datavorous/entrance-exam-dataset huggingface dataset:
Two particles $A$ and $B$ initially at rest, move towards each other, under mutual force of attraction. At an instance when the speed of $A$ is $v$ and speed of $B$ is $2 v$, the speed of centre of mass $(\mathrm{CM})$ is

Options:
A Zero
B $v$
C $2.5v$
D $4v$

The correct answer is:
Zero

Solution:
As initially both the particles were at rest therefore velocity of centre of mass was zero and there is no external force on the system so speed of centre of mass remains constant $i e$, it should be equal to zero.

Topic: Physics, Center of Mass Momentum and Collision, AP EAMCET, AP EAMCET 2008
Two particles $A$ and $B$ initially at rest, move towards each other under mutual force of attraction. At an instance when the speed of $A$ is $v$ and speed of $B$ is $3 v$, the speed of centre of mass is

Options:
A $v$
B $4 v$
C $2 v$
D Zero

The correct answer is:
Zero

Solution:
Only internal force is acting on the system. Therefore $\vec{F}_{\text {ext }}=0$. It is also given that the initial velocity of COM is zero, therefore it will remain at rest.

Topic: Physics, Center of Mass Momentum and Collision, NEET, NEET 2023 (Manipur)
Two particles $A$ and $B$ initially at rest, move towards each other, under mutual force of attraction. At an instance when the speed of $A$ is $v$ and speed of $B$ is $2 v$, the speed of centre of mass $(\mathrm{CM})$ is

Options:
A Zero
B $v$
C $2.5v$
D $4v$

The correct answer is:
Zero

Solution:
As initially both the particles were at rest therefore velocity of centre of mass was zero and there is no external force on the system so speed of centre of mass remains constant $i e$, it should be equal to zero.

Topic: Physics, Center of Mass Momentum and Collision, TS EAMCET, TS EAMCET 2008
Two particles A and B of equal mass M are moving with the same speed v as shown in figure. They collide completely inelastic and move as a single particle C. The angle θ that the path of C makes with the X-axis is given by-

Options:
A tan⁡θ=3- 21- 2
B tan⁡θ=1- 22 1+ 3
C tan⁡θ=1- 31+ 2
D tan⁡θ=3+ 21- 2

The correct answer is:
tan⁡θ=3+ 21- 2

Solution:
After inelastic collision mass of the combination is 2m

v' be the velocity of the combined mass

Using conservation of momentum along X and Y-axis
2mv′sin⁡θ=mv2+mv32
2mv′cos⁡θ=mv2-mv2
2sinθ2cosθ=12+3212-12=2+31-2
tan θ=2+3 1- 2

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
Two particles A and B of equal mass M are moving with the same speed v as shown in figure. They collide completely inelastic and move as a single particle C. The angle θ that the path of C makes with the X-axis is given by-

Options:
A tan⁡θ=3- 21- 2
B tan⁡θ=1- 22 1+ 3
C tan⁡θ=1- 31+ 2
D tan⁡θ=3+ 21- 2

The correct answer is:
tan⁡θ=3+ 21- 2

Solution:
After inelastic collision mass of the combination is 2m

v' be the velocity of the combined mass

Using conservation of momentum along X and Y-axis
2mv′sin⁡θ=mv2+mv32
2mv′cos⁡θ=mv2-mv2
2sinθ2cosθ=12+3212-12=2+31-2
tan θ=2+3 1- 2

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2017 (09 Apr Online)
Two particles $\mathrm{A}$ and $\mathrm{B}$ of equal masses are suspended from two massless springs of spring constant $\mathrm{k}_1$ and $\mathrm{k}_2$, respectively. If the maximum velocities, during oscillation, are equal, the ratio of amplitude of $A$ and $B$ is

Options:
A
$\sqrt{\frac{\mathrm{k}_1}{\mathrm{k}_2}}$

B
$\frac{\mathrm{k}_2}{\mathrm{k}_1}$

C
$\sqrt{\frac{\mathrm{k}_2}{\mathrm{k}_1}}$

D
$\frac{\mathrm{k}_1}{\mathrm{k}_2}$

The correct answer is:
$\sqrt{\frac{\mathrm{k}_2}{\mathrm{k}_1}}$

Solution:
$\mathrm{V}_{\max }=\omega \mathrm{A}=\mathrm{m} \omega^2 \mathrm{I}=\mathrm{k}$
$\Rightarrow \omega^2=\frac{\mathrm{k}}{\mathrm{m}} \Rightarrow \omega=\sqrt{\frac{\mathrm{k}}{\mathrm{m}}} \quad \Rightarrow \omega \propto \sqrt{\mathrm{K}} \quad$ or $\quad \frac{\omega_1}{\omega_2}=\sqrt{\frac{\mathrm{k}_1}{\mathrm{k}_2}}$
$\mathrm{V}_{\mathrm{A}} \max =\mathrm{V}_{\mathrm{B}} \max \quad \Rightarrow\left(\sqrt{\frac{\mathrm{k}_1}{\mathrm{~m}}}\right)\left(\mathrm{A}_1\right)=\left(\sqrt{\frac{\mathrm{k}_2}{\mathrm{~m}}}\right)\left(\mathrm{A}_2\right) \Rightarrow \frac{\mathrm{A}_1}{\mathrm{~A}_2}=\sqrt{\frac{\mathrm{k}_2}{\mathrm{k}_1}}$

Topic: Physics, Oscillations, JEE Main, JEE Main 2003
Two particles A and B of equal masses are suspended from two massless springs of spring constants K1 and K2 respectively.If the maximum velocities during oscillations are equal, the ratio of the amplitude of A and B is

Options:
A K2 K1
B K1 K2
C K1 K2
D K2 K1

The correct answer is:
K2 K1

Solution:
A1ω1=A2ω2

A1k1m=A2k2m

A1 A2=k2k1

Topic: Physics, Oscillations, JEE Main, JEE Main 2021 (17 Mar Shift 2)
Two particles A and B of same mass have their de-Broglie wavelengths in the ratio XA :XB=K :1. Their potential energies UA :UB=1 :K2. The ratio of their total energies EA :EB is

Options:
A K2:1
B 1 :K2
C K :1
D 1 :K

The correct answer is:
1 :K2

Solution:
According to question, XAXB=K1 ...(i)
UAUB=1K2
So, UA=y and UB=K2y
According to de-Broglie wavelength,
λ=h2mK1 .....(ii)
Here, K1 is kinetic energy.
So, from Equations (i) and (ii),
⇒ h2mK1Ah2mK1B=K1⇒K1BK1A=K1
⇒ K1BK1A=K2
So, K1B=K2x and K1A=x
∵ Total energy, E=K1+U
So, EA=K1A+UA
and EB=K1B+UB
or EAEB=K1A+UAK1B+UB=x+yK2x+K2y
⇒ So, EAEB=1K2 or EA :EB=1 :K2

Topic: Physics, Dual Nature of Matter, JEE Main
Two particles are executing simple harmonic motion of the same amplitude $\mathrm{A}$ and frequency $\omega$ along the $x$-axis. Their mean position is separated by distance $X_0\left(X_0>A\right)$. If the maximum separation between them is $\left(X_0+A\right)$, the phase difference between their motion is :

Options:
A
$\frac{\pi}{3}$

B
$\frac{\pi}{4}$

C
$\frac{\pi}{6}$

D
$\frac{\pi}{2}$

The correct answer is:
$\frac{\pi}{2}$

Solution:
$\phi_1=0$
$\phi_2=\frac{\pi}{2}$

Topic: Physics, Oscillations, JEE Main, JEE Main 2011
Two particles are executing simple harmonic motion of the same amplitude A and frequency ω along the x-axis. Their mean positions are separated by distance x0(x0>A). If the maximum separation between them is (x0+A), the phase difference between their motions is

Options:
A 2π3
B π4
C π6
D π2

The correct answer is:
2π3

Solution:
When the distance between the particles is maximum, their relative velocity will be zero. This means the particles are at the same distance from from their respective mean positions. Since the maximum distance is x0+A, it means the particles are at half the amplitude position. Also, if one of the particles is left of its mean position, then the other one would be to the right of the mean position.

x1=Asin⁡(ωt+ϕ1)

At t=0, let x1=A2, then

ωt+ϕ1=π6

x2=Asin⁡(ωt+ϕ2)

At t=0, x2=-A2

ωt+ϕ2=2π-π3

⇒ϕ2-ϕ1=2π3

Topic: Physics, Oscillations, JEE Main
Two particles are performing simple harmonic motion in a straight line about the same equilibrium point. The amplitude and time period for both particles are same and equal to A and T, respectively. At time t=0 one particle has displacement A while the other one has displacement -A2 and they are moving towards each other. If they cross each other at time t, then t is:

Options:
A 5T6
B T3
C T4
D T6

The correct answer is:
T6

Solution:
Angle covered to meet =60o=π3rad
Therefore, t= θ ω = π 3 × T 2π = T 6

Topic: Physics, Oscillations, JEE Main, JEE Main 2016 (09 Apr Online)
Two particles are projected from a point at the same instant with velocities whose horizontal components and vertical components are u1,v1 and u2,v2 respectively. The time interval between their passing through the other common point of their path (other than origin) is

Options:
A 2gv1u1-v2u2u1+u2
B 2gv12+v22u1+u2
C 2gu12+u22v1+v2
D 2gv1u2-v2u1u1+u2

The correct answer is:
2gv1u2-v2u1u1+u2

Solution:
The time for particle 1 to pass through common point of their path
t1=xu1
The time for particle 2 to pass through common point of their path
t2=xu2
Difference of time t=t1-t2
Time t=xu1-xu2=x1u1-1u2
y=xtanθ-gx22ux2
y=xv1u1-gx22u12=xv2u2-gx22u22
v1u1-v2u2=xg21u12-1u22
v1u1-v2u2=xg21u1+1u21u1-1u2
t=x1u1-1u2=2gv1u2-v2u1u1+u2

Topic: Physics, Motion In Two Dimensions, JEE Main
Two particles are projected from the same point with the same speed u such that they have the same range R, but different maximum heights, h1 and h2. Which of the following is correct?

Options:
A R2=h1h2
B R2=4 h1h2
C R2=2 h1h2
D R2=16 h1h2

The correct answer is:
R2=16 h1h2

Solution:
We get same range for angle of projections θ and 90-θ .
h=u2sin2θ2g
⇒h1=u2sin2θ2g,h2=u2cos2θ2g
h1h2=u4sin2θcos2θ4g2=4u4sin2θcos2θ16g2
=u2sin2θg2116=R216
∴    R2=16h1h2

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2019 (12 Apr Shift 2)
Two particles are projected vertically upwards from the surface of the earth with velocities v1=2gR3 and v2=4gR3 respectively. If the maximum heights attained by the two particles are h1 and h2 respectively, then calculate the ratio h1h2.

Options:

The correct answer is:
.25

Solution:
Loss of KE= Gain in PE

â12mv2=mghRh+R

CASE I.

â12m2gR3=mgh1Rh1+R ...(i)

â12m4gR3=mgh2Rh2+R ...(ii)

Solving (i)

â13=h1h1+R

âh1=R2 ...(iii)

Solving (ii)

â23=h2h2+R ...(iv)

âh2=2R

From (iii) and (iv)

âh1h2=R/22R

â´ h1h2=14=0.25

Topic: Physics, Gravitation, JEE Main
Two particles are simultaneously thrown from the top of two towers as shown. Their velocities are 2 m s-1 and 14 m s-1. Horizontal and vertical separations between these particles are 22 m and 9 m respectively. Then the minimum separation between the particles in the process of their motion in meters is (g = 10 m s-2)

Options:

The correct answer is:
6

Solution:
vx=82m s-1= relative velocity along x-axis
∴x=22-82t
vy=62m/s= relative velocity along the y-axis
∴y=9-62t
r=x2+y2
For minimum r,
drdt=0⇒t=23102 s

r=x2+y2

substituting the value of t in r, rmin=6 m

Topic: Physics, Motion In Two Dimensions, JEE Main
Two particles have masses $\mathrm{m} \& 4 \mathrm{~m}$ and their kinetic energies are in the ratio $2: 1$. What is the ratio of their linear momenta?

Options:
A $\frac{1}{\sqrt{2}}$
B $\frac{1}{2}$
C $\frac{1}{4}$
D $\frac{1}{16}$

The correct answer is:
$\frac{1}{\sqrt{2}}$

Solution:
$\frac{K E_1}{K E_2}=\frac{\frac{p_1^2}{2 m}}{\frac{p_2^2}{2 \times 4 m}}=\frac{2}{1} \Rightarrow \frac{p_1}{p_2}=\frac{1}{\sqrt{2}}$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
Two particles having de-Broglie wavelengths Î» 1 and Î» 2 , while moving along mutually perpendicular directions, undergo perfectly inelastic collision. The de-Broglie wavelength Î» , of the final particle is

Options:
A Î»=Î»1Î»2Î»12+Î»22
B λ= λ 1 λ 2
C λ= λ 1 + λ 2
D λ= λ 1 + λ 2 2

The correct answer is:
λ=λ1λ2λ12+λ22

Solution:
Let us assume that one of the particles is moving along the positive x direction and the other is moving along the positive y direction. The net momentum of the system is

Pânet=Pi^+Pj^

hÎ»=hÎ»1i^+hÎ»2j^
1Î»=1Î»12+1Î»22
Î»=Î»1Î»2Î»12+Î»22

Topic: Physics, Dual Nature of Matter, JEE Main
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 ___.

Options:

The correct answer is:
1

Solution:
p122×4=p222×16

p1p2=12

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2021 (25 Feb Shift 2)
Two particles having masses m and 4m are separated by distance l. The distance of the centre of mass from m is x1 and x2 is the distance of point at which gravitational field intensity is zero. Find the value of x1x2

Options:
A 1
B 2
C 2.4
D 3.6

The correct answer is:
2.4

Solution:
x1=4ml5m=0.8l
Gmx22=4Gm(l-x2)2
2x2+x2=l
x2=l3
x1x2=2.4

Topic: Physics, Gravitation, JEE Main
Two particles having position vectors $\mathbf{r}_1=(3 \hat{\mathbf{i}}+5 \hat{\mathbf{j}})$ metres and $\mathbf{r}_2=(-5 \hat{\mathbf{i}}-3 \hat{\mathbf{j}})$ metres are moving with velocities $\mathbf{v}_1=(4 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s}$ and $\mathbf{v}_2=(a \hat{\mathbf{i}}+7 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s}$. If they collide after $2 \mathrm{~s}$, then the value of $a$ is :

Options:
A 2
B 4
C 6
D 8

The correct answer is:
8

Solution:
$\mathbf{r}_1=(3 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}) \mathrm{m}$
$\mathbf{r}_2=(-5 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}) \mathrm{m}$
$\mathbf{v}_1=(4 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s}$ and
$\mathbf{v}_2=(a \hat{\mathbf{i}}+7 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s}$
Time taken in collision, $t=\frac{r}{v}$ ...(i)
where, $r=$ resultant position vector
$=\mathbf{r}_1-\mathbf{r}_2$
$=8 \hat{\mathbf{i}}+8 \hat{\mathbf{j}}$
$v=$ resultant velocity
$\mathbf{v}_2-\mathbf{v}_1=(a-4) \hat{\mathbf{i}}+4 \hat{\mathbf{j}}$
From Eqs. (i), we get
$\therefore \quad 2=\frac{8 \hat{\mathbf{i}}+8 \hat{\mathbf{j}}}{(a-4) \hat{\mathbf{j}}+3 \hat{\mathbf{j}}} \quad[\because t=2 \mathrm{~s}]$
$(a-4) \hat{\mathbf{i}}+4 \hat{\mathbf{j}}=4 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}$
$a-4=4$
$a=8$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
Two particles having position vectors $\overrightarrow{r_1}=(3 \hat{i}+5 \hat{j})$ metres and $\overrightarrow{r_2}=(-5 \hat{i}-3 \hat{j})$ metres are moving with velocities $\vec{v}_1=(4 \hat{i}+3 \hat{j}) \mathrm{m} / \mathrm{s}$ and $\vec{v}_2=(\alpha \hat{i}+7 \hat{j}) \quad \mathrm{m} / \mathrm{s}$. If they collide after 2 seconds, the value of ' $\alpha$ ' is I

Options:
A 2
B 4
C 6
D 8

The correct answer is:
8

Solution:
It is clear from figure that the displacement vector $\Delta \dot{r}$ between particles p1 fand p2 is $\Delta \vec{r}=\vec{r}_2-\vec{r}_r=-8 \hat{i}-8 \hat{i}$
$|\Delta \hat{r}|=\sqrt{(-8)^2+(-8)^2}=8 \sqrt{2}$ ....(i)
Now, as the particles are moving in same direction ($\because \overrightarrow{v_1}$ and $\overrightarrow{v_2}$ are $\left.+v e\right)$, the relative velocity is given by
$\begin{aligned}
& \vec{v}_{\text {/el }}=\overrightarrow{v_2}-\vec{v}_1=(\alpha-4) \hat{i}+4 \hat{j} \\
& v_{\text {/al }}=\sqrt{(\alpha-4)^2+16}
\end{aligned}$ ....(ii)
Substituting the values of $\dot{v}_m$ and $|\Delta \vec{v}|$ from equation (i) and (ii) and $t=2 \mathrm{~s}$, then on solving we get $a=8$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
Two particles move at right angle to each other. Their de Broglie wavelengths are λ1 and λ2 respectively. The particles suffer perfectly inelastic collision. The de Broglie wavelength λ of the final particle, is given by:

Options:
A 2λ=1λ1+1λ2
B λ=λ1+λ22
C λ=λ1λ2
D 1λ2=1λ12+1λ22

The correct answer is:
1λ2=1λ12+1λ22

Solution:
P1→=hλ1i^,    P2→=hλ2j^
P→=P1→+P2→=hλ1i^+hλ2j^
P→=h1λ12+1λ22
hλ=h1λ12+1λ22
1λ2=1λ12+1λ22

Topic: Physics, Dual Nature of Matter, JEE Main
Two particles move at right angle to each other. Their de Broglie wavelengths are λ1 and λ2 respectively. The particles suffer perfectly inelastic collision. The de Broglie wavelength λ of the final particle, is given by:

Options:
A 2λ=1λ1+1λ2
B λ=λ1+λ22
C λ=λ1λ2
D 1λ2=1λ12+1λ22

The correct answer is:
1λ2=1λ12+1λ22

Solution:
P1→=hλ1i^,    P2→=hλ2j^
P→=P1→+P2→=hλ1i^+hλ2j^
P→=h1λ12+1λ22
hλ=h1λ12+1λ22
1λ2=1λ12+1λ22

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2019 (08 Apr Shift 1)
Two particles move at right angle to each other. Their de Broglie wavelengths are \( \lambda_{1} \) and \( \lambda_{2} \) respectively. The particles suffer perfectly inelastic collision. The de Broglie wavelength \( \lambda \) of the final particle, is given by:

Options:
A \( \frac{2}{\lambda}=\frac{1}{\lambda_{1}}+\frac{1}{\lambda_{2}} \)
B \( \lambda=\frac{\lambda_{1}+\lambda_{2}}{2} \)
C \( \lambda=\sqrt{\lambda_{1} \lambda_{2}} \)
D \( \frac{1}{\lambda^{2}}=\frac{1}{\lambda_{1}^{2}}+\frac{1}{\lambda_{2}^{2}} \)

The correct answer is:
\( \frac{1}{\lambda^{2}}=\frac{1}{\lambda_{1}^{2}}+\frac{1}{\lambda_{2}^{2}} \)

Solution:
P1→=hλ1i^,    P2→=hλ2j^
P→=P1→+P2→=hλ1i^+hλ2j^
P→=h1λ12+1λ22
hλ=h1λ12+1λ22
1λ2=1λ12+1λ22

Topic: Physics, Dual Nature of Matter, JEE Main
Two particles of equal mass have velocities $\overrightarrow{\mathbf{v}}_1=4 \hat{\mathbf{i}}$ and $\overrightarrow{\mathbf{v}}_2=4 \hat{\mathbf{j}} \mathrm{ms}^{-1}$. First particle has an acceleration $\overrightarrow{\mathbf{a}}_1=(5 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}) \mathrm{ms}^{-2}$ while the acceleration of the other particle is zero. The centre of mass of the two particles moves in a path of

Options:
A straight line
B parabola
C circle
D ellipse

The correct answer is:
straight line

Solution:
$$
\begin{aligned}
& \mathbf{v}_{\mathrm{CM}}=\frac{m_1 \mathbf{v}_1+m_2 \mathbf{v}_2}{m_1+m_2} \\
& =\frac{\mathbf{v}_1+\mathbf{v}_2}{2}\left(\text { as } m_1=m_2\right) \\
& \text { and } \mathbf{a}_{\mathrm{CM}}=\frac{m_1 \mathbf{a}_1+m_2 \mathbf{a}_2}{m_1+m_2}=\frac{\mathbf{a}_1+0}{2}=\frac{\mathbf{a}_1}{2}
\end{aligned}
$$
The centre of mass of two particles will move with the mean velocity of two particles having common acceleration $\frac{\mathbf{a}}{2}$. Hence, path of CM will be a straight line.

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
Two particles of equal mass have velocities v→1=4i^ ms−1 and v→2=4j^  ms−1. First particle has an acceleration a→1=5i^+5j^  ms−2, while the acceleration of the other particle is zero. The centre of mass of the two particles moves in a path of

Options:
A Straight line
B Parabola
C Circle
D Ellipse

The correct answer is:
Straight line

Solution:
Since, m 1 = m 2 =m
v → cm = m 1 v → 1 + m 2 v → 2 m 1 + m 2 = m 2m v → 1 + v → 2
= 1 2 ×(4 i ^ +4 j ^ )
=2( i ^ + j ^ )
Similarly a → cm = m 1 a 1 → + m 2 a 2 → m 1 + m 2 = m 2m ( a 1 → + a 2 → )
= 1 2 ×(5 i ^ +5 j ^ +0)
= 5 2 ( i ^ + j ^ )
Both v → cm and a → cm are parallel
Centre of mass will move in a straight line because velocity and acceleration of centre of mass are in same direction.

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
Two particles of equal mass m are moving with velocities vi^ and vi^+j^2 respectively. If the particles collide completely inelastically, then the energy lost in the process is

Options:
A 3mv24
B 2mv23
C mv23
D mv28

The correct answer is:
mv28

Solution:
Conserving momentum

mvi^+mv2i^+v2j^=2mv1i^+v2j^

On solving

v1=3v4 and v2=v4

Change in K.E.

12mv2+12mv222-12(2m)9v216+v216

=3mv24-5mv28=mv28

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
Two particles of equal mass $m$ go round a circle of radius $\mathrm{R}$ under the action of their mutual gravitational attraction. The speed of each particle is

Options:
A $\frac{1}{2} \sqrt{\frac{G m}{R}}$
B $\sqrt{\frac{4 G m}{R}}$
C $\sqrt{\frac{G m}{2 R}}$
D $\frac{1}{2 R} \sqrt{\frac{1}{G m}}$

The correct answer is:
$\frac{1}{2} \sqrt{\frac{G m}{R}}$

Solution:
From given condition
$\frac{\mathrm{Gmm}}{(2 \mathrm{R})^{2}}=\frac{\mathrm{mv}^{2}}{\mathrm{R}}, \mathrm{v}=\frac{\mathrm{Gm}}{4 \mathrm{R}}=\frac{1}{2} \sqrt{\frac{\mathrm{Gm}}{\mathrm{R}}}$

Topic: Physics, Gravitation, JEE Main
Two particles of equal mass m have respective initial velocities ui^ and ui^+j^2 . They collide completely inelastically. The energy lost in the process is:

Options:
A 13mu2
B 18mu2
C 34mu2
D 23mu2

The correct answer is:
18mu2

Solution:
Conserving momentum
mui^+mu2i^+u2j^=2mu1i^+u2j^
On solving
u1=3u4 and u2=u4
Change in K.E.
12mu2+12mu222-122m9u216+u216
=3mu24-5mu28=mu28

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2020 (09 Jan Shift 1)
Two particles of equal mass m move in a circle of radius r under the action of their mutual gravitational attraction. The speed of each particle will be :

Options:
A Gm2r
B 4Gmr
C Gmr
D Gm4r

The correct answer is:
Gm4r

Solution:
Let speed of both the particles be v.

Then, gravitational force between the particles will provide the required centripetal force.

Therefore,

Gmm2r2=mv2r

⇒v=Gm4r

Topic: Physics, Gravitation, JEE Main, JEE Main 2023 (29 Jan Shift 1)
Two particles of mass 1 kg and 3 kg move towards each other under their mutual force of attraction. No other force acts on them. When the relative velocity of approach of two particles is 2 m s-1, their centre of mass has a velocity of 0.5 m s-1. When the relative velocity of approach becomes 3 m s-1, the velocity of the centre of mass will be

Options:
A 0.75 m s-1

B 0.5 m s-1

C 10 m s-1

D 1.25 m s-1

The correct answer is:
0.5 m s-1

Solution:
Net external force on the system is zero. So, the velocity of the centre of mass will remain constant, i.e., 0.5 m s-1.

Topic: Physics, Center of Mass Momentum and Collision, NEET
Two particles of mass $m_{1}$ and $m_{2}$; approach each other due to their mutual gravitational attraction only. Then,

Options:
A accelerations of both the particles are equal
B acceleration of the particle of mass $m_{1}$ is proportional to $\overline{m_{1}}$
C acceleration of the particle of mass $m_{1}$ is proportional to $m_{2}$
D acceleration of the particle of mass $m_{1}$ is inversely proportional to $m_{1}$

The correct answer is:
acceleration of the particle of mass $m_{1}$ is proportional to $m_{2}$

Solution:
The gravitational force acting between the two masses $m_{1}$ and $m_{2}$ is given by
$$
F_{O}=\frac{G m_{1} m_{2}}{r^{2}}
$$
Force on mass $m_{1}$.
$$
F_{1}=\frac{G m_{1} m_{2}}{r^{2}}=m_{1} a_{1}
$$
where, $a_{1}=$ acceleration
$$
\Rightarrow \quad a_{1}=\frac{G m_{2}}{I^{2}} \Rightarrow a_{1} \propto m_{2}
$$
and similarly, $a_{2} \propto m_{1}$

Topic: Physics, Gravitation, JEE Main
Two particles of masses 2 kg and 3 kg are projected horizontally in opposite directions from the top of a tower of height 39.2 m with velocities 5 m s-1 and 10 m s-1 respectively. The horizontal range of the centre of mass of two particles is [g=9.8 m s-2]

Options:
A 82 m in the direction of 2 kg
B 82 m in the direction of 3 kg
C 8 m in the direction of 2 kg
D 8 m in the direction of 3 kg

The correct answer is:
82 m in the direction of 3 kg

Solution:
Range of C.M =Vcm2hg
Vcm=m1v1+m2v2m1+m2
V com = −2×5+3×10 2+3 =4m/s
Range =4 2×39.2 9.8 =8 2

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
Two particles of masses are in the ratio $1: 2$. The ratio of distance of the center of mass of the system of these two particles from the lighter particle to that from the heavier particle is

Options:
A 4:1
B 1:2
C 2:1
D 1:4

The correct answer is:
2:1

Solution:
Mass is inversely proportional To the distance
$\begin{aligned} & \text { So }, \frac{\mathrm{m}_1}{\mathrm{~m}_2}=\frac{\mathrm{d}_2}{\mathrm{~d}_1} \\ & =2: 1\end{aligned}$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
Two particles of masses m1 and m2 are initially at rest at an infinite distance apart. If they approach each other under their mutual interaction given by F=-Kr2. Their speed of approach at the instant when they are at a distance d apart is

Options:
A 2Kd1m1+1m2
B 2Kd1m1-1m2
C 2Kdm1m2m1+m2
D 2Kdm1m2m1-m2

The correct answer is:
2Kd1m1+1m2

Solution:
Let μ be the reduced mass of the system and v be the approach speed of the particles, then

12μv2=∫∞dF→.dr→

⇒12m1m2m1+m2v2=-∫∞dKr2dr

∴v=2Kd1m1+1m2

Topic: Physics, Work Power Energy, JEE Main
Two particles of masses m1 and m2 initially at rest start moving towards each other under their mutual force of attraction. The speed of the centre of mass at any time t, when they are at a distance r apart, is

Options:
A zero
B Gm1m2r2·1m1t
C Gm1m2r2·1m2t
D Gm1m2r2·1m1+m2t

The correct answer is:
zero

Solution:
As initially both the particles were at rest therefore velocity of centre of mass was zero and there is no external force on the system so speed of centre of mass remains constant i.e. it should be equal to zero.

Topic: Physics, Center of Mass Momentum and Collision, NEET
Two particles of masses m1, m2 move with initial velocities u1 and u2. On collision, one of the particles get excited to higher level, after absorbing energy ε. If final velocities of particles be v1 and v2 then we must have:

Options:
A m12u1+m22u2-Îµ=m12v1+m22v2
B 12m1u12+12m2u22=12m1v12+12m2v22-Îµ
C 12m1u12+12m2u22-Îµ=12m1v12+12m2v22
D 1 2 m 2 u 1 2 + 1 2 m 1 u 1 2 + ε = 1 2 m 2 v 1 2 + 1 2 m 1 v 1 2

The correct answer is:
12m1u12+12m2u22-ε=12m1v12+12m2v22

Solution:
From conservation of energy initial total energy
Ei=12m1u12+12m2u22
After collision total final energy
Ef=12m1v12+12m2v22+ε

∴12m1u12+12m2u22=12m1v12+12m2v22+ε
12m1u12+12m2u22-ε=12m1v12+12m2v22

Topic: Physics, Center of Mass Momentum and Collision, JEE Main