 # Center of mass problems with solutions

Question 1.A system consisting of two objects has a total momentum of (18 kgm/sec)i and its center of mass has the velocity of (3 m/s)i.One of the object has the mass 4 kg and velocity (1.5 m/s)i.The mass and velocity of the other objects are
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

Solution.

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 Vcm=(m1v1+m2v2)/(m1+m2)

or
3i=(4*1.5i + 2v)/6
or 18i=6i+2v
v=6i m/sec

hence a is correct

Question 2.A moving bullet hits a solid target resting on a frictionless surface and get embeded in it.What is conserved in it?
a. Momentum Alone
b KE alone
c. Both Momentum and KE
d. Neither KE nor momentum

Solution Since no external force is present,Momentum is conserved in the collision
Since the collison is in elastic ,KE is not conserved

Question 3. A stationary body of mass 3 kg explodes into three equal parts.Two of the pieces fly off at right angles to each other with the velocities 2i m/s and 3j m/s.If the explosion takes place in 10-3 sec.find out the average force action on the third piece in N
a.(-2i-3j)103
b. (2i+3j)103
c (2i-3j)10-3
d. none of these

Solution Net momentum before explosion zero
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) 103

hence a is correct

Question 4.Consider the following two statements.
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=m1v1+m2v2
Net Kinectic Energy=(1/2)m1v12+(1/2)m2v22

Let v1=v ,v2=-v and m1=m2=m
Then Net momentum=0 but Net Kinectic Energy is not equal to zero

Now lets v1= v2=0

Then Net Kinectic Energy=0 and Net momentum=0
Hence (c) is correct

Question 5.The Position vector of the center of mass of uniform semi circular ring of radius R and Mass M whose center coincided with the origin
a.r=(2R/π)j
b.r=(R/π)j
c.r=(4R/π)j
d. none of the these

Solution 5.

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

Xcm=(1/m)∫(Rcosθ)(Rdθ) integrating from 0 and π=0
Ycm=(1/m)∫(Rsinθ)(Rdθ) integrating from 0 and π
or Ycm=2R/π
Hence A is correct.

Question 6.Two electrons(e) are the at the point (-a,0) and (a,0).Their mass is m.They are released from rest.The acceleration of the center of mass of the system when the electron at 4a distance apart
a. e2/64mπε0a2
b. zero
c. e2/16mπε0a2
d. none of the above

Solution
Since there are no external force on the system,center of mass remains at rest so acceleration is zero
Question 7.A system consisting of two objects has a total momentum of (18 kgm/sec)i and its center of mass has the velocity of (3 m/s)i.One of the object has the mass 4 kg and velocity (1.5 m/s)i.The mass and velocity of the other objects are
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

Solution
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 Vcm=(m1v1+m2v2)/(m1+m2)

or
3i=(4*1.5i + 2v)/6
or 18i=6i+2v
v=6i m/sec

hence a is correct

Question 8An astronaut has just finished fixing a space telescope using a big instrument whose mass is one tenth as big as his mass. You realize you have no way to get back to your spaceship which is 10 meters away from you, so you throw the instrument as hard as you can in a direction away from the spaceship which causes you to move in the opposite direction, toward the spaceship. When you finally reach the space ship, how far away are you from the instrument? a.110m
b.100m
c.80m
d.90
Solution In this question, we define the system to be astronaut and the instrument, and the center of mass of the system is initially at rest a distance of 10 meters from your spaceship. Since there are no external forces acting on the system and the center of mass is initially at rest, the location of the center of mass of the system can never change
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
(Mast+Minst)xcm=Mastxast+Minstxinst

Now xcm=0
So
Mastxast=-Minstxinst

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