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Momentum And Center Of Mass

Learn displacement formula in physics with momentum and center of mass. Practice velocity of center of mass and displacement of an object.

Displacement of Center of Mass

• Sometimes there are situations when we want to calculate distance relative to different objects.
• Consider a situation having two snapshots of two cars at different instant of time, the snapshots are like as shown.

• Since there is no other object in the snapshot, thus we can't surely say which car moves and by what distance.

Case 1:

• If we consider car A is at rest, then we can surely say that car B moves by 5 m towards left.

Case 2:

• If we consider car B is at rest, then we can say that car A moves by 5 m towards right.

Three cars A, B and C are placed as shown in figure at time t = 0. After some time their position are as shown. Calculate the distance covered by car C if car A is at rest.

A 20 m

B 25 m

C 30 m

D 15 m

×

Since, car A is at rest, it means it does not change its position.

At t = 0, car C is 15 m ahead of car A and after some time it is 15 m behind of car A.

Thus, total distance covered by car C will be = 15 + 15 = 30 m

Three cars A, B and C are placed as shown in figure at time t = 0. After some time their position are as shown. Calculate the distance covered by car C if car A is at rest.

A

20 m

.

B

25 m

C

30 m

D

15 m

Option C is Correct

Calculation of Center of Mass of Two or more Objects

• As we know by definition of center of mass, the position of center of mass between two objects having mass m1 and m2 is given as

$$\vec{r} _{cm} = \dfrac{m_1 \vec{r}_1 + m_2 \vec{r}_2}{m_1 + m_2}$$

• Consider two objects with masses m1 and m2 respectively having distance d between them.

• To calculate center of mass from A, consider A as origin.

• Then by using above formula, co-ordinate of center of mass will be

$$m_1 = m_1 \;\;\;\;\;\;\;\; x_1 = 0$$

$$m_2 = m_2 \;\;\;\;\;\;\; x_2 = d$$

• Thus, $$x_{cm} = \dfrac{m_1x_1 + m _2 x_2}{m_1 + m_2}$$

$$= \dfrac{m_1 \times0 + m_2 \;d}{m_1 + m_2 }\, =\;\dfrac{m_2\; d}{m_1 + m_2 }$$

Two objects of masses m1 = 40 kg and m2 = 60 kg are placed d = 60 cm apart. Calculate the position of center of mass from A towards B.

A 24 cm

B 18 cm

C 36 cm

D 42 cm

×

To calculate position of center of mass from A, consider A as origin.

So, position of center of mass is given as

$$x_{cm} = \dfrac{m_1 x_1 + m_2x_2}{m_1 + m_2 }$$

Co-ordinate of center of mass

$$m_1 = 40 \;kg \; at \; x_1 = 0$$

$$m_2 = 60 \;kg \; at \; x_2 = 60$$

$$x_{cm}= \dfrac{40 × 0 + 60 × 60 }{40 + 60}$$

$$x_{cm} = \dfrac{3600}{100} = 36 \; cm$$

Two objects of masses m1 = 40 kg and m2 = 60 kg are placed d = 60 cm apart. Calculate the position of center of mass from A towards B.

A

24 cm

.

B

18 cm

C

36 cm

D

42 cm

Option C is Correct

Calculation of Displacement of m1 when both Blocks are Moving

• As we know by definition of center of mass, the position of center of mass between two objects having masses m1 and m2 is given as

$$\vec{x}_{cm} = \dfrac{m_1x_1 + m_2x_2}{m_1 + m_2}$$

• Consider a block of mass m1 is placed on the top of bigger block of mass m2, as shown in figure.
• All the surfaces are frictionless.

• After some time, the block reaches the ground.
• To calculate the distance covered by smaller block, consider it as origin.
• Since, no external force is acting on the system along the x - axis, the displacement of center of mass along x - axis is zero.
• But there is a force of gravity along y-axis so, we don't apply same concept along y-axis.
• Hence, we consider only x-coordinate and don't care about y-coordinate as acceleration of y does not affect the velocity of x-direction.
• To calculate displacement of wedge, consider it as origin in both events.
• Since, no external force is acting on x-axis so, center of mass is at rest.
• Hence, displacement of center of mass along x-axis is zero.

$$\vec{s}_{com} = 0$$

$$m_1\vec{s}_1+ m_2 \vec{s}_2 = 0$$  where $$\vec{s}_1 = - x, \, \vec{s}_2 = (\ell - x)$$

$$\Rightarrow - m_1x + m_2 [+ (\ell - x)] = 0$$

$$\Rightarrow - m_1x + m_2 (\ell - x) = 0$$

$$\Rightarrow m_1x = m_2 (\ell - x)$$

$$\Rightarrow m_1x = m_2\ell-m_2x$$

$$\Rightarrow (m_1 + m_2) x = m _2 \; \ell$$

$$\Rightarrow \; x = \left(\dfrac{m_2}{m_1 + m_2}\right) \ell$$

Two blocks of masses m1 = 2 kg and m2 = 8 kg. The block of mass m1 is placed on block of mass m2, as shown in figure.The length $$(\ \ell\ )$$ of block of mass m2 is 5 cm. Calculate the displacement of smaller block when both blocks are moving.

A 2 cm

B 4 cm

C 6 cm

D 5 cm

×

Since, no external force is acting on x-axis so, center of mass is at rest.

Hence, displacement of center of mass along x-axis is zero.

$$\vec{s}_{com} = 0$$

$$m_1\vec{s}_1+ m_2 \vec{s}_2 = 0$$  where $$\vec{s}_1 = - x, \, \vec{s}_2 = (\ell - x)$$

$$\Rightarrow - m_1x + m_2 [+ (\ell - x)] = 0$$

$$\Rightarrow - m_1x + m_2 (\ell - x) = 0$$

$$\Rightarrow m_1x = m_2 (\ell - x)$$

$$\Rightarrow m_1x = m_2\ell-m_2x$$

$$\Rightarrow (m_1 + m_2) x = m _2 \; \ell$$

$$\Rightarrow \; x = \left(\dfrac{m_2}{m_1 + m_2}\right) \ell$$

Given: $$m_1 = 2 \, kg , \,m_2 = 8 \, kg , \,\ell = 5 \, cm$$

$$x = \left(\dfrac{8}{2 + 8}\right) 5$$

$$x = \dfrac{8 }{10} × 5$$

$$x = 4 \; cm$$

Two blocks of masses m1 = 2 kg and m2 = 8 kg. The block of mass m1 is placed on block of mass m2, as shown in figure.The length $$(\ \ell\ )$$ of block of mass m2 is 5 cm. Calculate the displacement of smaller block when both blocks are moving.

A

2 cm

.

B

4 cm

C

6 cm

D

5 cm

Option B is Correct

Displacement of Ring

• Consider an arrangement of a ring and a mass, as shown in figure.
• Let the mass of ring be m1.
• The distance between ring and mass (m2) is $$\ell$$.
• To calculate displacement of ring, we consider only x–coordinate as no external force is acting along x–axis.
• Let the center of mass be the origin.

$$\vec{s}_{com} = 0$$

$$m_1 \vec{s}_1 + m_2 \vec{s}_2 = 0$$

where $$\vec{s}_1 = + x, \,\vec{s}_2 = - (\ell - x)$$

So, $$m_1x+m_2 [- (\ell - x)] = 0$$

$$\Rightarrow m_1 x - m_2 \ell + m_2 x = 0$$

$$\Rightarrow (m_1 + m_2 )x = m_2 \ell$$

$$x = \dfrac{m_2 \; \ell }{m_1 + m_2}$$

A ring of mass m1 = 1 kg and a mass m2 = 5 kg are attached by means of string of length $$\ell$$ = 60 cm, as shown in figure. After some time the ring and mass are displaced as shown. Calculate the displacement of ring.

A 20 cm

B 50 cm

C 30 cm

D 40 cm

×

To calculate displacement of ring, we consider only x–coordinate as no external force is acting along x–axis.

Let the center of mass be the origin.

$$\vec{s}_{com} = 0$$

$$m_1 \vec{s}_1 + m_2 \vec{s}_2 = 0$$

where $$\vec{s}_1 = + x, \;\;\;\;\vec{s}_2 = - (\ell - x)$$

So, $$m_1x+m_2 [- (\ell - x)] = 0$$

$$\Rightarrow m_1 x - m_2 \ell + m_2 x = 0$$

$$\Rightarrow (m_1 + m_2 )x = m_2 \ell$$

$$x = \dfrac{m_2 \; \ell }{m_1 + m_2}$$

Given: $$m_1 = 1 \, kg , \,m_2 = 5 \, kg , \,\ell = 60 \, cm$$

$$x = \dfrac{5 × 60 }{1 +5} = 50 \; cm$$

A ring of mass m1 = 1 kg and a mass m2 = 5 kg are attached by means of string of length $$\ell$$ = 60 cm, as shown in figure. After some time the ring and mass are displaced as shown. Calculate the displacement of ring.

A

20 cm

.

B

50 cm

C

30 cm

D

40 cm

Option B is Correct

Calculation of Displacement of Object using Center of Mass

• Consider two objects are under motion and no external force acting on them.
• So, when two objects are under motion due to internal force, then momentum of system does not change.
• If the velocity of center of mass is zero, then the velocity of center of mass remains zero.
• Thus, center of mass of the system remains at rest.
• Consider a situation in which A and B pull each other on a frictionless surface, as shown in figure.

• No external force is acting between A and B.
• Consider the masses of A and B as m1 and m2 respectively and distance between them as x1.
• Let us assume that A is situated at origin.
• Since, both A and B pull each other so, after some time the distance between them is reduced. Let it be x2

• Since, no external force is acting in this situation. So, the velocity of center of mass is zero as it remains at rest.
• This means that the displacement of center of mass is zero.

$$\vec{s}_{com} = 0$$

• After event, both A and B are displaced from their position.

• So, to calculate displacement of A,

by applying

$$\vec{s}_{com} = 0$$

$$m_1 \vec{s}_1 + m_2 \vec{s}_2 = 0$$

$$\vec{s}_{1} = x$$   and   $$\vec{s}_2 = - [x_1- (x_2 + x)]$$                [from figure]

$$m_1 \; x - m_2 [x_1 - (x_2 + x)] = 0$$

$$m_1 \; x - m_2x_1 + m_2 (x_2 + x) = 0$$

$$m_1 \; x - m_2 x_1 + m_2 x_2 + m_2 x = 0$$

$$(m_1 + m_2 ) x = m_ 2 (x_1 - x_2)$$

$$x = \dfrac{m_2 (x_1 - x_2 )}{m_1 + m_2 }$$

Two persons A of mass m1 = 30 kg and B of mass m2 = 70 kg, placed x1 = 8 m apart on a frictionless surface, are pulling each other. After some time the distance between them is reduced to x2 = 4 m. Calculate the displacement of A.

A 3.6 m

B 2.8 m

C 4.8 m

D 2.4 m

×

Since, no external force is acting in this situation. So, the velocity of center of mass is zero and hence, center of mass will be at rest.

To calculate displacement of A

Considering A at origin

Since, displacement of center of mass is zero.

$$\vec{s}_{com} = 0$$

$$m_1\vec{s}_1 + m_2 \vec{s}_2 = 0$$

$$m_1 x - m_2 [x_1 - (x_2 + x)] = 0 \;\;\;— (1)$$

Given: $$m_1 = 30 \, kg, \,m_2 = 70 \, kg , \,x_1 = 8 \, m, \, x_2 = 4 \, m$$

From (1),  $$30x-70\big[8-(4+x)\big]=0$$

$$30 \; x - 560 + 28 0 + 70 \; x = 0$$

$$10 0\; x = 280$$

$$x = 2.8 \; m$$

Two persons A of mass m1 = 30 kg and B of mass m2 = 70 kg, placed x1 = 8 m apart on a frictionless surface, are pulling each other. After some time the distance between them is reduced to x2 = 4 m. Calculate the displacement of A.

A

3.6 m

.

B

2.8 m

C

4.8 m

D

2.4 m

Option B is Correct

Displacement of Man / Trolley with respect to Center of Mass

• Consider two objects are under motion and no external force acting on them.
• So, when two objects are under motion due to internal force, then momentum of system does not change.
• If the velocity of center of mass is zero, then the velocity of center of mass remains zero.
• Thus, center of mass of the system remains at rest.
• Consider a situation in which a man A is standing on the moving boat, as shown in figure. The surface of boat has negligible friction.

• No external force is acting on system of boat and A.
• Let the mass of boat and man be m1 and m2 respectively and length of boat be $$\ell$$ with A situated on its left end.

• To calculate distance covered by man when he reaches at the other end, consider P as origin.
• Since, no external force is acting so, center of mass will be at rest.
• Thus, $$\vec{s}_{com} = 0$$

$$m_1 \vec{s}_1 + m_2 \vec{s}_2 = 0$$

Here, $$\vec{s}_1 = -x, \,\vec{s}_2 = + (\ell - x)$$

$$m_1(-x) + m_2 [+ (\ell -x)] = 0$$

$$-m_1 \; x + m_2 (\ell-x) = 0$$

$$m_1\; x = m_2\; (\ell -x)$$

$$m_1\; x + m_2 \; x = m _2 \; \ell$$

$$(m_1 + m_2) x = m _2 \; \ell$$

$$x = \dfrac{m_2 \; \ell }{m_1+ m_2}$$

A man of mass m2 = 40 kg is standing on the left end of friction less surface of the boat of mass m1 = 120 kg. Calculate the displacement of the boat. Given that the length of the boat $$\ell$$ = 3 m.

A 2 m

B 2.5 m

C .75 m

D 1 m

×

Since, no external force is acting so, center of mass will be at rest.

Thus, $$\vec{s}_{com} = 0$$

$$m_1 \vec{s}_1 + m_2 \vec{s}_2 = 0$$

From figure, $$\vec{s}_1 =-x , \,\vec{s}_2 = (\ell- x)$$

Given: $$m_1 = 120 \, kg , \,m_2 = 40 \, kg , \, \ell = 3 \, m$$

So, $$m_1 \vec{s}_1 + m _2 \vec{s}_2 = 0$$

$$- 120 \; x + 40 [+ (3 - x)] = 0$$

$$- 120 \; x + 40 (3-x) = 0$$

$$- 120 \; x + 120 - 40 \; x = 0$$

$$160\; x = 120$$

$$x = \dfrac {120}{160}$$

$$x = \,. 75 \, m$$

A man of mass m2 = 40 kg is standing on the left end of friction less surface of the boat of mass m1 = 120 kg. Calculate the displacement of the boat. Given that the length of the boat $$\ell$$ = 3 m.

A

2 m

.

B

2.5 m

C

.75 m

D

1 m

Option C is Correct

Displacement of Pieces of Object from Center of Mass

• When a particle moves in air, the path through which it travels represents its trajectory.
• A particle moving in air can have two trajectories – (a) Straight line   (b) Parabola
• If the initial velocity of particle is along acceleration, the trajectory of particle is straight line.
• If the initial velocity of particle makes some angle with acceleration, the trajectory of particle will be parabola.
• Now, suppose the object explodes in mid-air such that it breaks into many pieces.
• So, every piece of the object follows its own path.

(a) If $$\vec{v}$$ is parallel to $$\vec{a}$$ then straight line

(b) If $$\vec{v}$$ is not parallel to $$\vec{a}$$ then parabola.

• But the center of mass follows the same path as there is no explosion.

• Consider a  situation in which a bomb is dropped from some height, then it follows straight line.
• Now, suppose the bomb explodes in mid-air into two pieces of masses m1 and m2.
• Assume that m1 falls on ground at distance d from straight line.
• To calculate the position of particle of mass m2, we consider the center of mass to be at origin.
• The location of center of mass will remain same as there is no explosion.

• Considering center of mass as origin and applying definition of center of mass.

Since   $$m_1 = m_1 , \, m_2 = m_2$$

$$x_1 = -d , \, x _2 = d_2$$

$$x_{cm} = 0$$

$$o = \dfrac{m_1 (-d) + m_2 (d_2)}{m_1 + m_2 }$$

$$d_2 = \dfrac{m_1 \; d}{m_2}$$

An object explodes in the mid-air and breaks into pieces, A and B of masses m1 = 1 kg and m2 = 2 kg respectively. The distance (d) of object A from center of mass is  9 cm, as shown in figure. Calculate the distance of object B from center of mass.

A 6 cm

B 5 cm

C 4.5 cm

D 4 cm

×

Considering center of mass as origin and applying definition of center of mass.

Since   $$m_1 = m_1 , \, m_2 = m_2$$

$$x_1 = -d , \, x _2 = d_2$$

$$x_{cm} = 0$$

$$o = \dfrac{m_1 (-d) + m_2 (d_2)}{m_1 + m_2 }$$

$$d_2 = \dfrac{m_1 \; d}{m_2}$$

Given : $$m_1 = 1 \, kg , \,m_2 = 2 \, kg , \,d = 9 \, cm$$

$$d_2 = \dfrac{1 × 9}{2}$$

$$d_2 = 4.5 \; cm$$

An object explodes in the mid-air and breaks into pieces, A and B of masses m1 = 1 kg and m2 = 2 kg respectively. The distance (d) of object A from center of mass is  9 cm, as shown in figure. Calculate the distance of object B from center of mass.

A

6 cm

.

B

5 cm

C

4.5 cm

D

4 cm

Option C is Correct

Range of one Fragment when Projectile is Fired at Highest Point

• When a particle moves in air, the path through which it travels represents its trajectory.
• A particle moving in air can have two trajectories – (a) Straight line   (b) Parabola
• If the initial velocity of particle is along acceleration, the trajectory of particle is straight line.
• If the initial velocity of particle makes some angle with acceleration, the trajectory of particle will be parabola.
• Now, suppose the object explodes in mid-air such that it breaks into many pieces.
• So, every piece of the object follows its own path.

(a) If $$\vec{v}$$ is parallel to $$\vec{a}$$ then straight line

(b) If $$\vec{v}$$ is not parallel to $$\vec{a}$$ then parabola.

• But the center of mass follows the same path as there is no explosion.

• Now consider a situation in which projectile explodes in two parts at the highest point and one of them comes to rest immediately.
• Let the masses of both particles be m1 and m2.
• Since at the highest point of projectile, range is $$\dfrac{R}{2}$$ and the particle (m1) which comes to rest immediately, will have x–coordinate $$\dfrac{R}{2}$$
• Since, there is no gravity along x–axis, we take only x–axis into consideration.
• Now for center of mass since, there is no explosion so, it's position will be (R,0).
• To calculate the distance of particle having mass m2, use definition of center of mass.
• $$x_{cm} = \dfrac{m_1 x_1 + m_2 x_2 }{m_1 + m_2 }$$

Here $$m_1 = m_1 \, , \,m_2 = m_2$$

$$x_1 = R/2 \, , \, x_{cm} = R$$

$$R = \dfrac{m_1(R/2) + m_2 (x)}{m_1 + m_2 }$$

$$m_1 R + m_2 R = m_1 \dfrac{R}{2} + m_2 x$$

$$x = \dfrac{m_1 \dfrac{R}{2} + m_2 R}{m_2}$$

A projectile explodes in the air at its highest point into two parts A and B. The mass of A, m1 = 5 kg, and mass of B, m2 = 3 kg. The range of projectile R = 6 cm. The particle A falls immediately at highest point. Calculate the distance of particle B from origin.

A 11 cm

B 12 cm

C 15 cm

D 18 cm

×

Since, there is no gravity along x–axis, we take only x–axis into consideration.

Since at the highest point of projectile, range is $$\dfrac{R}{2}$$ and the particle (m1) which comes to rest immediately, will have x–coordinate $$\dfrac{R}{2}$$

Now for center of mass since, there is no explosion so, it's position will be (R,0).

To calculate the distance of particle having mass m2, use definition of center of mass.

$$x_{cm} = \dfrac{m_1 x_1 + m_2 x_2 }{m_1 + m_2 }$$

Here $$m_1 = m_1 \, , \,m_2 = m_2$$

$$x_1 = R/2 \, , \,x_{cm} = R$$

$$R = \dfrac{m_1(R/2) + m_2 (x)}{m_1 + m_2 }$$

$$m_1 R + m_2 R = m_1 \dfrac{R}{2} + m_2 x$$

$$x = \dfrac{m_1 \dfrac{R}{2} + m_2 R}{m_2}$$

Given: $$m_1 = 5 \, kg , \,m_2 = 3 \, kg , \,R =6 \, cm$$

$$x = \dfrac{5 \left(\dfrac{6}{2}\right) + 3 \;(6)}{3}$$

$$x = \dfrac{15 +18}{3 }$$

$$x = \dfrac{33}{3} = 11 \; cm$$

A projectile explodes in the air at its highest point into two parts A and B. The mass of A, m1 = 5 kg, and mass of B, m2 = 3 kg. The range of projectile R = 6 cm. The particle A falls immediately at highest point. Calculate the distance of particle B from origin.

A

11 cm

.

B

12 cm

C

15 cm

D

18 cm

Option A is Correct