- Center of Mass
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- Velocity of centre of mass
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- Acceleration of centre of mass
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- Kinetic energy of the system of particles
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- Linear Momentum of System of particles
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- Center Of Mass Frame of Reference
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- Center of Mass Problems with Solutions

- Included with Linear momentum
- Assignment 1
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- Assignment 2
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- Assignment 3
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- Assignment 4
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- Center of mass Problems for class 11

a. 2 kg, (6 m/s)i

b. 2 kg, (-6 m/s)i

c. 2 kg, (3 m/s)i

d. 2 kg, (-3 m/s)i

Total momentum=(18 kgm/sec)i

velocity of Center of mass=(3 m/s)i.

Mass of one object=4 kg

Velocity of this object=(1.5 m/s)i

let m be the mass of other object

And v be the velocity

Now we know total momentum =Total massX velocity of center of mass

(18 kgm/sec)i=(m+4)(3 m/s)i.

or m=2 kg

Now V

or

3i=(4*1.5i + 2v)/6

or 18i=6i+2v

v=6i m/sec

hence a is correct

a. Momentum Alone

b KE alone

c. Both Momentum and KE

d. Neither KE nor momentum

Since the collison is in elastic ,KE is not conserved

a.(-2i-3j)10

b. (2i+3j)10

c (2i-3j)10

d. none of these

Since momentum is conserved in explosion

Net momentum after collosion is zero

Momentum of first part after explosion=2i

Momentum of second part after explosion=3j

So momentum of third part after explosion=-(2i+3j) as net momentum is zero

Now Net change is momentum of this part =-(2i+3j)

Now we know that

Average force X time =Net change in momentum

Average force=-(2i+3j) 10

hence a is correct

STATEMENT 1 Linear momentum of a system of particles is zero.

STATEMENT 2 Kinetic energy of system of particles is zero.

(A) A does not imply B and B does not imply A.

(B) A implies B but B does not imply A

(C) A does not imply B but b implies A’

(D) A implies B and B implies A.

Net momentum=m

Net Kinectic Energy=(1/2)m

Let v

Then Net momentum=0 but Net Kinectic Energy is not equal to zero

Now lets v

Then Net Kinectic Energy=0 and Net momentum=0

Hence (c) is correct

a.r=(2R/π)j

b.r=(R/π)j

c.r=(4R/π)j

d. none of the these

Consider a differential element of lenght dl of the ring whose radius vector makes an angle θ with the x-axis .If the angle subtended by the length dl is dθ at the center then,dl=Rdθ

Let λ be the mass per unit length

Then mass of this element is dm=λRdθ

X

Y

or Y

Hence A is correct.

a. e

b. zero

c. e

d. none of the above

Since there are no external force on the system,center of mass remains at rest so acceleration is zero

a. 2 kg, (6 m/s)i

b. 2 kg, (-6 m/s)i

c. 2 kg, (3 m/s)i

d. 2 kg, (-3 m/s)i

Total momentum=(18 kgm/sec)i

velocity of Center of mass=(3 m/s)i.

Mass of one object=4 kg

Velocity of this object=(1.5 m/s)i

let m be the mass of other object

And v be the velocity

Now we know total momentum =Total massX velocity of center of mass

(18 kgm/sec)i=(m+4)(3 m/s)i.

or m=2 kg

Now V

or

3i=(4*1.5i + 2v)/6

or 18i=6i+2v

v=6i m/sec

hence a is correct

b.100m

c.80m

d.90

If we choose the initial location of the center of mass to be at x = 0, the center of mass will always be at x = 0.

The location of the center of mass of the system is determined from its definition

(M

Now x

So

M

Since the mass of the instrument is 1/10 of mass of the astronaut, the instrument will always be ten times as far away from the center of mass as you are, and it will always be on the opposite side of the center of mass from you,

So for 10 m move by astronaut, the instrument will move by 100m. Final distance of the instrument from the spaceship will be 110 m

Class 11 Maths Class 11 Physics Class 11 Chemistry