = 4020 km/h**

**Question 10** => A bullet of mass 50g is fired into bob of a pendulum of mass 450g. When bullet is fired into bob, it sticks with bob and due to force by bullet bob rises in height of 1.8m. Find out what would be the speed of bullet in this scenario?

**Solution** => As Per law of conversation of momentum, **Total Momentum of System before bullet collision with bob = Total momentum of system after bullet collision with bob** (*Equation 1*)

Let velocity of bullet = v

Let velocity of bullet and bob together = V**Momentum of System before bullet collision with bob** = (0.05 kg)v**Momentum of system after bullet collision with bob** = (0.45kg + 0.05kg)V

Using *Equation 1*

0.05v = (0.45 + 0.05)V

Solving this equation we get **V = v/10**

As Per Third Equation of Motion we have v^{2} = u^{2} + 2ah

So this specific scenario of bob we have u = 0, h = 1.8m, a = g = 10ms^{-2}, v = V

Putting values into equation we get => V^{2} = 0 + 2 X 10 X 1.8

V^{2} = 36 => **V = 6**

As V = v/10 So v = V X 10 = 6 X 10 = 60 ms^{-2}**So finally Velocity of bullet should be 60 ms ^{-2} for bob to be raised by 1.8 m height when bullet hit it.**

**Question 11** => A block of mass m moving at a velocity v collides head on with another block of mass 2m, which was at rest. If coefficient of restitution is 1/2, then find out velocities of both blocks after collision happened.

**Solution** => Suppose After collision block of mass m moves at velocity u_{1} and other mass 2m moves at velocity u_{2}.

Velocity of mass m after collision = u_{1}

Velocity of mass 2m after collision = u_{2}

As Per Law of Conservation of Momentum, **Momentum Before Collision = Momentum after collision**

Momentum Before Collision = mv

Momentum After Collision = mu_{1} + 2mu_{2}

Putting this together mv = mu_{1} + 2mu_{2}

So **v = u _{1} + 2u_{2}** (

*Equation 1*)

Coefificient of Restitution is defined as = (Velocity of Separation) / (Velocity of Approach)

1/2 = (u

_{2}– u

_{1}) / v

Solving this we get => v = 2(u

_{2}– u

_{1}) (

*Equation 2*)

Putting Together

*Equation 1*and

*Equation 2*

u

_{1}+ 2u

_{2}= 2(u

_{2}– u

_{1})

u

_{1}+ 2u

_{2}= 2u

_{2}– 2u

_{1}

Solving this we get

**u**

_{1}= 0Putting value of u

_{1}in

*Equation 2*

v = 2(u

_{2}– 0)

**u**

_{2}= v/2**So After Collision block of mass m will stops, while block of mass 2m moves at speed of v/2**

**Question 12** => A block of mass 1.2 kg moving at a speed of 0.20 ms^{-1} collides head-on with a similar block kept at rest. If coefficient of restitution is 1/5. Find how much kinetic energy was lost during the collision.

**Solution** => Let Speed of first block after collision = v_{1}

Let Speed of Second block after collision = v_{2}

By conversation of law, **Total Momentum before Collision = Total Momentum after Collision**

Total Momentum before Collision = 1.2 kg X 0.20 ms^{-1} = 0.24 kgms^{-1}

Total Momentum after Collision = 1.2 kg X v_{1} + 1.2 kg X v_{2} = 1.2v_{1} + 1.2v_{2}

Putting both of these together

1.2v_{1} + 1.2v_{2} = 0.24**v _{1} + v_{2} = 0.2** (

*Equation 1*)

Coefficient of Restitution = (Velocity of Separation) / (Velocity of Approach)

1/5 = (v

_{2}– v

_{1}) / 0.20

**v**(

_{2}– v_{1}= 0.04*Equation 2*)

Solving

*Equation 1*and

*Equation*2

**v**

v

_{1}= 0.08 ms^{-1}v

_{2}= 0.12 ms^{-1}Total Kinetic Energy Lost During Collision = Total Kinetic Energy Before Collision – Total Kinetic Energy After Collision

Total Kinetic Energy Before Collision = 1.2 X (0.20)

^{2}/ 2 = 0.024 Joule

Total Kinetic Energy After Collision = 1.2 X (0.08)

^{2}/ 2 + 1.2 X (0.12)

^{2}/ 2 = 0.01248 Joule

Total Kinetic Energy Lost During Collision = 0.024 – 0.01248 = 0.01152 Joule

**Total Kinetic Energy Lost During Collision = 0.01152 Joule**

**Question 13** => Consider the following two statements –

(A) Linear momentum of the system remains constant

(B) Centre of mass of the system remains at rest

- A implies B and B implies A
- A does not imply B and B does not imply A
- A implies B but does not imply A
- B implies A but A does not imply B

**Solution** => **4**

**Question 14** => Consider the following two statements –

(A) Linear momentum of a system of particles is zero

(B) Kinetic energy of a system of particles is zero

- A implies B and B implies A
- A does not imply B and B does not imply A
- A implies B but B does not imply A
- B implies A but A does not imply B

**Solution** => 4

**Question 15** => Consider the following two statements –

(A) The linear momentum of a particle is independent of the frame of reference

(B) The kinetic energy of a particle is independent of the frame of reference

- Both A and B are true
- A is true but B is false
- A is false but B is true
- Both A and B are false

**Solution** => 4