By conservation of momentum, since no external impulses force acts on the system
$p_i=p_f$
Or
$0=m_rv_r+m_bv_b$ ---(1)
$v_r=\frac{-m_b}{m_r}v_b$
From equation (1)
$0=m_r\frac{\Delta x_r}{\Delta t}+m_b\frac{\Delta x_b}{\Delta t}$
$0=m_r\Delta x_r+m_b\Delta x_b$
Which shows
$|\frac{\Delta x_r}{\Delta x_b}|=\frac{m_b}{m_r}=\frac{6}{4000}=\frac{1}{667}$
Answer is (b)
Maximum energy loss occurs in head on collision
Let $P_0$ be initial momentum
$P_1$ and $P_2$ are final momentum
Applying the law of conservation of Linear momentum
$P_0=P_1+P_2$ ---(1)
Applying the law of conservation of energy
$\frac{P_0^2}{2m}=\frac{P_1^2}{2m}+\frac{P_2^2}{2M}$ ---(2)
Solving both equation (1) and (2)
$P_1=\frac{m-M}{m+M}P_0$
Final kinetic energy of the Neutron
$=\frac{(m-M)^2P_0^2}{2m(M+m)^2}$
$=\frac{K_0(m-M)^2}{(m+M)^2}$
Loss in kinetic energy
$=K_0-\frac{K_0(m-M)^2}{(m+M)^2}=\frac{4mMK_0}{(M+m)^2}$
Taking Railway car + jack + ball as the system, all the forces are internal, So linear momentum will remain conserved throughout the motion and on every collision
$M\mathbf{v_1}+m\mathbf{v_2}=0$ ---(1)
Now Velocity of the railway car before first collision
$M\mathbf{v}+mu\mathbf{i}=0$
Or
$\mathbf{v}=-\frac {-m}{M} u \mathbf{i}$
Now Velocity of the railway car after first collision
The effect of first collision will be simply to reverse the both velocity vector
$\mathbf{v}=\frac{mu\mathbf{i}}{M}$
Now Velocity of the railway car after second collision.
Since ball comes to rest, so as per equation (1)
Velocity of railway car=0
Since no external force acting, Center of mass will remain at rest .
P ->C
Q -> A
R -> B
S -> A
All are correct
Now time taken for first collision,
Displacement =L
Relative velocity with respect to floor of the rail car=$u+\frac{m}{M}u$
So time taken
$t=\frac{L}{(1+\frac{m}{M})u}$
Similarly for second collision
Displacement =L
Relative velocity with respect to floor of the rail car=$u+\frac{m}{M}u$
So time taken
$t=\frac{L}{(1+\frac{m}{M})u}$
So total time for second collision from the start
$t=\frac{2L}{(1+\frac{m}{M})u}$
Now It is seen that railcar first moves to the left a distance
$=(\frac{m}{M}u)(\frac{L}{u+\frac{m}{M}u})=\frac{m}{m+M}L $
Then moves to the right an equal distance. So it comes to rest at the starting point only
Now since no external force, position of center of mass will not change
Answer is (d)
Speed of the mass m just before the string becomes taut
$v=\sqrt{2gh}$
After the string becomes both the masses will start moving with same velocity. Let it $v_1$
Then
Applying law of conservation of linear momentum
$mv=(M+m)v_1$
$v_1=\frac{m}{M+m}\sqrt{2gh}$
Answer is (a)
Acceleration of the system after the jerk
$=\frac{-(M-m)}{M+m}g$
Now
$S=v_1t+\frac{1}{2}at^2$
$0=(\frac{m}{M+m}\sqrt{2gh})t-\frac{1}{2}\frac{(M-m)}{M+m}gt^2$
Or
$t=\frac{2m}{M-m}\sqrt{\frac{2h}{g}}$
$KE_{Fraction}=\frac{\frac{1}{2}mv^2-\frac{1}{2}(M+m)v_1^2}{\frac{1}{2}mv^2}$
Substituting the values from previous answer
$KE_{FRACTION}=\frac{M}{m+M}$
Now
$M\mathbf{v}=\frac{M}{4}\mathbf{v_1}+\frac{3M}{4}\mathbf{v_2}$
Or
$\mathbf{v_2}=\frac{4}{3}\mathbf{v} - 13\mathbf{v_1}$
$v_2^2=\mathbf{v_2}. \mathbf{v_2}=\frac {16}{9}(\mathbf{v} .\mathbf{v} )- \frac {8}{9}(\mathbf{v}.\mathbf{v_1})+\frac {1}{9} (\mathbf{v_1}.\mathbf{v_1})$
Or
$v_2^2=\frac{16}{9}(200)^2-\frac{8}{9}(200)(400)(cos{6}0)+\frac{1}{9}(400)^2=\frac{48*10^4}{9}$
Or
$v_2=231 \ m/s$
Now Initial Kinetic energy
$=\frac{1}{2}Mv^2=\frac{1}{2}M(200)^2=20000M$
Final kinetic energy
$=\frac{1}{2}(\frac{M}{4})(400)^2+\frac{1}{2}(\frac{3M}{4})(231)^2=\frac{213361M}{8}$
$Q=\frac{213361M}{8}-20000M=\frac{53361M}{8}$
Answer is (a) and ( b)
Given
$\boldsymbol{p(t)}=\mathbf{i}+t(1-at)\mathbf{j}$
$\mathbf{F}=\frac{d\mathbf{p}}{dt}=(1-2at)\mathbf{j}$ ---(2)
It is clear from equation (2),that force is acting always in the direction of $\mathbf{j}$ vector, And $\mathbf{i}$ vector is perpendicular to $\mathbf{j}$ vector.
So for $\mathbf{F}\bot \mathbf{p}$, $\mathbf{p}$ should be along $\mathbf{i}$ vector
t(1-at) =0
or t=0 and 1/a
Substituting these values in equation (2)
$\mathbf{F_1}=\mathbf{j}$ and $\mathbf{F_2}=-\mathbf{j}$
(a)
Given
$\boldsymbol{p(t)}=\mathbf{i}+t(1-at)\mathbf{j}$
$\boldsymbol{p(t+\Delta t)}=\mathbf{i}+(t+\Delta t)[1-a(t+\Delta t)]\mathbf{j}$ ---(2)
Now impulse is defined as
$\mathbf{I}=\boldsymbol{p(t+\Delta t)} -\boldsymbol{p(t)}$
Substituting the values from equation (1) and (2)
$\mathbf{I}=\Delta t(1-a\Delta t+2a\Delta t)\mathbf{j}$
Answer is (d)
From standard relation
$m\frac{d\mathbf{v}}{dt}=\mathbf{F}+\frac{dm}{dt}\mathbf{u}$
Now taking upward direction as positive
$m\frac{dv}{dt}=-mg-mbv+\lambda mu$
Or
$u\lambda-g-bv=\frac{dv}{dt}$
$\frac{dv}{u\lambda-g-bv}=dt$
Integrating
${log}_e{(}u\lambda-g-bv)=-bt+c$
Now at t=0,v=0
So
${log}_e{\frac{u\lambda-g-bv}{u\lambda-g}}=-bt $
$\frac{u\lambda-g-bv}{u\lambda-g}=e^{-bt}$
$1-\frac{bv}{u\lambda-g}=e^{-bt}$
$1-e^{-bt}=\frac{bv}{u\lambda-g}$
Or
$v=\frac{u\lambda-g}{b}-\frac{e^{-bt}(u\lambda-g)}{b}$
answer is (c)
$v=\frac{u\lambda-g}{b}-\frac{e^{-bt}(u\lambda-g)}{b}$
As t-> $\infty$
$v=\frac{u\lambda-g}{b}$
Case1
Considering the velocities to the right as positive, the initial momentum of the system is
$\frac{W+3w}{g}v_0$
Final momentum of the car is
$\frac{W}{g}(v_0+\Delta v)$
While that of 3 men is
$\frac{3w}{g}(v_0+\Delta v-u)$
Since no external force acts on the system, Linear momentum will remain conserved
$\frac{W+3w}{g}v_0=\frac{W}{g}(v_0+\Delta v)+\frac{3w}{g}(v_0+\Delta v-u)$ --(1)
Solving equation (1)
$\Delta v=\frac{3wu}{W+3w}$
Case 2
Now let s examine the motion by one man
The initial momentum of the system is
$\frac{W+3w}{g}v_0$
Final momentum of the car is
$\frac{W+2w}{g}(v_0+\Delta v)$
While that of men is
$\frac{w}{g}(v_0+\Delta v-u)$
Since no external force acts on the system, Linear momentum will remain conserved
$\frac{W+3w}{g}v_0=\frac{W+2w}{g}(v_0+\Delta v)+\frac{w}{g}(v_0+\Delta v-u)$ - (2)
Solving equation (2)
$\Delta v=\frac{wu}{W+3w}$
Similarly for the jump of second men
$\Delta v=\frac{wu}{W+2w}$
Third men
$\Delta v=\frac{wu}{W+w}$
So final velocity
$v=v_0+\frac{wu}{W+3w}+\frac{wu}{W+2w}+\frac{wu}{W+w}$
So all the four options are correct
(a), (b), (c)
When the disks breaks off the body M, its velocity towards right (along x axis) equals the velocity of the body M. Let it is $v_x$
Let $v_y$ be the disk velocity in upward direction along y axis at the time breaking off
There is no net external force in x direction, so linear momentum will remain conserved for the system
$mv=(M+m)v_x$
Or
$v_x=\frac{mv}{M+m}$ ---(1)
Applying law of conservation of energy for the system in the field of gravity
$\frac{1}{2}mv^2=\frac{1}{2}(M+m)v_x^2+\frac{1}{2}mv_y^2+mgh_1$ ---(2)
Substituting the value of $v_x$ from equation (1) in equation (2)
Where $h_1$ is the height of break off point from initial level
$v_y^2=v^2-\frac{mv^2}{M+m}-2gh_1$
Let $h_2$ be the height raised after the break off point, then
$v_y^2=2gh_2$ --(3)
From equation (2) and (3)
$h_1+h_2=\frac{Mv^2}{2g(M+m)}$
Answer is (b)
Applying law of conservation of linear momentum
$0=\frac{M}{3}(v_0\mathbf{i}+v_0\mathbf{j})+\frac{M}{3}(v_0\mathbf{j})+\frac{M}{3}\mathbf{v}$
Or
$\mathbf{v}=-v_0\mathbf{i}-2v_0\mathbf{j}$
(a) is the correct answer
We know that
$m\frac{dv}{dt}=F+\frac{dm}{dt}u$
Here F=0 and u=0
So
$m\frac{dv}{dt}=0$
So velocity remain constant
Change in momentum of bullet=.004*600=2.4 kgm/sec
Now force=Change in momentum /sec
=6*2.4=14.4 N
Energy given to each bullet=$\frac{1}{2} \times .004 \times (600)^2=720 \ J$
For six bullet=6*720=4320 J
Power developed=Workdone/sec=energy given/sec=4320 J/sec=4.32 KW
Let takes positive x-axis as downward along the inclined plane
Initial momentum of the system =m v_{0}
Final momentum of the system = p_{1}
Change in momentum of the system= $\mathbf{p_1} - m \mathbf{v_0}$
Now taking the system (Canon + shell) as whole along axis
Final momentum - Initial momentum = Impulse on the system
$pcos{\alpha}-mv_0=mgsin{\alpha}\Delta t$
Or $\Delta t=\frac{pcos{\alpha}-mv_0}{mgsin{\alpha}}$
Initial acceleration of flat car
$=\frac {F}{m_0}$
Let at any point of time t, velocity of the flat car is v
Now from the general equation of variable mass system
$m\frac{d\mathbf{v}}{dt}=\mathbf{F}+\mathbf{u}\frac{dm}{dt}$ -(1)
$m= m_0+ \mu t$
Now $\mathbf{u}=-\mathbf{v}$
So equation 1 can be written as
$m\frac{dv}{dt}=F-v\frac{dm}{dt}$
Or
$\frac{d(mv)}{dt}=F$
Or $v=\frac {Ft}{m}$
$v=\frac{Ft}{m_0+\mu t}$
Acceleration
$=\frac{dv}{dt}$
$=\frac{F}{m_0(1+\frac{\mu t}{m_0})^2}$
Momentum of the car
=mv
=Ft
Now from the general equation of variable mass system
$m\frac{d\mathbf{v}}{dt}=\mathbf{F}+\mathbf{u}\frac{dm}{dt}$ -(1)
Now $\mathbf{u}=-\mathbf{v}-\mathbf{v_0}$
Or
$(m_0+\mu t)\frac{dv}{dt}=F-\mu(v+v_0)$
Integrating
$V=\frac{(F-\mu v_0)\mu t}{m_0+\mu t}$
From law of conservation of linear momentum
$mv=(m+M)V$
or $V=\frac {mv}{(m+M)}$
KE of the composite block =$\frac{1}{2}(m+M)V^2$
$ =\frac{1}{2}(m+M)\frac{m^2v^2}{(m+M)^2}$
$=\frac{m^2v^2}{2(m+M)}$
For height
(M+m)gh= KE of the composite block
$(M+m)gh=\frac{m^2v^2}{2(m+M)}$
Or $h=\frac{m^2v^2}{2g(m+M)^2}$