datavorous/entrance-exam-dataset huggingface dataset:
A particle of mass m is moving along the x-axis with initial velocity ui^. It collides elastically with a particle of mass 10m at rest and then moves with half its initial kinetic energy (see figure). If sin θ1=nsin θ2 then value of n is ________.

Options:

The correct answer is:
10

Solution:
From momentum conservation in perpendicular direction of initial motion.
mv1sinθ1=10mv2sinθ2
Given that 12mu212=12mv12 …….(2)
From equation (1) & (2)
sinθ1=10sinθ2
n=10

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2020 (02 Sep Shift 2)
A particle of mass M is moving in a circle of fixed radius R in such a way that its centripetal acceleration at time t is given by n2Rt2, where n is a constant. The power delivered to the particle by the force acting on it, is :

Options:
A 12 Mn2R2t2
B Mn2R2t
C MnR2t2
D MnR2t

The correct answer is:
Mn2R2t

Solution:
ac=v2R=n2Rt2
v=nRT
at=dvdt=nR
Power =Matv
=M nR nRt
=Mn2R2t

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2016 (10 Apr Online)
A particle of mass m is moving in a circular path of constant radius r such that its centripetal acceleration ac is varying with time t as ac=k2rt2 , where k is a constant. The power delivered to the particle by the force acting on it is -

Options:
A Zero
B 2mk2r2t
C mk2r2t
D 2mk2rt

The correct answer is:
mk2r2t

Solution:
ac=k2rt2

Or v2r=k2rt2 Or v=krt
Therefore, tangential acceleration, at=dvdt=kr

Therefore, tangential force, Ft=mat=mkr
Only tangential force does work.
Power=Ftv=mkrkrt ⇒Power=mk2r2t

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2022 (28 Jun Shift 1)
A particle of mass m is moving in a circular path of constant radius r such that its centripetal acceleration ac is varying with time t as ac=k2 rt2 , where k is a constant. The power delivered to the particle by the force acting on it is

Options:
A 2ÏÂ mk2Â r2
B mk2 r2t
C mK4 r2t53
D Zero

The correct answer is:
mk2 r2t

Solution:
ac=k2 rt2
or v2r= k2 rt2
or v=krt
Therefore, tangential acceleration, at =dvdt= kr
or Tangential force,
Ft=mat=mkr
Only tangential force does work.
Power=Ft v=mkrkrt
or Power=mk2 r2 t

Topic: Physics, Work Power Energy, JEE Main
A particle of mass $\mathrm{m}$ is moving in a straight line with momentum p. Starting at time $t=0,$ a force $F=k$ t acts in the same direction on the moving particle during time interval T so that its momentum changes from $\mathrm{p}$ to $3 \mathrm{p}$. Here $k$ is a constant. The value of $\mathrm{T}$ is

Options:
A $2 \sqrt{\frac{k}{p}}$
B $2 \sqrt{\frac{\mathrm{p}}{k}}$
C $\sqrt{\frac{2 k}{p}}$
D $\sqrt{\frac{2 p}{\mathrm{k}}}$

The correct answer is:
$2 \sqrt{\frac{\mathrm{p}}{k}}$

Solution:
From Newton's second law

$\frac{\mathrm{dp}}{\mathrm{dt}}=\mathrm{F}=\mathrm{kt}$

Integrating both sides we get, $\int_{\mathrm{p}}^{3 \mathrm{p}} \mathrm{dp}=\int_{0}^{\mathrm{T}} \mathrm{ktdt} \Rightarrow[\mathrm{p}]_{\mathrm{p}}^{3 \mathrm{p}}=\mathrm{k}\left[\frac{\mathrm{t}^{2}}{2}\right]_{0}^{\mathrm{T}}$

$\Rightarrow 2 \mathrm{p}=\frac{\mathrm{kT}^{2}}{2} \Rightarrow \mathrm{T}=2 \sqrt{\frac{\mathrm{p}}{\mathrm{k}}}$

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2019 (11 Jan Shift 2)
A particle of mass m is moving in time t on a trajectory given by,

r→=10αt2i^+5βt-5j^

where α and β are dimensional constants.

The angular momentum of the particle becomes the same as it was for t=0 at time t=_____ seconds.

Options:

The correct answer is:
10

Solution:
r→=10αt2i^+5βt-5j^

v→=20α+i^+5βj^

L→=mr→×v→

=m10αt2i^+5βt-5j^×20αi^+5βj^

L→=m50αβt2k^-100αβt2-5tk^

At t=0, L→=0→

50αβt2-100αβt2-5t=0

t-2t-5=0

t=10 sec

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2021 (25 Jul Shift 1)
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 the collision, the first mass is stopped completely while the second one splits into two particles each of mass m, which move at an angle 45o with respect to the original direction. The speed of each of the moving particle will be

Options:
A 2v
B v2
C 22v
D v22

The correct answer is:
22v

Solution:
As 2m splits into equal parts, v1=v2 and θ=450
From conservation of linear momentum,
m2v+2mv=2mv1cos⁡45o
⇒v1=v2=22v

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2019 (09 Apr Shift 2)
A particle of mass m is performing the linear simple harmonic motion. Its equilibrium is at x=0, force constant is K and amplitude of SHM is A. The maximum power supplied by the restoring force to the particle during SHM will be:

Options:
A K32A2m
B 2K32A2m
C K32A23m
D K32A22m

The correct answer is:
K32A22m

Solution:
Power supplied by the restoring force F=Kx is:
P=FV
P=KxwA2-x2
P=KKmA2x2-x4
P will be max when dpdx=0
⇒A22x-4x3=0
x=A2
Pmax=P|atx=A2=KKmA2A22-A24
Pmax=K32A22m

Topic: Physics, Oscillations, JEE Main
A particle of mass m is projected at an angle of 60o with a velocity of 20 m/s relative to the ground from a plank of same mass m which is placed on a smooth surface initially plank was at rest. The minimum length of the plank for which the ball will fall on the plank itself is

Options:
A 4 0 3 m
B 2 0 3 m
C 1 0 3 m
D 6 0 3 m

The correct answer is:
4 0 3 m

Solution:
Since the external force on the system is zero, location of COM will not shift.
â m1x1=m2x2
m x=mL-x
â x=L2
so plank will shift by L2 towards left.
Now range R should be <_ L2
â L=2R
âL=2 u2sin30g=40 3

Topic: Physics, Motion In Two Dimensions, JEE Main
A particle of mass m is projected at t = 0 from a point O on the ground with a speed v0 at an angle of 45Â° to the horizontal. The magnitude of the angular momentum of the particle about O at the time t = v0/g is close to [Take 2=1.4]

Options:
A 0.25mv03/g
B 0.35mv03/g
C 0.50mv03/g
D 0.60mv03/g

The correct answer is:
0.35mv03/g

Solution:
Using 1st-equation of motion v=u+at along horizontal direction,
v x = v 0 / 2 + 0 Ã t = 0.7v as a = 0
and along vertical direction
v y = v 0 / 2 - g v 0 / g = - 0.3v 0
So, the velocity of particle at time t = (v0/g) in vector form will be
v â = i â 0.7v 0 - j â 0.3v 0 ....(i)
Now from 2nd equation of motion s = ut + 1 2 at 2 along horizontal direction,
x = v 0 2 Ã v 0 g = 0.7 v 0 2 g
and along vertical direction,
y = v 0 2 Ã v 0 g - 1 2 g v 0 g 2 = 0.2 v 0 2 g
So the position vector r â at the t ( = v0/g) will be
r â = i â 0.7 v 0 2 g + j â 0.2 v 0 2 g ....(ii)
Now as by definitionLâ=râÃpâ=mrâÃvâ
So L â = m | i â j â k â 0.7v 0 2 g 0.2v 0 2 g 0 0.7v 0 - 0.3v 0 0 |
= m k â - 0.21 v 0 3 g - 0.14 v 0 3 g
i.e., L â = - 0.35mv 0 3 / g k â
i.e., angular momentum has magnitude 0.35 mv 0 3 / g and is directed along negative z-axis.

Topic: Physics, Rotational Motion, JEE Main
A particle of mass m is projected with a velocity v0 at an angle Î± with the horizontal. The coefficient of restitution for any of its impact with the smooth ground is e. The time after which the particle will stop bouncing off the ground is

Options:
A T=2v0sinα(1+e)g(1-e)
B T=2v0sinα(1-e)g(1+e)
C T=2v0sinαg(1+e)
D T=2v0sinαg(1-e)

The correct answer is:
T=2v0sinαg(1-e)

Solution:
After each collision, the particle loses some part of its vertical velocity while its horizontal velocity remains unchanged.

T=2v0sinαg+2ev0sinαg+2e2v0sinαg+..........

⇒T=2v0sinαg1+e+e2+.........

T=2v0sinαg(1-e)

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A particle of mass $m$ is projected with velocity $v$ making an angle of $45^{\circ}$ with the horizontal. When the particle lands on the level ground the magnitude of the change in its momentum will be

Options:
A zero
B $2 m v$
C $m v / \sqrt{2}$
D $m v \sqrt{2}$

The correct answer is:
$m v \sqrt{2}$

Solution:
Momentum change $=2 \mathrm{mv} \sin \theta$

Topic: Physics, Center of Mass Momentum and Collision, NEET, NEET 2008 (Mains)
A particle of mass m is released from height h on a smooth curved surface which ends into a vertical loop of radius r, as shown. If h=2r, then

Options:
A the particle reaches the top of the loop with zero velocity
B the particle reaches the top of the loop with a non-zero velocity
C the particle breaks off at a height h=r from base
D the particle breaks off at a height r<h<2r

The correct answer is:
the particle breaks off at a height r<h<2r

Solution:
To reach the top of loop, particle must have a minimum speed of v=5gr at the bottom of the loop.
but v2=2gh=4gr , so the block won't reach the highest point of the track.

Topic: Physics, Laws of Motion, JEE Main
A particle of mass $m$ is rotating in a plane in circularpath of radius $r$. Its angular momentum is $L$. The centripetal force acting on the particle is

Options:
A $\frac{L^{2}}{m r}$
B $\frac{L^{2} m}{r}$
C $\frac{L^{2}}{m^{2} r^{2}}$
D $\frac{L^{2}}{m r^{3}}$

The correct answer is:
$\frac{L^{2}}{m r^{3}}$

Solution:
Centripetal force, $F=\frac{m v^{2}}{r}=\frac{m}{r} \frac{L^{2}}{m^{2} r^{2}}=\frac{L^{2}}{m r^{3}}$
$\left[\right.$ As $\left.L=m v r \therefore v=\frac{L}{m r}\right]$

Topic: Physics, Rotational Motion, JEE Main
A particle of mass $M$ is situated at the centre of spherical shell of mass and radius $a$. The magnitude of the gravitational potential at a point situated at $\frac{a}{2}$ distance from the centre, will be

Options:
A $\frac{2 G M}{a}$
B $\frac{3 G M}{a}$
C $\frac{4 G M}{a}$
D $\frac{G M}{a}$

The correct answer is:
$\frac{3 G M}{a}$

Solution:
$V_P=V_{\text {sphere }}+V_{\text {partical }}$
$$
\begin{aligned}
& V_P=\frac{G M}{a}+\frac{G M}{a / 2} \\
& V=\frac{G M}{a}+\frac{2 G M}{a} \\
& V_P=\frac{3 G M}{a}
\end{aligned}
$$

Topic: Physics, Gravitation, NEET, NEET 2011 (Mains)
A particle of mass m is suspended from a ceiling through a string of length L. The particle moves in a horizontal circle of radius r such that r=L2. The speed of particle will be :

Options:
A rg
B 2rg
C rg2
D 2rg

The correct answer is:
rg

Solution:
sinθ=12

θ=45°

tanθ=mv2rmg=v2rg=1

v2=rg

v=rg

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2021 (26 Aug Shift 2)
A particle of mass m is suspended from a ceiling through a string of length L. The particle moves in a horizontal circle of radius r. The speed (v) of the particle and the tension (T) in the string is

Options:
A v=2rgL2-r21/2 and T=3mgLL2-r21/2
B v=rgL2-r21/4 and T=mgLL3-r31/3
C v=3rgL2-r21/2and T=mgLL2-r21/2
D v=rgL2-r21/4 and T=mgLL2-r21/2

The correct answer is:
v=rgL2-r21/4 and T=mgLL2-r21/2

Solution:
The situation is shown in the diagram. The angle θ made by the string with the vertical is given by
sinθ=rL ...(i)

The forces on the particle are
(1) the tension T along the string and
(2) the weight mg vertically downward.
The particle is moving in a circle with constant speed v. Thus, the radial acceleration towards the centre has magnitudev2r. Resolving the forces along the radial direction and applying Newton's second law,
T sin​θ = mv2r ...(ii)
As there is no acceleration in the vertical direction, net force in the vertical direction is zero,
T cos​θ = mg ...(iii)
Dividing (ii) by (iii),
tan θ = v 2 rg
or v = rg tan θ
And from (iii),
T = mg cos θ
Using (i),
v = r g L 2 - r 2 1/4 and T = mgL L 2 - r 2 1/2 â

Topic: Physics, Laws of Motion, JEE Main
A particle of mass $m$ is thrown upwards from the surface of the earth, with a velocity $u$. The mass and the radius of the earth are, respectively, $M$ and $R . G$ is gravitational constant and $g$ is acceleration due to gravity on the surface of the earth. The minimum value of $u$ so that the particle does not return back to earth, is

Options:
A $\sqrt{\frac{2 G M}{R}}$
B $\sqrt{\frac{2 G M}{R^2}}$
C $\sqrt{2 g R^2}$
D $\sqrt{\frac{2 G M}{R^2}}$

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

Solution:
Escape velocity
$$
\begin{aligned}
& y_e=\sqrt{2 g R} \\
& y_e=\sqrt{\frac{2 G M}{R^2} R} \quad\left(\because G M=g R^2\right) \\
& y_e=\sqrt{\frac{2 G M}{R}}
\end{aligned}
$$

Topic: Physics, Gravitation, NEET, NEET 2011 (Mains)
A particle of mass $m$ moves along line $P C$ with velocity $v$ as shown. What is the angular momentum of the particle about $\mathrm{P} ?$

Options:
A
mvL

B
mvl

C
mvr

D
zero

The correct answer is:
zero

Solution:
Zero, line of motion through the point P.

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2002
A particle of mass m moves along the internal smooth surface of a vertical cylinder of the radius R. Find the force with which the particle acts on the cylinder wall if at the initial moment of time its velocity equals v0. and forms an angle Î± with the horizontal.

Options:
A mV02Rsin2α
B mV022Rcos2α
C 2mV02Rcos2α
D mV02Rcos2α

The correct answer is:
mV02Rcos2α

Solution:
The direction of the velocity of the particle may be shown as in the diagram. This velocity V0 is resolved in two components. Here V0cosα component is tangential horizontal and V0sinα is along the surface but vertically downwards. Here, if the particle exerts a force N on the surface of the cylinder, the cylinder will also apply equal and opposite force on the particle

and this will provide a centripetal force to the particle.
So, N=mV0cosα2/R
=mV02cos2α/R
or, N=mV02Rcos2α

Topic: Physics, Laws of Motion, JEE Main
A particle of mass m moves along the quarter section of the circular path whose centre is at the origin. The radius of the circular path is a. A force Fâ=yi^-xj^ N acts on the particle, where x,y denote the coordinates of the position of the particle. Calculate the work done by this force in taking the particle from point A(a, 0) to point B(0, a) along the circular path.

Options:
A πa2 J
B πa22 J
C -Ïa22 J
D Zero

The correct answer is:
-Ïa22 J

Solution:
Work done by force F;
w=∫F→.dr=∫yi^-xj^.(dxi^+dyj^)
=∫ydx-xdy
∵x2+y2=a2
∴xdx+ydy=0
⇒W=∫y-ydyx-xdy=-∫x2+y2xdy
=-∫0aa2a2-y2dy=-πa22

Topic: Physics, Work Power Energy, JEE Main
A particle of mass $\mathrm{m}$ moves around the origin in a potential $\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{r}^{2}$, where $\mathrm{r}$ is the distance from the origin. Applying the Bohr model in this case, the radius of the particle in its $\mathrm{n}^{\text {th }}$ orbit in terms of $\mathrm{a}=\sqrt{\mathrm{h} /(2 \pi \mathrm{m} \omega)}$ is

Options:
A $a \sqrt{n}$
B an
C $a n^{2}$
D $\operatorname{an} \sqrt{\mathrm{n}}$

The correct answer is:
$a \sqrt{n}$

Solution:
$\mathrm{mvr}=\mathrm{n} \frac{\mathrm{h}}{2 \pi}$
$\mathrm{v}=\mathrm{r} \omega$
$\mathrm{mr}^{2} \omega=\mathrm{n} \frac{\mathrm{h}}{2 \pi}$
$\mathrm{r}=\sqrt{\frac{\mathrm{nh}^{2}}{2 \pi \mathrm{m} \omega}}$
$\mathrm{r}=\sqrt{\mathrm{n}} \sqrt{\frac{\mathrm{h}^{2}}{2 \pi \mathrm{m} \omega}}$
$\mathrm{r}=\sqrt{\mathrm{n}} \cdot \mathrm{a}$

Topic: Physics, Atomic Physics, JEE Main
A particle of mass m moves in a circular orbit in a central potential field Ur=12kr2. If Bohr's quantization conditions are applied, radii of possible orbitals and energy levels vary with quantum number n as:

Options:
A rn∝n2, En∝1n2
B rn∝n, En∝n
C rn∝n, En∝n
D rn∝n, En∝1n

The correct answer is:
rn∝n, En∝n

Solution:
Fr=-dUdr=-kr
For circular motion
|Fr|=kr=mv2r⇒kr2=mv2 ........(i)
Bohr's quantization ⇒mvr=nh2π ....(ii)
From (i) and (ii)
m2v2m=kr2
⇒1mnh2πr2=kr2⇒n2h24π2mk=r4
⇒r=h24π2mk1/4 n1/2
r∝n
From equation (i) U∝n
KE=12mv2; PE=12kr2
E=K+U=12mv2+12kr2=kr2∝n

Topic: Physics, Atomic Physics, JEE Main
A particle of mass m moves in a circular orbit in a central potential field Ur=12kr2. If Bohr's quantization conditions are applied, radii of possible orbitals and energy levels vary with quantum number n as:

Options:
A rn∝n2, En∝1n2
B rn∝n, En∝n
C rn∝n, En∝n
D rn∝n, En∝1n

The correct answer is:
rn∝n, En∝n

Solution:
Fr=-dUdr=-kr
For circular motion
|Fr|=kr=mv2r⇒kr2=mv2 ........(i)
Bohr's quantization ⇒mvr=nh2π ....(ii)
From (i) and (ii)
m2v2m=kr2
⇒1mnh2πr2=kr2⇒n2h24π2mk=r4
⇒r=h24π2mk1/4 n1/2
r∝n
From equation (i) U∝n
KE=12mv2; PE=12kr2
E=K+U=12mv2+12kr2=kr2∝n

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2019 (12 Jan Shift 1)
A particle of mass m moves in a circular orbit in a central potential field U(r)=U0r4. If Bohr's quantization conditions are applied, radii of possible orbitals rn vary with n1α, where α is _______ .

Options:

The correct answer is:
3

Solution:
F=-dUdr=-4U0r3=mv2r

mv2=4U0r4

V∝r2

r3∝n

⇒r∝n1/3=3

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2021 (17 Mar Shift 2)
A particle of mass \( m \) moves in a circular orbit in a central potential field \( U(r)=\frac{1}{2} k r^{2} . \) If Bohr's quantization conditions are applied, radii of possible orbitals and energy levels vary with quantum number \( n \) as:

Options:
A \( r_{n} \propto n^{2}, E_{n} \propto \frac{1}{n^{2}} \)
B \( r_{n} \propto \sqrt{n}, E_{n} \propto n \)
C \( r_{n} \propto n, E_{n} \propto n \)
D \( r_{n} \propto \sqrt{n}, E_{n} \propto \frac{1}{n} \)

The correct answer is:
\( r_{n} \propto \sqrt{n}, E_{n} \propto n \)

Solution:
Fr=-dUdr=-kr
For circular motion
|Fr|=kr=mv2r⇒kr2=mv2 ........(i)
Bohr's quantization ⇒mvr=nh2π ....(ii)
From (i) and (ii)
m2v2m=kr2
⇒1mnh2πr2=kr2⇒n2h24π2mk=r4
⇒r=h24π2mk1/4 n1/2
r∝n
From equation (i) U∝n
KE=12mv2; PE=12kr2
E=K+U=12mv2+12kr2=kr2∝n

Topic: Physics, Atomic Physics, JEE Main
A particle of mass m moves in a circular orbit under the central potential field, Ur=-Cr, where C is a positive constant. The correct radius - velocity graph of the particle's motion is :

Options:
A
B
C
D

The correct answer is:

Solution:
U=-Cr

F=-dUdr=-Cr2

F=mv2r

Cr2=mv2r

v2∝1r

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2021 (18 Mar Shift 2)
A particle of mass m moves in the XY plane with a velocity v along the straight line AB. If the angular momentum of the particle with respect to origin O is LA when it is at A and LB when it is at B, then

Options:
A LA>LB
B LA=LB
C The relationship between LA and LB depends upon the slope of the line AB
D LA<LB

The correct answer is:
LA=LB

Solution:
From the definition of angular momentum,
L=r×p=rmvsin⁡∅(-k^)
Therefore, the magnitude of L is
L=mvrsin⁡∅=mvd
where d=rsin⁡∅ is the distance of closest approach of the particle to the origin. As d is same for both the particles, hence LA=LB.

Topic: Physics, Rotational Motion, JEE Main
A particle of mass $m$ moves in the $X Y$ plane with a velocity $v$ along the straight line $\mathrm{AB}$. If the angular momentum of the particle with respect to origin $\mathrm{O}$ is $L_A$ when it is at A and $L_B$ when it is at B, then:

Options:
A $L_A=L_B$
B the relationship between $L_A$ and $L_B$ depends upon the slope of the line $\mathrm{AB}$
C $L_A < L_B$
D $L_A>L_{B^*}$

The correct answer is:
$L_A=L_B$

Solution:
Moment of Momentum is angular momentum. OP is the same whether the mass is at $\mathrm{A}$ or $\mathrm{B}$

Topic: Physics, Rotational Motion, JEE Main
A particle of mass $m$ moves on a straight line with its velocity increasing with distance according to the equation $v=\alpha \sqrt{x}$, where $\alpha$ is a constant. The total work done by all the forces applied on the particle during its displacement from $x=0$ to $x=\mathrm{d}$, will be :

Options:
A $\frac{\mathrm{m}}{2 \alpha^2 \mathrm{~d}}$
B $\frac{\mathrm{md}}{2 \alpha^2}$
C $2 \mathrm{~m} \alpha^2 \mathrm{~d}$
D $\frac{m \alpha^2 \mathrm{~d}}{2}$

The correct answer is:
$\frac{m \alpha^2 \mathrm{~d}}{2}$

Solution:
$\begin{aligned} & \mathrm{v}=\alpha \sqrt{\mathrm{x}} \\ & \text { at } \mathrm{x}=0: \mathrm{v}=0 \\ & \& \text { at } \mathrm{x}=\mathrm{d} ; \mathrm{v}=\alpha \sqrt{\mathrm{d}} \\ & \text { W.D }=\mathrm{K}_{\mathrm{f}}-\mathrm{K}_{\mathrm{i}} \\ & \text { W.D }=\frac{1}{2} \mathrm{~m}(\alpha \sqrt{\mathrm{d}})^2-\frac{1}{2} \mathrm{~m}(0)^2 \\ & \Rightarrow \text { W.D }=\frac{\mathrm{m} \alpha^2 \mathrm{~d}}{2}\end{aligned}$

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2024 (09 Apr Shift 1)
A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving in the y direction with speed v. If the collision is perfectly inelastic, the percentage loss in the energy during the collision is close to:

Options:
A 62%
B 44%
C 50%
D 56%

The correct answer is:
56%

Solution:
The initial momentum of system is Piâ=m2Vi^+2mV j^
According to question as
On perfectly inelastic collision the particles stick to each other.
Pfâ=3m Vfâ
By conservation of linear momentum
Pfâ=Piâ â3m Vfâ=m2V i^+2mVj^
âVfâ=2V3i^+j^â | V f | =223V
â´ loss in KE. of system =Kinitial-Kfinal
=12m2V2+122mV2-123m 2 2 Vâ¡ 3 2
=2mV2+mV2-43mV2=3mV2-4mV23=53mV2
% Loss in KE =100ÃâKKi=53mv23mV2=59Ã100=56%

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2015 (04 Apr)
A particle of mass m moving in the x-direction with speed 2v is hit by another particle of mass 2m moving in the y-direction with speed v. If the collision is perfectly inelastic, what is the percentage loss in the energy during the collision (to the closest integer value)?

Options:

The correct answer is:
56

Solution:
The initial momentum of system is Piâ=m2Vi^+2mV j^
According to question as
On perfectly inelastic collision the particles stick to each other.
Pfâ=3m Vfâ
By conservation of linear momentum
Pfâ=Piâ â3m Vfâ=m2V i^+2mVj^
âVfâ=2V3i^+j^â | V f | =223V
â´ loss in KE. of system =Kinitial-Kfinal
=12m2V2+122mV2-123m 2 2 Vâ¡ 3 2
=2mV2+mV2-43mV2=3mV2-4mV23=53mV2
% Loss in KE =100ÃâKKi=53mv23mV2=59Ã100=56%

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A particle of mass m moving with a speed u strikes a smooth horizontal surface at an angle 45Â°. The particle rebounds at an angle Ï with speed v, if coefficient of restitution is 13, then the angle Ï is

Options:
A 30o
B 45o
C 60o
D 37o

The correct answer is:
30o

Solution:
In y direction
vsin⁡ϕ=eusin⁡45o ….i
In x direction
vcos⁡ϕ=ucos⁡45o ….ii
Equation i÷ii
tan⁡ϕ=13; ϕ=30o

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A particle of mass 'm', moving with a velocity 'v' makes an elastic collision in one dimension with a stationary particle of mass 'm'. During the collision, they remain in contact with each other for an extremely small time 'T': Their force of contact, with time is shown in the figure. Then F0=

Options:
A 2mvT
B 4mv3T
C mvT
D 3mv4T

The correct answer is:
4mv3T

Solution:
In an elastic head-on collision between two particles, the velocity of the particle gets exchanged, therefore net change of momentum of a particle is, ∆p=m∆v=mv-0=mv

From newton's second law, the rate of change of momentum is force.

F=dpdt⇒Fdt=dp⇒FT=∆p

Here, F is force of contact during the collision for a small-time T that causes the ∆p change in momentum of particle.

Thus, the area under the graph of F-t curve gives the change in momentum.

Area=12×T4×F0+3T4-T4×F0+12×T-3T4×F0=34TF0

Therefore,

Area=∆p⇒mv=34TF0⇒F0=4mv3T

Topic: Physics, Center of Mass Momentum and Collision, AP EAMCET, AP EAMCET 2021 (19 Aug Shift 1)
A particle of mass $m$ moving with horizontal speed $6 \mathrm{~m} / \mathrm{sec}$ as shown in figure. If $m\lt \lt M$ then for one dimensional elastic collision, the speed of lighter particle after collision will be

Options:
A $2 \mathrm{~m} / \mathrm{sec}$ in original direction
B $2 \mathrm{~m} / \mathrm{sec}$ opposite to the original direction
C $4 \mathrm{~m} / \mathrm{sec}$ opposite to the original direction
D $4 \mathrm{~m} / \mathrm{sec}$ in original direction

The correct answer is:
$2 \mathrm{~m} / \mathrm{sec}$ in original direction

Solution:
$\begin{aligned}
& v_1=\left(\frac{m_1-m_2}{m_1+m_2}\right) u_1+\frac{2 m_2 u_2}{m_1+m_2} \\
& \text { Substituting } m_2=0, v_1=-u_1+2 u_2 \\
& \Rightarrow v_1=-6+2(4)=2 m / s
\end{aligned}$
i.e the lighter particle will move in original direction with the speed of $2 \mathrm{~m} / \mathrm{s}$.

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A particle of mass m moving with velocity v collides with a stationary particle of mass 2m. After collision, they stick together and continue to move together with velocity

Options:
A v3
B v4
C v
D v2

The correct answer is:
v3

Solution:
The law of conservation of momentum is given as

m1u+m2u'=m1+m2v'

The given data is

u=vu'=0 m s-1m1=mm2=2m

Using the conservation of momentum equation,

mv+2m(0)=(m+2m)v'⇒v'=mv3m=v3

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2023 (10 Apr Shift 1)