Center of Mass and Collision
133 Questions
Start JEE Mains Test
2022
Q51
JEE Mains
Numerical
10 Mar 2026
A batsman hits back a ball of mass 0.4 kg straight in the direction of the bowler without changing its initial speed of 15 ms$-$1. The impulse imparted to the ball is ___________ Ns.
Correct Answer: 12
Explanation:
$l = m\Delta v$
$ = 0.4 \times 2 \times 15 = 12$ Ns
2021
Q52
JEE Mains
MCQ
10 Mar 2026
A body of mass M moving at speed V0 collides elastically with a mass 'm' at rest. After the collision, the two masses move at angles $\theta$1 and $\theta$2 with respect to the initial direction of motion of the body of mass M. The largest possible value of the ratio M/m, for which the angles $\theta$1 and $\theta$2 will be equal, is :
A.
4
B.
1
C.
3
D.
2
2021
Q53
JEE Mains
MCQ
10 Mar 2026
Three objects A, B and C are kept in a straight line on a frictionless horizontal surface. The masses of A, B and C are m, 2m and 2m respectively. A moves towards B with a speed of 9 m/s and makes an elastic collision with it. Thereafter B makes a completely inelastic collision with C. All motions occur along same straight line. The final speed of C is :
A.
6 m/s
B.
9 m/s
C.
4 m/s
D.
3 m/s
2021
Q54
JEE Mains
MCQ
10 Mar 2026
Two billiard balls of equal mass 30g strike a rigid wall with same speed of 108 kmph (as shown) but at different angles. If the balls get reflected with the same speed then the ratio of the magnitude of impulses imparted to ball 'a' and ball 'b' by the wall along 'X' direction is :
A.
1 : 1
B.
$\sqrt 2 $ : 1
C.
2 : 1
D.
1 : $\sqrt 2 $
2021
Q55
JEE Mains
MCQ
10 Mar 2026
A bullet of '4 g' mass is fired from a gun of mass 4 kg. If the bullet moves with the muzzle speed of 50 ms$-$1, the impulse imparted to the gun and velocity of recoil of gun are :
A.
0.2 kg ms$-$1, 0.1 ms$-$1
B.
0.4 kg ms$-$1, 0.05 ms$-$1
C.
0.2 kg ms$-$1, 0.05 ms$-$1
D.
0.4 kg ms$-$1, 0.1 ms$-$1
2021
Q56
JEE Mains
MCQ
10 Mar 2026
If the Kinetic energy of a moving body becomes four times its initial Kinetic energy, then the percentage change in its momentum will be :
A.
100%
B.
200%
C.
300%
D.
400%
2021
Q57
JEE Mains
MCQ
10 Mar 2026
An object of mass m1 collides with another object of mass m2, which is at rest. After the collision the objects move with equal speeds in opposite direction. The ratio of the masses m2 : m1 is :
A.
1 : 1
B.
3 : 1
C.
2 : 1
D.
1 : 2
2021
Q58
JEE Mains
MCQ
10 Mar 2026
Two identical blocks A and B each of mass m resting on the smooth horizontal floor are connected by a light spring of natural length L and spring constant K. A third block C of mass m moving with a speed v along the line joining A and B collides with A. The maximum compression in the spring is
A.
$\sqrt {{{mv} \over K}} $
B.
$\sqrt {{m \over {2K}}} $
C.
$\sqrt {{{mv} \over {2K}}} $
D.
$v\sqrt {{m \over {2K}}} $
2021
Q59
JEE Mains
MCQ
10 Mar 2026
A large block of wood of mass M = 5.99 kg is hanging from two long massless cords. A bullet of mass m = 10 g is fired into the block and gets embedded in it. The (block + bullet) then swing upwards, their centre of mass rising a vertical distance h = 9.8 cm before the (block + bullet) pendulum comes momentarily to rest at the end of its arc. The speed of the bullet just before collision is :
(take g = 9.8 ms-2)
(take g = 9.8 ms-2)
A.
831.4 m/s
B.
811.4 m/s
C.
841.4 m/s
D.
821.4 m/s
2021
Q60
JEE Mains
MCQ
10 Mar 2026
Four equal masses, m each are placed at the comers of a square of length (l) as shown in the figure. The moment of inertia of the system about an axis passing through A and parallel to DB would be :
A.
$\sqrt 3 $ ml2
B.
2 ml2
C.
ml2
D.
3 ml2
2021
Q61
JEE Mains
MCQ
10 Mar 2026
Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R.
Assertion A : Body 'P' having mass M moving with speed 'u' has head-on collision elastically with another body 'Q' having mass 'm' initially at rest. If m << M, body 'Q' will have a maximum speed equal to '2u' after collision.
Reason R : During elastic collision, the momentum and kinetic energy are both conserved.
In the light of the above statements, choose the most appropriate answer from the options given below :
Assertion A : Body 'P' having mass M moving with speed 'u' has head-on collision elastically with another body 'Q' having mass 'm' initially at rest. If m << M, body 'Q' will have a maximum speed equal to '2u' after collision.
Reason R : During elastic collision, the momentum and kinetic energy are both conserved.
In the light of the above statements, choose the most appropriate answer from the options given below :
A.
A is correct but R is not correct.
B.
A is not correct but R is correct.
C.
Both A and R are correct and R is the correct explanation of A.
D.
Both A and R are correct but R is NOT the correct explanation of A.
2021
Q62
JEE Mains
MCQ
10 Mar 2026
A circular hole of radius $\left( {{a \over 2}} \right)$ is cut out of a circular disc of radius 'a' as shown in figure. The centroid of the remaining circular portion with respect to point 'O' will be :
A.
${1 \over 6}a$
B.
${2 \over 3}a$
C.
${5 \over 6}a$
D.
${10 \over 11}a$
Let $\sigma$ is the surface mass density of disc.
2021
Q63
JEE Mains
Numerical
10 Mar 2026
A bullet of 10 g, moving with velocity v, collides head-on with the stationary bob of a pendulum and recoils with velocity 100 m/s. The length of the pendulum is 0.5 m and mass of the bob is 1 kg. The minimum value of v = ____________ m/s so that the pendulum describes a circle. (Assume the string to be inextensible and g = 10 m/s2)
Correct Answer: 400
Explanation:
$V' = \sqrt {5gR} = \sqrt {5 \times 10 \times 0.5} $
V' = 5 m/s
m1 V = m2 $\times$ 5 $-$ m1 $\times$ 100
${{10} \over {1000}} \times V = 5 - {{10} \over {1000}} \times 100$
V = 400 m/s
V' = 5 m/s
m1 V = m2 $\times$ 5 $-$ m1 $\times$ 100
${{10} \over {1000}} \times V = 5 - {{10} \over {1000}} \times 100$
V = 400 m/s
2021
Q64
JEE Mains
Numerical
10 Mar 2026
A body of mass 2 kg moving with a speed of 4 m/s. makes an elastic collision with another body at rest and continues to move in the original direction but with one fourth of its initial peed. The speed of the two body centre of mass is ${x \over {10}}$ m/s. Then the value of x is ___________.
Correct Answer: 25
Explanation:
pi = pf
2 $\times$ 4 = 2 $\times$ 1 + m2 $\times$ v2
m2v2 = 6 ..... (i)
by coefficient of restitution
$1 = {{{v_2} - 1} \over 4} \Rightarrow {v_2} = 5$ m/s
by (i)
m2 $\times$ 5 = 6
m2 = 1.2 kg
${v_{cm}} = {{{m_1}{v_1} + {m_2}{v_2}} \over {{m_1} + {m_2}}}$
${v_{cm}} = {{2 \times 1 + 1.2 \times 5} \over {2 + 1.2}} = {8 \over {3.2}} = {{25} \over {10}}$
x = 25
2 $\times$ 4 = 2 $\times$ 1 + m2 $\times$ v2
m2v2 = 6 ..... (i)
by coefficient of restitution
$1 = {{{v_2} - 1} \over 4} \Rightarrow {v_2} = 5$ m/s
by (i)
m2 $\times$ 5 = 6
m2 = 1.2 kg
${v_{cm}} = {{{m_1}{v_1} + {m_2}{v_2}} \over {{m_1} + {m_2}}}$
${v_{cm}} = {{2 \times 1 + 1.2 \times 5} \over {2 + 1.2}} = {8 \over {3.2}} = {{25} \over {10}}$
x = 25
2021
Q65
JEE Mains
Numerical
10 Mar 2026
The position of the centre of mass of a uniform semi-circular wire of radius 'R' placed in x-y plane with its centre at the origin and the line joining its ends as x-axis is given by $\left( {0,{{xR} \over \pi }} \right)$. Then, the value of | x | is ______________.
Correct Answer: 2
Explanation:
Centre of mass of half ring is located at a distance ${{2R} \over \pi }$ from centre of the ring on its axis of symmetry so position of centre of mass in the given question will be $\left( {0,{{xR} \over \pi }} \right)$
$\Rightarrow$ | x | = 2
$\Rightarrow$ | x | = 2
2021
Q66
JEE Mains
Numerical
10 Mar 2026
A rod of mass M and length L is lying on a horizontal frictionless surface. A particle of mass 'm' travelling along the surface hits at one end of the rod with a velocity 'u' in a direction perpendicular to the rod. The collision is completely elastic. After collision, particle comes to rest. The ratio of masses $\left( {{m \over M}} \right)$ is ${1 \over x}$. The value of 'x' will be ____________.
Correct Answer: 4
Explanation:
The given situation can be shown as
Before collision,
As the collision is perfectly elastic, therefore momentum is conserved, i.e.
pinitially = pfinally
$\Rightarrow$ mu = Mv ..... (i)
Angular momentum will also be conserved about point O.
$ \Rightarrow mv\,.\,{L \over 2} = {{M{L^2}} \over {12}}\omega $
$ \Rightarrow \omega = {{6mv} \over {ML}}$ .... (ii)
$\because$ Coefficient of restitution,
$e = {{{\mathop{\rm Relative}\nolimits} \,velocity\,after\,collision} \over {{\mathop{\rm Relative}\nolimits} \,velocity\,before\,collision}}$
$ \Rightarrow 1 = {{v + {{\omega L} \over 2}} \over u}$
$ \Rightarrow v + {{\omega L} \over 2} = u$ .... (iii)
From Eqs. (ii) and (iii), we get
$v + {{3mu} \over M} = u$
$ \Rightarrow {{mu} \over M} + {{3mu} \over M} = u$ [using Eq. (i)]
$ \Rightarrow {{4mu} \over M} = u \Rightarrow {m \over M} = {1 \over 4}$ .... (iv)
According to question,
ratio of masses $\left( {{m \over M}} \right) = {1 \over x}$.
Comparing it with Eq. (iv), we get x = 4.
Before collision,
As the collision is perfectly elastic, therefore momentum is conserved, i.e.
pinitially = pfinally
$\Rightarrow$ mu = Mv ..... (i)
Angular momentum will also be conserved about point O.
$ \Rightarrow mv\,.\,{L \over 2} = {{M{L^2}} \over {12}}\omega $
$ \Rightarrow \omega = {{6mv} \over {ML}}$ .... (ii)
$\because$ Coefficient of restitution,
$e = {{{\mathop{\rm Relative}\nolimits} \,velocity\,after\,collision} \over {{\mathop{\rm Relative}\nolimits} \,velocity\,before\,collision}}$
$ \Rightarrow 1 = {{v + {{\omega L} \over 2}} \over u}$
$ \Rightarrow v + {{\omega L} \over 2} = u$ .... (iii)
From Eqs. (ii) and (iii), we get
$v + {{3mu} \over M} = u$
$ \Rightarrow {{mu} \over M} + {{3mu} \over M} = u$ [using Eq. (i)]
$ \Rightarrow {{4mu} \over M} = u \Rightarrow {m \over M} = {1 \over 4}$ .... (iv)
According to question,
ratio of masses $\left( {{m \over M}} \right) = {1 \over x}$.
Comparing it with Eq. (iv), we get x = 4.
2021
Q67
JEE Mains
Numerical
10 Mar 2026
The projectile motion of a particle of mass 5 g is shown in the figure.
The initial velocity of the particle is $5\sqrt 2 $ ms-1 and the air resistance is assumed to be negligible. The magnitude of the change in momentum between the points A and B is x $\times$ 10-2 kgms-1. The value of x, to the nearest integer, is __________.
The initial velocity of the particle is $5\sqrt 2 $ ms-1 and the air resistance is assumed to be negligible. The magnitude of the change in momentum between the points A and B is x $\times$ 10-2 kgms-1. The value of x, to the nearest integer, is __________.
Correct Answer: 5
Explanation:
$|\overrightarrow u | = |\overrightarrow v |$ .... (1)
$\overrightarrow u = u\cos 45\widehat i + u\sin 45\widehat j$ ..... (2)
$\overrightarrow v = v\cos 45\widehat i - v\sin 45\widehat j$ ..... (3)
$|\Delta \overrightarrow P | = |m(\overrightarrow v - \overrightarrow u )|$ .... (4)
$\Delta P = 2mu\sin 45^\circ $
$ = 2 \times 5 \times {10^{ - 3}} \times 5\sqrt 2 \times {1 \over {\sqrt 2 }}$
= 50 $\times$ 10$-$3
= 5 $\times$ 10$-$2
2021
Q68
JEE Mains
Numerical
10 Mar 2026
A ball of mass 10 kg moving with a velocity 10$\sqrt 3 $ m/s along the x-axis, hits another ball of mass 20 kg which is at rest. After the collision, first ball comes to rest while the second ball disintegrates into two equal pieces. One piece starts moving along y-axis with a speed of 10 m/s. The second piece starts moving at an angle of 30$^\circ$ with respect to the x-axis. The velocity of the ball moving at 30$^\circ$ with x-axis is x m/s. The configuration of pieces after collision is shown in the figure below. The value of x to the nearest integer is ____________.
Correct Answer: 20
Explanation:
Velocity of 10 kg ball = ${v_{10}} = 10\sqrt {3\widehat i} $
initial total momentum of system = $10 \times 10\sqrt {3\widehat i} $
Final total momentum of system
$ = 10 \times 10\widehat j + 10 \times x(\cos 30^\circ \widehat i - \sin 30^\circ \widehat j)$
Now by conservation of momentum
$10 \times 10\sqrt {3\widehat i} = 10 \times 10\widehat j + 10 \times x\left( {{{\sqrt 3 } \over 2}\widehat i - {1 \over 2}\widehat j} \right)$
$ \Rightarrow $ x = 20
initial total momentum of system = $10 \times 10\sqrt {3\widehat i} $
Final total momentum of system
$ = 10 \times 10\widehat j + 10 \times x(\cos 30^\circ \widehat i - \sin 30^\circ \widehat j)$
Now by conservation of momentum
$10 \times 10\sqrt {3\widehat i} = 10 \times 10\widehat j + 10 \times x\left( {{{\sqrt 3 } \over 2}\widehat i - {1 \over 2}\widehat j} \right)$
$ \Rightarrow $ x = 20
2021
Q69
JEE Mains
Numerical
10 Mar 2026
The disc of mass M with uniform surface mass density $\sigma$ is shown in the figure. The centre of mass of the quarter disc (the shaded area) is at the position ${x \over 3}{a \over \pi },{x \over 3}{a \over \pi }$ where x is _____________. (Round off to the Nearest Integer).
[a is an area as shown in the figure]
[a is an area as shown in the figure]
Correct Answer: 4
Explanation:
$dm = \sigma {1 \over 2}R \times Rd\theta $
$dm = {{\sigma {R^2}d\theta } \over 2}$
${x_{cm}} = {{\int {x\,dm} } \over {\int {dm} }} = {{\int\limits_0^{\pi /2} {{{\sigma {R^2}} \over 2}d\theta \left( {{{2R} \over 3}\cos \theta } \right)} } \over {\int\limits_0^{\pi /2} {{{\sigma {R^2}} \over 2}d\theta } }}$
$ = {{2R} \over 3}{{\int\limits_0^{\pi /2} {\cos \theta d\theta } } \over {\int\limits_0^{\pi /2} {d\theta } }}$
$ = {{2R} \over 3}\left( {{2 \over \pi }} \right)$
$ = {{4R} \over {3\pi }}$
So the value of x = 4
2021
Q70
JEE Mains
Numerical
10 Mar 2026
A ball of mass 10 kg moving with a velocity $10\sqrt 3 $ m s$-$1 along X-axis, hits another ball of mass 20 kg which is at rest. After collision, the first ball comes to rest and the second one disintegrates into two equal pieces. One of the pieces starts moving along Y-axis at a speed of 10 m/s. The second piece starts moving at a speed of 20 m/s at an angle $\theta$ (degree) with respect to the X-axis.
The configuration of pieces after collision is shown in the figure.
The value of $\theta$ to the nearest integer is ____________.
The configuration of pieces after collision is shown in the figure.
The value of $\theta$ to the nearest integer is ____________.
Correct Answer: 30
Explanation:
Conserving momentum along x-axis
$\overrightarrow {{p_i}} $ = $\overrightarrow {{p_f}} $
10 $\times$ 10$\sqrt 3 $ $=$ 10 $\times$ 20 cos$\theta$
$ \Rightarrow $ cos$\theta$ = ${{\sqrt 3 } \over 2}$
$ \therefore $ $\theta$ = 30$^\circ$
2021
Q71
JEE Mains
Numerical
10 Mar 2026
Two particles having masses 4 g and 16 g respectively are moving with equal kinetic energies. The ratio of the magnitudes of their linear momentum is n : 2. The value of n will be ___________.
Correct Answer: 1
Explanation:
$\because$ relation b/w kinetic energy & momentum is
$P = \sqrt {2mKE} $ ($\because$ KE = same)
$ \Rightarrow $ ${{{p_1}} \over {{p_2}}} = \sqrt {{{{m_1}} \over {{m_2}}}} $
$ \Rightarrow $ ${n \over 2} = \sqrt {{4 \over {16}}} $
$ \Rightarrow $ $n = 1$
$P = \sqrt {2mKE} $ ($\because$ KE = same)
$ \Rightarrow $ ${{{p_1}} \over {{p_2}}} = \sqrt {{{{m_1}} \over {{m_2}}}} $
$ \Rightarrow $ ${n \over 2} = \sqrt {{4 \over {16}}} $
$ \Rightarrow $ $n = 1$
2021
Q72
JEE Mains
Numerical
10 Mar 2026
Two solids A and B of mass 1 kg and 2 kg respectively are moving with equal linear momentum. The ratio of their kinetic energies (K.E.)A : (K.E.)B will be ${{A \over 1}}$, so the value of A will be ________.
Correct Answer: 2
Explanation:
Given that, ${{{M_1}} \over {{M_2}}} = {1 \over 2}$
Also, p1 = p2 = p
$ \Rightarrow $ M1V1 = M2V2 = p
Also, we know that
$K = {{{p^2}} \over {2M}} \Rightarrow {K_1} = {{{p^2}} \over {2{M_1}}}$ & ${K_2} = {{{p^2}} \over {2{M_2}}}$
$ \Rightarrow {{{K_1}} \over {{K_2}}} = {{{p^2}} \over {2{M_1}}} \times {{2{M_2}} \over {{p^2}}} \Rightarrow {{{K_1}} \over {{K_2}}} = {{{M_2}} \over {{M_1}}} = {2 \over 1}$
$ \Rightarrow {A \over 1} = {2 \over 1}$
$ \Rightarrow $ $ \therefore $ A = 2
Also, p1 = p2 = p
$ \Rightarrow $ M1V1 = M2V2 = p
Also, we know that
$K = {{{p^2}} \over {2M}} \Rightarrow {K_1} = {{{p^2}} \over {2{M_1}}}$ & ${K_2} = {{{p^2}} \over {2{M_2}}}$
$ \Rightarrow {{{K_1}} \over {{K_2}}} = {{{p^2}} \over {2{M_1}}} \times {{2{M_2}} \over {{p^2}}} \Rightarrow {{{K_1}} \over {{K_2}}} = {{{M_2}} \over {{M_1}}} = {2 \over 1}$
$ \Rightarrow {A \over 1} = {2 \over 1}$
$ \Rightarrow $ $ \therefore $ A = 2
2021
Q73
JEE Mains
Numerical
10 Mar 2026
A ball with a speed of 9 m/s collides with another identical ball at rest. After the collision, the direction of each ball makes an angle of 30$^\circ$ with the original direction. The ratio of velocities of the balls after collision is x : y, where x is __________.
Correct Answer: 1
Explanation:
The situation is shown below
Using conservation of linear momentum in y-direction,
pi = pf
As, pi = 0
and pf = mv1 sin30$^\circ$ $-$ mv2 sin30$^\circ$
$\Rightarrow$ 0 = m $\times$ ${1 \over 2}$v1 $-$ m $\times$ ${1 \over 2}$v2
$\Rightarrow$ v1 = v2 or v1 : v2 = 1 : 1
Since, v1 : v2 = x : y (given)
$\therefore$ x = 1
Using conservation of linear momentum in y-direction,
pi = pf
As, pi = 0
and pf = mv1 sin30$^\circ$ $-$ mv2 sin30$^\circ$
$\Rightarrow$ 0 = m $\times$ ${1 \over 2}$v1 $-$ m $\times$ ${1 \over 2}$v2
$\Rightarrow$ v1 = v2 or v1 : v2 = 1 : 1
Since, v1 : v2 = x : y (given)
$\therefore$ x = 1
2020
Q74
JEE Mains
MCQ
10 Mar 2026
Particle A of mass m1
moving with velocity $\left( {\sqrt3\widehat i + \widehat j} \right)m{s^{ - 1}}$ collides with another particle B of mass m2
which is at rest initially. Let $\overrightarrow {{V_1}} $
and $\overrightarrow {{V_2}} $
be the velocities of particles A and B after collision
respectively. If m1
= 2m2
and after
collision $\overrightarrow {{V_1}} = $$\left( {\widehat i + \sqrt 3 \widehat j} \right)$ , the angle between $\overrightarrow {{V_1}} $ and $\overrightarrow {{V_2}} $ is :
collision $\overrightarrow {{V_1}} = $$\left( {\widehat i + \sqrt 3 \widehat j} \right)$ , the angle between $\overrightarrow {{V_1}} $ and $\overrightarrow {{V_2}} $ is :
A.
105o
B.
15o
C.
-45o
D.
60o
2020
Q75
JEE Mains
MCQ
10 Mar 2026
Blocks of masses m, 2m, 4m and 8m are arranged in a line on a frictionless floor. Another block of
mass m, moving with speed v along the same line (see figure) collides with mass m in perfectly
inelastic manner. All the subsequent collisions are also perfectly inelastic. By the time the last block
of mass 8m starts moving the total energy loss is p% of the original energy. Value of 'p' is close to :
A.
37
B.
77
C.
87
D.
94
2020
Q76
JEE Mains
MCQ
10 Mar 2026
A block of mass 1.9 kg is at rest at the edge of a table, of height 1 m. A bullet of mass 0.1 kg
collides with the block and sticks to it. If the velocity of the bullet is 20 m/s in the horizontal
direction just before the collision then the kinetic energy just before the combined system strikes
the floor, is [Take g = 10 m/s2
. Assume there is no rotational motion and loss of energy after the
collision is negligable.]
A.
23 J
B.
21 J
C.
20 J
D.
19 J
2020
Q77
JEE Mains
MCQ
10 Mar 2026
A block of mass m = 1 kg slides with velocity v = 6 m/s on a frictionless horizontal surface and
collides with a uniform vertical rod and sticks to it as shown. The rod is pivoted about O and swings
as a result of the collision making angle $\theta $ before momentarily coming to rest. If the rod has mass
M = 2 kg, and length $l$ = 1 m, the value of $\theta $ is approximately :
(take g = 10 m/s2)
(take g = 10 m/s2)
A.
63o
B.
69o
C.
55o
D.
49
2020
Q78
JEE Mains
MCQ
10 Mar 2026
A particle of mass m with an initial velocity $u\widehat i$
collides perfectly elastically with a mass 3 m at
rest. It moves with a velocity $v\widehat j$ after collision,
then, v is given by :
A.
$v = \sqrt {{2 \over 3}} u$
B.
$v = {u \over {\sqrt 3 }}$
C.
$v = {u \over {\sqrt 2 }}$
D.
$v = {1 \over {\sqrt 6 }}u$
2020
Q79
JEE Mains
MCQ
10 Mar 2026
A rod of length L has non-uniform linear mass
density given by $\rho $(x) = $a + b{\left( {{x \over L}} \right)^2}$ , where a
and b are constants and 0 $ \le $ x $ \le $ L. The value
of x for the centre of mass of the rod is at :
density given by $\rho $(x) = $a + b{\left( {{x \over L}} \right)^2}$ , where a
and b are constants and 0 $ \le $ x $ \le $ L. The value
of x for the centre of mass of the rod is at :
A.
${3 \over 2}\left( {{{a + b} \over {2a + b}}} \right)L$
B.
${4 \over 3}\left( {{{a + b} \over {2a + 3b}}} \right)L$
C.
${3 \over 4}\left( {{{2a + b} \over {3a + b}}} \right)L$
D.
${3 \over 2}\left( {{{2a + b} \over {3a + b}}} \right)L$
2020
Q80
JEE Mains
MCQ
10 Mar 2026
A particle of mass m is projected with a speed
u from the ground at an angle
$\theta = {\pi \over 3}$ w.r.t.
horizontal (x-axis). When it has reached its
maximum height, it collides completely
inelastically with another particle of the same
mass and velocity $u\widehat i$ . The horizontal distance
covered by the combined mass before reaching
the ground is:
A.
$2\sqrt 2 {{{u^2}} \over g}$
B.
${{3\sqrt 3 } \over 8}{{{u^2}} \over g}$
C.
${{3\sqrt 2 } \over 4}{{{u^2}} \over g}$
D.
${5 \over 8}{{{u^2}} \over g}$
2020
Q81
JEE Mains
MCQ
10 Mar 2026
Two particles of equal mass m have respective
initial velocities $u\widehat i$ and $u\left( {{{\widehat i + \widehat j} \over 2}} \right)$.
They collide completely inelastically. The energy lost in the process is :
initial velocities $u\widehat i$ and $u\left( {{{\widehat i + \widehat j} \over 2}} \right)$.
They collide completely inelastically. The energy lost in the process is :
A.
${1 \over 3}m{u^2}$
B.
${1 \over 8}m{u^2}$
C.
${3 \over 4}m{u^2}$
D.
$\sqrt {{2 \over 3}} m{u^2}$
2020
Q82
JEE Mains
MCQ
10 Mar 2026
A particle of mass m is dropped from a height
h above the ground. At the same time another
particle of the same mass is thrown vertically
upwards from the ground with a speed of $\sqrt {2gh} $. If they collide head-on completely
inelastically, the time taken for the combined
mass to reach the ground, in units of $\sqrt {{h \over g}} $ is :
A.
$\sqrt {{1 \over 2}} $
B.
${1 \over 2}$
C.
$\sqrt {{3 \over 2}} $
D.
$\sqrt {{3 \over 4}} $
2020
Q83
JEE Mains
MCQ
10 Mar 2026
As shown in figure, when a spherical cavity
(centered at O) of radius 1 is cut out of a uniform
sphere of radius R (centered at C), the centre of
mass of remaining (shaded) part of sphere is at
G, i.e, on the surface of the cavity. R can be
detemined by the equation :
A.
(R2 + R – 1) (2 – R) = 1
B.
(R2 – R – 1) (2 – R) = 1
C.
(R2 – R + 1) (2 – R) = 1
D.
(R2 + R + 1) (2 – R) = 1
2020
Q84
JEE Mains
MCQ
10 Mar 2026
The coordinates of centre of mass of a uniform
flag shaped lamina (thin flat plate) of mass 4kg.
(The coordinates of the same are shown in
figure) are :
A.
(0.75m, 1.75m)
B.
(1m, 1.75m)
C.
(0.75m, 0.75m)
D.
(1.25m, 1.50m)
2020
Q85
JEE Mains
MCQ
10 Mar 2026
Three point particles of masses 1.0 kg, 1.5 kg and 2.5 kg are placed at three corners of a right angle triangle of sides 4.0 cm, 3.0 cm and 5.0 cm as shown in the figure. The center of mass of the system is at a point:
A.
2.0 cm right and 0.9 cm above 1 kg mass
B.
0.9 cm right and 2.0 cm above 1 kg mass
C.
0.6 cm right and 2.0 cm above 1 kg mass
D.
1.5 cm right and 1.2 cm above 1 kg mass
2020
Q86
JEE Mains
Numerical
10 Mar 2026
The centre of mass of solid hemisphere of radius 8 cm is x from the centre of the flat surface. Then
value of x is __________.
Correct Answer: 3
Explanation:
As we know c.o.m. or hemisphere = ${{3R} \over 8}$
$ \therefore $ x = ${{3R} \over 8}$ = ${{3 \times 8} \over 8}$ = 3 cm
$ \therefore $ x = ${{3R} \over 8}$ = ${{3 \times 8} \over 8}$ = 3 cm
2020
Q87
JEE Mains
Numerical
10 Mar 2026
Two bodies of the same mass are moving with the same speed, but in different directions in a
plane. They have a completely inelastic collision and move together thereafter with a final speed
which is half of their initial speed. The angle between the initial velocities of the two bodies (in
degree) is ________.
Correct Answer: 120
Explanation:
Momentum conservation along x,
mv0 × cos $\theta $ × 2 = 2m $ \times $ $\left( {{{{v_0}} \over 2}} \right)$
$ \Rightarrow $ cos $\theta $ = ${1 \over 2}$
$ \Rightarrow $ $\theta $ = 60o
$ \therefore $ 2$\theta $ = 120o
2020
Q88
JEE Mains
Numerical
10 Mar 2026
A square shaped hole of side l = ${a \over 2}$ is carved
out at a distance d = ${a \over 2}$ from the centre ‘O’ of
a uniform circular disk of radius a. If the
distance of the centre of mass of the
remaining portion form O is
$ - {a \over X}$ , value of X (to
the nearest integer) is :
Correct Answer: 23
Explanation:
Xcom = ${{\sigma \pi {a^2} \times 0 - \sigma .{{{a^2}} \over 4}.d} \over {\sigma \pi {a^2} - \sigma .{{{a^2}} \over 4}}}$
= ${{ - {{{a^2}} \over 4}.d} \over {\pi {a^2} - {{{a^2}} \over 4}}}$
= ${{ - d} \over {4\pi - 1}}$ = ${{ - a} \over {2\left( {4\pi - 1} \right)}}$
$ \therefore $ $ - {a \over X}$ = ${{ - a} \over {2\left( {4\pi - 1} \right)}}$
$ \Rightarrow $ X = ${2\left( {4\pi - 1} \right)}$ = ${\left( {8\pi - 2} \right)}$ = 23.12
So, nearest integer value of X = 23
= ${{ - {{{a^2}} \over 4}.d} \over {\pi {a^2} - {{{a^2}} \over 4}}}$
= ${{ - d} \over {4\pi - 1}}$ = ${{ - a} \over {2\left( {4\pi - 1} \right)}}$
$ \therefore $ $ - {a \over X}$ = ${{ - a} \over {2\left( {4\pi - 1} \right)}}$
$ \Rightarrow $ X = ${2\left( {4\pi - 1} \right)}$ = ${\left( {8\pi - 2} \right)}$ = 23.12
So, nearest integer value of X = 23
2020
Q89
JEE Mains
Numerical
10 Mar 2026
A particle of mass m is moving along the x-axis
with initial velocity $u\widehat i$. It collides elastically
with a particle of mass 10 m at rest and then
moves with half its initial kinetic energy (see
figure). If $\sin {\theta _1} = \sqrt n \sin {\theta _2}$ then value of n is
________.
Correct Answer: 10
Explanation:
${1 \over 2}mv_1^2 = {1 \over 2}\left( {{1 \over 2}m{u^2}} \right)$
$ \Rightarrow $ $v_1^2 = {{{u^2}} \over 2}$
$ \Rightarrow $ ${v_1} = {u \over {\sqrt 2 }}$ ......(1)
Also collision is elastic : ki = kf
${{1 \over 2}m{u^2} = {1 \over 2}mv_1^2 + {1 \over 2}\left( {10m} \right)v_2^2}$
$ \Rightarrow $ ${{1 \over 2} \times {1 \over 2}m{u^2} = {1 \over 2}\left( {10m} \right)v_2^2}$
$ \Rightarrow $ ${v_2^2 = {{{u^2}} \over {20}}}$
$ \Rightarrow $ ${v_2} = {u \over {\sqrt {20} }}$
By momentum conservation along y :
m1v1sin $\theta $1 = m2v2sin $\theta $2
$ \Rightarrow $ mv1sin $\theta $1 = 10mv2sin $\theta $2
$ \Rightarrow $ ${u \over {\sqrt 2 }}$sin $\theta $1 = 10 $ \times $ ${u \over {\sqrt {20} }}$sin $\theta $2
$ \Rightarrow $ sin $\theta $1 = $\sqrt {10} $ sin $\theta $2
$ \therefore $ n = 10
2020
Q90
JEE Mains
Numerical
10 Mar 2026
A body A, of mass m = 0.1 kg has an initial
velocity of 3$\widehat i$ ms-1 . It collides elastically with
another body, B of the same mass which has
an initial velocity of 5$\widehat j$ ms-1. After collision,
A moves with a velocity $\overrightarrow v = 4\left( {\widehat i + \widehat j} \right)$. The
energy of B after collision is written as ${x \over {10}}$. The value of x is ___________.
Correct Answer: 1
Explanation:
By conservation of linear momentum :
(0.1)(3$\widehat i$) + (0.1)(5$\widehat j$) = (0.1)(4)($\widehat i$ + $\widehat j$) + (0.1)$\overrightarrow v $
$ \Rightarrow $ $\overrightarrow v = - \widehat i + \widehat j$
$ \therefore $ $\left| {\overrightarrow v } \right|$ = $\sqrt 2 $
KEB = ${1 \over 2} \times 0.1 \times {\left( {\sqrt 2 } \right)^2}$ = ${1 \over {10}}$ J
$ \therefore $ x = 1
(0.1)(3$\widehat i$) + (0.1)(5$\widehat j$) = (0.1)(4)($\widehat i$ + $\widehat j$) + (0.1)$\overrightarrow v $
$ \Rightarrow $ $\overrightarrow v = - \widehat i + \widehat j$
$ \therefore $ $\left| {\overrightarrow v } \right|$ = $\sqrt 2 $
KEB = ${1 \over 2} \times 0.1 \times {\left( {\sqrt 2 } \right)^2}$ = ${1 \over {10}}$ J
$ \therefore $ x = 1
2019
Q91
JEE Mains
MCQ
10 Mar 2026
Three particles of masses 50 g, 100 g and 150 g are placed at the vertices of an equilateral triangle of side
1 m (as shown in the figure). The (x, y) coordinates of the centre of mass will be :
A.
$\left( {{{\sqrt 3 } \over 4}m,{5 \over {12}}m} \right)$
B.
$\left( {{7 \over {12}}m,{{\sqrt 3 } \over 8}m} \right)$
C.
$\left( {{7 \over {12}}m,{{\sqrt 3 } \over 4}m} \right)$
D.
$\left( {{{\sqrt 3 } \over 8}m,{7 \over {12}}m} \right)$
2019
Q92
JEE Mains
MCQ
10 Mar 2026
A man (mass = 50 kg) and his son (mass = 20 kg) are standing on a frictionless surface facing each other. The
man pushes his son so that he starts moving at a speed of 0.70 ms–1 with respect to the man. The speed of the
man with respect to the surface is :
A.
0.28 ms–1
B.
0.47 ms–1
C.
0.20 ms–1
D.
0.14 ms–1
2019
Q93
JEE Mains
MCQ
10 Mar 2026
Two particles, of masses M and 2M, moving,
as shown, with speeds of 10 m/s and 5 m/s,
collide elastically at the origin. After the
collision, they move along the indicated
directions with speeds u1 and u2, respectively.
The values of u1 and u2 are nearly :
A.
6.5 m/s and 3.2 m/s
B.
6.5 m/s and 6.3 m/s
C.
3.2 m/s and 6.3 m/s
D.
3.2 m/s and 12.6 m/s
2019
Q94
JEE Mains
MCQ
10 Mar 2026
A particle of mass 'm' is moving with speed '2v'
and collides with a mass '2m' moving with
speed 'v' in the same direction. After collision,
the first mass is stopped completely while the
second one splits into two particles each of
mass 'm', which move at angle 45° with respect
to the origianl direction.
The speed of each of the moving particle will
be :-
A.
2 $\sqrt2$v
B.
v / (2 $\sqrt2$ )
C.
v / $\sqrt2$
D.
$\sqrt2$v
$ \therefore 2{{mv} \over {\sqrt 2 }} = 4mv$
2019
Q95
JEE Mains
MCQ
10 Mar 2026
A wedge of mass M = 4m lies on a frictionless
plane. A particle of mass m approaches the
wedge with speed v. There is no friction
between the particle and the plane or between
the particle and the wedge. The maximum
height climbed by the particle on the wedge is
given by :-
A.
${{{v^2}} \over {g}}$
B.
${{2{v^2}} \over {7g}}$
C.
${{{v^2}} \over {2g}}$
D.
${{2{v^2}} \over {5g}}$
2019
Q96
JEE Mains
MCQ
10 Mar 2026
A ball is thrown vertically up (taken as +z-axis)
from the ground. The correct momentum-height
(p-h) diagram is :
A.
B.
C.
D.
2019
Q97
JEE Mains
MCQ
10 Mar 2026
A body of mass 2 kg makes an eleastic collision
with a second body at rest and continues to move
in the original direction but with one fourth of its
original speed. What is the mass of the second
body ?
A.
1.2 kg
B.
1.0 kg
C.
1.8 kg
D.
1.5 kg
2019
Q98
JEE Mains
MCQ
10 Mar 2026
A uniform rectangular thin sheet ABCD of
mass M has length a and breadth b, as shown
in the figure. If the shaded portion HBGO is
cut-off, the coordinates of the centre of mass
of the remaining portion will be :-
A.
$\left( {{{5a} \over 3},{{5b} \over 3}} \right)$
B.
$\left( {{{2a} \over 3},{{2b} \over 3}} \right)$
C.
$\left( {{{5a} \over 12},{{5b} \over 12}} \right)$
D.
$\left( {{{3a} \over 4},{{3b} \over 4}} \right)$
2019
Q99
JEE Mains
MCQ
10 Mar 2026
A body of mass m1 moving with an unknown
velocity of ${v_1}\mathop i\limits^ \wedge $, undergoes a collinear collision
with a body of mass m2 moving with a velocity
${v_2}\mathop i\limits^ \wedge $ . After collision, m1 and m2 move with
velocities of ${v_3}\mathop i\limits^ \wedge $ and ${v_4}\mathop i\limits^ \wedge $ , respectively.
If m2 = 0.5 m1 and v3 = 0.5 v1, then v1 is :-
A.
${v_4} - {{{v_2}} \over 2}$
B.
${v_4} - {{{v_2}} \over 4}$
C.
${v_4} - {v_2}$
D.
${v_4} + {v_2}$
2019
Q100
JEE Mains
MCQ
10 Mar 2026
Four particles A, B, C and D with masses
mA = m, mB = 2m, mC = 3m and mD = 4m are
at the corners of a square. They have
accelerations of equal magnitude with
directions as shown. The acceleration of the
centre of mass of the particles is :
A.
${a \over 5}\left( {\mathop i\limits^ \wedge + \mathop j\limits^ \wedge } \right)$
B.
${a \over 5}\left( {\mathop i\limits^ \wedge - \mathop j\limits^ \wedge } \right)$
C.
${a }\left( {\mathop i\limits^ \wedge + \mathop j\limits^ \wedge } \right)$
D.
Zero