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
Since no external force is present,Momentum is conserved in the collision
Since the collision is in elastic ,KE is not conserved
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
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
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.
Since there are no external force on the system,center of mass remains at rest so acceleration is zero
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
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
$KE_i=\frac{1}{2}mv_0^2$ --(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
$=\frac {v_0 \sqrt 2[\sqrt {v_0^2+2gh}-v_0]}{g}$
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
Ans. a
Total linear momentum = (M_{1} + M_{2})v_{cm}
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}
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)
Ans. (a)
$x_{cm}=\frac{\int x d m}{\int d m}=\frac{\int{x(Ax+B)dx}}{\int{(Ax+B)dx}}$
$x_{cm}=\frac{L(2AL+3B)}{(3AL+6B)}$
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 $
$x_{cm} = \frac {-a^3b}{R^3-a^3}$
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})
Ans. (a)
v_{cm} = (0.5i + 3j/8)t
So total linear momentum = (8i + 6j)t
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
$R_{cm} = constant$
$\Delta R_{cm}=0$
$\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
$L=\frac{-mx}{M+m}$
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}$
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
$=m_1\mathbf{v_1}+m_2\mathbf{v_2}+\mathbf{g}t$
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
$\mathbf{r_c}=\frac{m_1\mathbf{v_1}+m_2\mathbf{v_2}}{m_1+m_2}t+\frac{1}{2}\mathbf{g}t^2$
(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
$\frac{d\mathbf{p}_{total}}{dt}=\mathbf{F_{ext}}$
$\frac{d\mathbf{p}_{total}}{dt}=m_1\mathbf{g+}m_2\mathbf{g}$
$X_{cm}=\frac{m_{1}x_1+m_2x_2+m_3x_3+m_4x_4+..........+m_{10}x_{10}}{m_1+m_2+m_3+m_4+.......m_{10}}$
$=\frac{m_0L+2m_02L+3m_03L+.............10m_010L}{m_0+2m_0+3m_0+......10m_0}$
$=\frac{L(1+2^2+3^2+4^2+......10^2)}{(1+2+3+4+.....+10)}$
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