Multiple Choice Questions
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 mass × velocity of center of mass
(18 kgm/sec)i=(m+4)(3 m/s)i.
or m=2 kg
Now V_{cm}=(m_{1}v_{1}+m_{2}v_{2})/(m_{1}+m_{2})
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 embedded 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 collision 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)10
^{3}
(b) (2i+3j)10
^{3}
(c) (2i-3j)10
^{-3}
(d) none of these
Solution
Net momentum before explosion zero
Since momentum is conserved in explosion
Net momentum after collision 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^{3}
hence a is correct
Question 4
Consider the following two statements.
STATEMENT A
Linear Momentum of a system of particles is zero.
STATEMENT B 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.
Solution
Net momentum=m_{1}v_{1}+m_{2}v_{2}
Net Kinetic Energy=(1/2)m_{1}v_{1}^{2}+(1/2)m_{2}v_{2}^{2}
Let v_{1}=v ,v_{2}=-v and m_{1}=m_{2}=m
Then Net momentum=0 but Net Kinetic Energy is not equal to zero
Now lets v_{1}= v_{2}=0
Then Net Kinetic 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
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_{cm}=(1/m)∫(Rcosθ)(Rdθ) integrating from 0 and π=0
Y_{cm}=(1/m)∫(Rsinθ)(Rdθ) integrating from 0 and π
or Y_{cm}=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) e
^{2}/64mπε
_{0}a
^{2}
(b) zero
(c) e
^{2}/16mπε
_{0}a
^{2}
(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 V_{cm}=(m_{1}v_{1}+m_{2}v_{2})/(m_{1}+m_{2})
or
3i=(4*1.5i + 2v)/6
or 18i=6i+2v
v=6i m/sec
hence a is correct
Question 8
An 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
(M_{ast}+M_{inst})x_{cm}=M_{ast}x_{ast}+M_{inst}x_{inst}
Now x_{cm}=0
So
M_{ast}x_{ast}=-M_{inst}x_{inst}
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
Question 9
A grenade of mass M is falling with a velocity v
_{0} at height when it explodes into two equal fragments that initially move horizontally in center of mass frame. The explosion is such that
$Q=KE_f - KE_i=mv_0^2$
Determine where the fragment will fall on the ground relative to the point directly below the grenade at the time of explosion
(a) $ \frac {v_0 [\sqrt {v_0^2+2gh}-v_0]}{g}$
(b) $ \frac {v_0 \sqrt 2[\sqrt {v_0^2+2gh}-v_0]}{g}$
(c) $ \frac {2v_0 [\sqrt {v_0^2+2gh}-v_0]}{g}$
(d) None of the above
Solution
$KE_i=\frac{1}{2}mv_0^2$ --(1)
From above Q equation,
$KE_f=\frac{3}{2}mv_0^2$ ---(2)
Centre of mass has velocity $v_0$ at the time of explosion and since both the piece move horizontally with C frame after explosion. There vertical component of velocity is $v_0$ and let v be velocity horizontal component and since momentum is constant. Both the piece will have same vertical and horizontal velocity in opposite direction.
So total energy
$KE_f=\frac{1}{2}(\frac{M}{2})v_0^2+\frac{1}{2}(\frac{M}{2})v^2+\frac{1}{2}(\frac{M}{2})v_0^2+\frac{1}{2}(\frac{M}{2})v^2$ ---(3)
From equation (2) and (3)
$v=v_0\sqrt2$
Now time taken to fall
$h=v_0t+\frac{1}{2}gt^2$
$gt^2+2v_0t-2h=0$
Solving the quadratic equation and taking positive root
t=\frac {[\sqrt {v_0^2+2gh}-v_0]}{g}$
So Distance travelled in horizontal direction
$=\frac {v_0 \sqrt 2[\sqrt {v_0^2+2gh}-v_0]}{g}$
Paragraph Based Questions
Two masses constrained to move in a horizontal plane collide. Given initially that M
_{1} = 85gm and M
_{2} = 200gm,
v_{1} = 6.4
i cm/sec and
v_{2} = -6.7
i - 2
j cm/sec.
i and
j are the unit vectors across x and y axis respectively
Question 10
Find the
Velocity of Center of mass
(a) -2.8
i - 1.4
j
(b) 2.8
i + 1.4
j
(c) 2
i + 1.4
j
(d) none of the above
Solution
Ans. (a)
v_{cm} = (M_{1}v_{1} +M_{2}v_{2})/M_{1} + M_{2}
=[85(6.4i) + 200(-6.7i - 2j)]/285
=-2.8i - 1.4j
Question 11
Find the total
Linear Momentum of the system
(a) -798
i - 400
j
(b) -796
i + 400
j
(c) 600
i + 400
j
(d) none of the above
Solution
Ans. a
Total linear momentum = (M_{1} + M_{2})v_{cm}
Question 12
Find the velocity in the reference of frame in which centre of mass is at rest
(a) 9.2
i - 1.4
j ,- 3.9
i - 0.6
j
(b) 9.2
i + 1.4
j ,- 3.9
i - 0.6
j
(c) -9.2
i + 1.4
j, - 3.9
i - 0.6
j
(d) none of the above
Solution
Ans. (b)
The velocity in the reference of frame in which center of mass is at rest is given by
v'_{1} = v_{1} - v_{cm}
v'_{2} = v_{2} - v_{cm}
Question 13
Let
w_{1} and
w_{2} are final velocity
now we know |
w_{1}| = 9.2 and
w_{2} = 4.4
i + 1.9
j cm/sec
find the direction of
w_{1}
(a) 26° with respect to x axis
(b) 84° with respect to x axis
(c) -84° with respect to x axis
(d) none of the above
Solution
Ans. (c)
Law of conservation of linear momentum
M_{1}v_{1} +M_{2}v_{2} = M_{1}w_{1} + M_{2}w_{2}
So
-798i - 400j=85 w_{1} + 200(4.4i + 1.9j)
Multiple Choice Questions
Question 14
A straight rod of length L has one of its end at origin and other at (L, 0). If the mass per unit length of rod is Ax + B .Find the centre of mass
(a) L(2AL + 3B)/(3AL + 6B)
(b) L(AL + 3B)/(AL + 6B)
(c) L(AL - B)/(AL + 2B)
(d) none of the above
Solution
Ans. (a)
$x_{cm}=\frac{\int x d m}{\int d m}=\frac{\int{x(Ax+B)dx}}{\int{(Ax+B)dx}}$
$x_{cm}=\frac{L(2AL+3B)}{(3AL+6B)}$
Question 15
A uniform solid sphere has a spherical hole in it. Find the centre of mass
(a) x
_{cm} = -a
^{3}b/(R
^{3}-a
^{3}), y
_{cm} = 0, z
_{cm} = 0
(b) x
_{cm} = a
^{3}b/(R
^{3}-a
^{3}), y
_{cm} = 0, z
_{cm} = 0
(c) x
_{cm} = b
^{3}a/(R
^{3}-a
^{3}), y
_{cm} = 0, z
_{cm} = 0
(d) none of the above
Solution
Ans. (a)
Imagine the total hole filled with matter so as to produce uniform sphere of radius R and density $\rho$.
The filled hole can then be represented by point $\frac {4}{3} \pi a^3 \rho$ at (b, 0, 0)
The remained of sphere of mass $\frac {4}{3} \pi(R^3 - a^3) \rho$ at ($x_{cm}$, 0, 0)
The centre of mass of these two part must be at centre of mass of sphere
So $\frac {4}{3} \pi(R^3 - a^3) \rho x_{cm} + \frac {4}{3} \pi a^3 \rho b = 0 $
$x_{cm} = \frac {-a^3b}{R^3-a^3}$
Paragraph Based Questions
(B) Two mass m
_{1}= 10 and m
_{2} = 6 are joined by a rigid bar of negligible mass as shown in Figure
Being initially at rest, they are subjected to force F
_{1} = 8
i N and F
_{2} = 6
j N
Question 16
Find the coordinate of their centre of mass as a function of time
(a)
i(1.5 - 0.25t
^{2}) +
j(1.88 - 0.188t)
(b)
i(1.5 + 0.25t
^{2}) +
j(1.88 + 0.188t
^{2})
(c) (1.5 + 0.188t
^{2})
i
(d) none of the above
Solution
Ans. (b)
v_{cm} at t = 0
Initial co-ordinates of Center of mass
x = 24/16 = 1.5
y = 3*10/16 = 15/8
Position Vector=1.5i+1.88 j
now
(M_{1} + M_{2})a_{cm} = F_{1} + F_{2}
16a_{cm} = 8i + 6j
a_{cm} = .5i + (3/8)j
So displacement in center of mass due to this acceleration
$\mathbf{s}=\mathbf{v}t + \frac {1}{2}\mathbf{a}t^2$
s=(1/2)(0.5i + 3j/8)*t^{2}
= (0.25t^{2}) i+ (.188t^{2}) j
So position vector of Center of mass
= (0.25t^{2}) i+ (.188t^{2}) j+ 1.5i + 1.88j
= i(1.5 + 0.25t^{2}) + j(1.88 + 0.188t^{2})
Question 17
Find the total linear momentum as function of time
(a) (8
i + 6
j)t
(b) (8
i -6
j)t
(c) (2
i +6
j)t
(d) none of the above
Solution
Ans. (a)
v_{cm} = (0.5i + 3j/8)t
So total linear momentum = (8i + 6j)t
(C) A mass of man m standing on a block of mass M. The system is at rest. The man moves relative in x direction to the block with velocity v' and then stops.
Question 18
Find the
Velocity of Center of mass
(a) v'
(b) 0
(c) $\frac {mv'}{M+m}$
(d) None of the above
Solution
As no external force, There will be no change in center of mass velocity
Initially center of mass is at rest and it will remain in rest
Question 19
Find the displacement of block relative to ground if the displacement of the man with respect to block is x
i
(a) $\frac {-mx}{M+m}\mathbf{i}$
(b) $\frac {mx}{M+m}\mathbf{i}$
(c) x
i
(d) none of the above
Solution
$R_{cm} = constant$
$\Delta R_{cm}=0$
$\sum{m_i\Delta r_i=0} $ --- (1)
Let L be the displacement of block relative to ground
Then Man displacement with respect to ground=L+x
Now applying equation (1)
$ML+m(L+x)=0$
Or
$L=\frac{-mx}{M+m}$
Question 20
Find the horizontal component of the force with when acted on the block M during the motion
(a) $\frac {-mM}{M+m} \frac {dv'}{dt}\mathbf{i}$
(b) $-\frac {M+m}{Mm} \frac {dv'}{dt}\mathbf{i}$
(c) $m \frac {dv'}{dt} \mathbf{i}$
$
(d) none of the above
Solution
As the net external force on the men-block system is zero, therefore momentum of the system should not change
$0=m(v'+v_2)+Mv_2$
Or
$v_2=-\frac{mv'}{m+M}$
Now as $v_2$ is along horizontal direction. So net force of the block
$ =M\frac{dv_2}{dt}=-\frac{Mm}{M+m}\frac{dv'}{dt}$
(D)A system is composed of two particles having mass m
_{1} and m
_{2} respectively. They are connected by a light spring of spring constant K. At time t=0, the system is thrown in the air such that m
_{1} has initial velocity
v_{1} and m
_{2} has initial velocity
v_{2} . Neglect any air drag in the motion
Question 21
Find the total momentum of the system at time t
(a)$m_1\mathbf{v_1}+m_2\mathbf{v_2}+(m_1+m_2)\mathbf{g}t$
(b)$m_1\mathbf{v_1}+m_2\mathbf{v_2}+\mathbf{g}t$
(c)$m_1\mathbf{v_1}+m_2\mathbf{v_2}$
(d) None of these
Solution
We know that
$\mathbf{F}=\frac{d\mathbf{p}}{dt}$
Now total force on the system
$\mathbf{F}=(m_1\mathbf{g}+m_2\mathbf{g})$
So Momentum at time t
$=m_1\mathbf{v_1}+m_2\mathbf{v_2}+\mathbf{g}t$
Question 22
Find the radius vector of the center of the mass relative to initial position.
(a) $\frac{1}{2}\mathbf{g}t^2$
(b)$\frac{m_1\mathbf{v_1}+m_2\mathbf{v_2}}{m_1+m_2}t+\frac{1}{2}\mathbf{g}t^2$
(c) $(m_1\mathbf{v_1}+m_2\mathbf{v_2})t$
(d) None of these
Solution
We know that center of mass will move under the action of force of gravity
So
$\mathbf{r_c}=\mathbf{v_c}t+\frac{1}{2}\mathbf{g}t^2$
Or
$\mathbf{r_c}=\frac{m_1\mathbf{v_1}+m_2\mathbf{v_2}}{m_1+m_2}t+\frac{1}{2}\mathbf{g}t^2$
Multiple Choice Questions
Question 23
Two masses m
_{1} and m
_{2} separated by thin rod of length L. The system is shown below
The rod mass can be neglected
The system is thrown into the air. It is assumed only gravitational forces are acting and frictional forces are assumed to be null
Which of the following statement is true?
(a) The center of mass of the system will follow a parabolic path
(b) $\frac{d\mathbf{p}_{total}}{dt}=m_1\mathbf{g+}m_2\mathbf{g}$
(c) The acceleration of the center of mass will be vertically downward and equal to g
(d) None of these
Solution
(a),(b),(c)
Equation of center of mass
$(m_1+m_2)\mathbf{a_{cm}}=\mathbf{F_{ext}}$
$(m_1+m_2)\mathbf{a_{cm}}=m_1\mathbf{g}+m_2\mathbf{g}$
$\mathbf{a_{cm}}=\mathbf{g}$
Since initial velocity of Center of mass is not zero, so it will follow a parabolic path
Also
$\frac{d\mathbf{p}_{total}}{dt}=\mathbf{F_{ext}}$
$\frac{d\mathbf{p}_{total}}{dt}=m_1\mathbf{g+}m_2\mathbf{g}$
Subjective Questions
Question 24
10 objects of mass m
_{0},2 m
_{0},3 m
_{0}………………….10 m
_{0} are placed on the x axis at
(L,0),(2L,0)……(10L,0).Find the
Center of mass of the system
Solution
$X_{cm}=\frac{m_{1}x_1+m_2x_2+m_3x_3+m_4x_4+..........+m_{10}x_{10}}{m_1+m_2+m_3+m_4+.......m_{10}}$
$=\frac{m_0L+2m_02L+3m_03L+.............10m_010L}{m_0+2m_0+3m_0+......10m_0}$
$=\frac{L(1+2^2+3^2+4^2+......10^2)}{(1+2+3+4+.....+10)}$
Now we know that
$1+2^2+.....n^2=\frac{n(n+1)(2n+1)}{6}$
$1+2+ .....+n=\frac {n(n+1)}{2}$
$=\frac{L \times 10 \times 11 \times 21 \times 2}{10 \times 11 \times 6}$
=7L
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