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Multiple choice Questions
Question 1
A 5kg rifle fires a 6g bullet with Speed of 500m/s. Find the ratio of the distance the rifle move backward while the bullet is in the barrel to the distance the bullet moves forward?
(a) 1/600
(b) 1/666
(c) 2/327
(d) none of the above Solution
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
$|\frac{\Delta x_r}{\Delta x_b}|=\frac{m_b}{m_r}=\frac{6}{4000}=\frac{1}{667}$
Question 2
A neutron of mass m is collides elastically with a nucleus of mass M, which is initially at rest. Neutron kinetic energy is k0. Find the maximum kinetic energy it can lose
(a) $\frac {2mMK_0}{(M+m)^2}$
(b) $\frac {4mMK_0}{(M+m)^2}$
(c) $\frac {mMK_0}{(M+m)^2}$
(d) none of the above Solution
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
$=K_0-\frac{K_0(m-M)^2}{(m+M)^2}=\frac{4mMK_0}{(M+m)^2}$
Linked Type Comprehensions
(A)A person Jack stands at left end of a railway car sitting on a railway track. There is no frictional force between the track and the railway car.
Person + Railway track Mass =M
Length of the railway car        =L
Jack throws a ball of mass m with velocity u towards the right end of the railway car where it collides elastically with the wall and travel back to the other end and strike the wall inelastically  and coming to rest. i is the unit vector across right end. Question 3 Match the column Column A
(P) Velocity of the railway car after the first collision
(Q) Velocity of the railway car after the second collision
(R) Velocity of the railway car before first collision
(S) Velocity of the Center of the mass of system( Railway car + jack + ball) Column B
(A) 0
(B)$\frac {-m}{M} u \mathbf{i}$
(C)$\frac {m}{M} u \mathbf{i}$
(D)$\frac {-m}{M+m} u \mathbf{i}$ Solution
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
Question 4
Which of the following statement is true
(a) The ball will take $\frac {L}{(1+ \frac {m}{M})u}$ time to strike the right end
(b) The ball will $\frac {2L}{(1+ \frac {m}{M})u}$ time to strike the left end
(c) There will be no net displacement of the railway car at the end of time $\frac {2L}{(1+ \frac {m}{M})u}$
(d) There will no change in center of the mass position of the system (Railway car + jack + ball) Solution
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
(B) After falling from rest through a distance h, a body of mass m begin to raise a body of Mass M (M >m) that is connected to it by means of a light inextensible string passing over a fixed smooth pulley Question 5
Find the velocity of the mass after the jerk
(a)$\frac{M}{m}\sqrt{2gh}$
(b) $\frac{M}{m+M}\sqrt{2gh}$
(c) $ \frac{m}{M}\sqrt{2gh}$
(d)$\frac{m}{m+M}\sqrt{2gh}$ Solution
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
$mv=(M+m)v_1$
$v_1=\frac{m}{M+m}\sqrt{2gh}$
Question 6
Find the time taken for the body of mass M to return to its original position
(a)$\frac{2m}{M-m}\sqrt{\frac{2h}{g}}$
(b)$\frac{2M}{M-m}\sqrt{\frac{2h}{g}}$
(c)$\frac{2m}{M}\sqrt{\frac{2h}{g}}$
(d) None of these Solution
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
$t=\frac{2m}{M-m}\sqrt{\frac{2h}{g}}$
Question 7
Find the fraction of the KE lost after the first Jerk
(a) $\frac{M}{m+M}$
(b) $\frac{m}{m+M}$
(c) $\frac{m}{M}$
(d) None of these Solution
$KE_{Fraction}=\frac{\frac{1}{2}mv^2-\frac{1}{2}(M+m)v_1^2}{\frac{1}{2}mv^2}$
Substituting the values from previous answer
$KE_{FRACTION}=\frac{M}{m+M}$
(C)A missile of mass M moving with velocity v= 200m/sec explodes in mid air breaking into two parts of mass M/4, 3M/4. If the smaller piece flies off at and angle of 60° with respect to the original direction with speed 400m/s.
Question 8
Find the velocity of the second piece
(a) 100 m/s
(b) 250 m/s
(c) 300m/s
(d) 231 m/s
Question 9
Q is defined as
$Q=KE_f - KE_i$
What is the value of Q in this case?
(a) $\frac {23361M}{8}$
(b)$\frac {13361M}{8}$
(c)$\frac {53361M}{8}$
(d) none of the above Solution 8-9
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
$=\frac{1}{2}(\frac{M}{4})(400)^2+\frac{1}{2}(\frac{3M}{4})(231)^2=\frac{213361M}{8}$
$Q=\frac{213361M}{8}-20000M=\frac{53361M}{8}$
(D) A particle of mass m moves such a way that momentum vector varies as
$\boldsymbol{p(t)}=\mathbf{i}+t(1-at)\mathbf{j}$
Where i and j are unit vector across x and y axis
Question 10
Find the Force vector on the mass where $\mathbf{F}\bot \mathbf{p}$
(a) j
(b) –j
(c) i +j
(d) i -j Solution
Answer is (a) and ( b)
Given
$\boldsymbol{p(t)}=\mathbf{i}+t(1-at)\mathbf{j}$
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 $\mathbf{F_2}=-\mathbf{j}$
Question 11
Find the impulse on the mass over the period t and t+ $\Delta t$
(a) $\Delta t(1-a\Delta t+2a\Delta t)\mathbf{j}$
(b) $\Delta t(1+a\Delta t-2a\Delta t)\mathbf{j}$
(c) $\Delta t(a\Delta t+2a\Delta t)\mathbf{j}$
(d) $\Delta t(1-2a\Delta t)\mathbf{j}$ Solution
(a)
Given
$\boldsymbol{p(t)}=\mathbf{i}+t(1-at)\mathbf{j}$
Now impulse is defined as
$\mathbf{I}=\boldsymbol{p(t+\Delta t)} -\boldsymbol{p(t)}$
Substituting the values from equation (1) and (2)
$\mathbf{I}=\Delta t(1-a\Delta t+2a\Delta t)\mathbf{j}$
(E) A rocket ascends from rest in a uniform gravitational field by ejecting exhaust with constant speed u. The rate of expulsion is $\frac {dm}{dt} = \lambda m$ where m is the instantaneous mass of the rocket and λ is constant. The rocket is also retarded by air resistance with a force (mbv) where b is constant
Question 12
Find the velocity of the rocket as a function of time
(a) $v=\frac{g}{b}-\frac{e^{-bt}(u\lambda-g)}{b}$
(b) $v=1-\frac{e^{-bt}(u\lambda-g)}{b}$
(c) $v=\frac{u\lambda-g}{b}-\frac{e^{-bt}}{b}$
(d) $v=\frac{u\lambda-g}{b}-\frac{e^{-bt}(u\lambda-g)}{b}$ Solution
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
$v=\frac{u\lambda-g}{b}-\frac{e^{-bt}(u\lambda-g)}{b}$
Question 13
Find the terminal velocity of the rocket
(a) $v=\frac{g}{b}-\frac{(u\lambda-g)}{b}$
(b) $v=\frac{g}{b}$
(c) $v=\frac{u\lambda-g}{b}$
d) $v=\frac{g-u\lambda}{b}$ Solution
answer is (c)
$v=\frac{u\lambda-g}{b}-\frac{e^{-bt}(u\lambda-g)}{b}$
As t-> $\infty$
$v=\frac{u\lambda-g}{b}$
Multiple Choice Questions
Question 14
A flat car of weight W roll without resistance along a horizontal surface .Initially a car together with a 3 men of weight w each is moving to the right with v0
Two cases are considered Case1- All the three move simultaneously with velocity u relative to floor of car toward left. And jump off at the left Case2- All the men run with Relative Velocity u and jump in succession towards left
Which of the following is true
(a) increase in velocity of flat car will be independent of the initial v0 of the system in both cases
(b) final velocity of flat car will be more in case2
(c) increment in velocity in case1 is given by $\frac {3wu}{W + 3w}$
(d) Increment in velocity in case2 after the second man jump is $\frac {wu}{W + 2w}$ Solution
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
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
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
Question 15 A body of mass M with small disc of mass m placed on its rest on a smooth horizontal plane.
The disc is set in motion in horizontal motion with velocity v.
which of the following is true
(a) the block M will starts moving towards due to constant force exerted by the disc m
(b) the height to which disc rise after breaking at point A from initial level is $\frac {Mv^2}{2g(M+m)}$
(c) velocity of mass M when disc break from is $\frac {mv}{M+m}$
(d) none of the above Solution
(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)
$h_1+h_2=\frac{Mv^2}{2g(M+m)}$
Question 16 An object of mass M is at the origin and is at rest. Suddenly an explosion happens in the object and object is divided three equal parts.
The first part goes with the velocity (v0i+v0j) .
The second part goes with the velocity v0j
What is the velocity vector for the third part?
(a) (v0i+v0j)
(b) (-v0i-2v0j)
(c) (v0i-v0j)
(d) None of these Solution
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
$\mathbf{v}=-v_0\mathbf{i}-2v_0\mathbf{j}$
Question 17
A flat car filled with sand is moving with velocity v. The flat car is leaking sand at the rate $\frac {dm}{dt}$ Find the velocity of the flat car after ½ amount has leaked
(a) v
(b) v/2
(c) v/4
(d) $v{log}_e{2}$ Solution
(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
Subjective Questions
Question 18
A machine gun fires six bullets per sec into a target. The mass of each bullet is 4 g.The speeds is 600m/s.Find the average force required to hold the gun in position. What is the power delivered to the bullet? Solution
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
Question 19
Cannon of Mass m starts sliding freely down a smooth inclined plane at an angle α to the horizontal. After the canon has reached v0 , a shot was fired .The shell leaves the canon in horizontal direction with a momentum p1. The cannon stop as consequences.
Assume the mass of the shell is negligible as compared to the canon
(a)The change in momentum of the system
(b)Calculate the duration of the shot Solution
Let takes positive x-axis as downward along the inclined plane
Initial momentum of the system =m v0
Final momentum of the system = p1
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 $\Delta t=\frac{pcos{\alpha}-mv_0}{mgsin{\alpha}}$
Question 20
A flat car of mass m0 starts moving to the right due to constant horizontal force F at t=0.Sand spills on the flat car from the stationary hopper .The velocity of loading is μ kg/sec.
(a) Calculate the Initial acceleration of the flat car
(b) Velocity and acceleration as a function of time
(c) Momentum of the car at any instant
(d) If the sand hopper moves with velocity v0 in the left direction, what will be the velocity of the car at any instant? Solution
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 $v=\frac {Ft}{m}$
$v=\frac{Ft}{m_0+\mu t}$
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
$V=\frac{(F-\mu v_0)\mu t}{m_0+\mu t}$
Question 21
A bullet of mass m  is fired horizontally with a velocity v on a wooden block of mass M suspended from a support and gets embedded in it.
(a) Calculate the final velocity of the composite block
(b) Calculate the KE of the composite block
(c) Find the height reached by the composite block Solution
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 $h=\frac{m^2v^2}{2g(m+M)^2}$
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