Three equal masses m are kept at vertices (A, B, C) of an equilateral triangle of side a in free space. At t = 0 , they are given an initial velocity VA = V0AC , VB = V0BA and VC = V0CB . Here, AC, CB and BA are unit vectors along the edges of the triangle. If the three masses interact gravitationally, then the magnitude of the net angular momentum of the system at the point of collision is:

Since the three masses interact purely through mutual internal gravitational forces, the net external torque acting on the system about any central reference point is zero: Let us compute the total angular momentum about the centroid of the equilateral triangle: From trigonometry, the perpendicular distance from the centroid to the velocity vector along any edge is: The initial angular momentum for one mass about the centroid is L1 = m V0 r. Since all three particles move cyclically in the same direction, their angular momenta reinforce cleanly: By conservation of angular momentum, this configuration value remains unchanged up to the point of collision. Mutual internal central forces can never alter the angular momentum of a system. Hence, the solution completely reduces to measuring the static configuration values at t=0 about the center of mass symmetry. Class 11 Physics: System of Particles and Rotational Motion Class 11 Physics: Gravitation

Three equal masses m are kept at vertices (A, B, C) of an equilateral triangle of side a in free space. At t = 0 , they are given an initial velocity VA = V0AC , VB = V0BA and VC = V0CB . Here, AC, CB

Since the three masses interact purely through mutual internal gravitational forces, the net external torque acting on the system about any central reference point is zero: Let us compute the total angular momentum about the centroid of the equilateral triangle: From trigonometry, the perpendicular distance from the centroid to the velocity vector along any edge is: The initial angular momentum for one mass about the centroid is L1 = m V0 r. Since all three particles move cyclically in the same

Three equal masses m are kept at vertices (A, B, C) of an equilateral triangle of side a in free space. At t = 0 , they are given an initial velocity VA = V0AC , VB = V0BA and VC = V0CB . Here, AC, CB and BA are unit vectors along the edges of the triangle. If the three masses interact gravitationally, then the magnitude of the net angular momentum of the system at the point of collision is:

Since the three masses interact purely through mutual internal gravitational forces, the net external torque acting on the system about any central reference point is zero: Let us compute the total angular momentum about the centroid of the equilateral triangle: From trigonometry, the perpendicular distance from the centroid to the velocity vector along any edge is: The initial angular momentum for one mass about the centroid is L1 = m V0 r. Since all three particles move cyclically in the same direction, their angular momenta reinforce cleanly: By conservation of angular momentum, this configuration value remains unchanged up to the point of collision. Mutual internal central forces can never alter the angular momentum of a system. Hence, the solution completely reduces to measuring the static configuration values at t=0 about the center of mass symmetry. Class 11 Physics: System of Particles and Rotational Motion Class 11 Physics: Gravitation

Three equal masses m are kept at vertices (A, B, C) of an equilateral triangle of side a in free space. At t = 0 , they are given an initial velocity VA = V0AC , VB = V0BA and VC = V0CB . Here, AC, CB

Since the three masses interact purely through mutual internal gravitational forces, the net external torque acting on the system about any central reference point is zero: Let us compute the total angular momentum about the centroid of the equilateral triangle: From trigonometry, the perpendicular distance from the centroid to the velocity vector along any edge is: The initial angular momentum for one mass about the centroid is L1 = m V0 r. Since all three particles move cyclically in the same

Three equal masses m are kept at vertices (A, B, C) of an equilateral triangle of side a in free space. At t = 0 , they…

Q60 jee_main_2024_29_january_evening Angular Momentum of a Particle

A body of mass

5text kg

moving with a uniform speed

3sqrt2text ms^-1

in X–Y plane along the line

y = x + 4

. The angular momentum of the particle about the origin will be ______

textkg m^2texts^-1

.

Numerical Answer. Answer: 60 to 60

Solution

### Related Formula The magnitude of the angular momentum

L

of a particle of mass

m

moving with velocity

v

is:

L = m v d

where: *

d

is the perpendicular distance from the axis of rotation (origin) to the line of motion of the particle. ### Core Logic Given parameters: * Mass,

m = 5text kg

* Velocity,

v = 3sqrt2text ms^-1

* Line of motion:

y = x + 4 implies x - y + 4 = 0

### Step 1: Calculate Perpendicular Distance The perpendicular distance

d

from the origin

(0,0)

to the line

Ax + By + C = 0

is:

d = frac|A(0) + B(0) + C|sqrtA^2 + B^2

For the line

x - y + 4 = 0

:

d = frac|4|sqrt1^2 + (-1)^2 = frac4sqrt2 = 2sqrt2text m

### Step 2: Calculate Angular Momentum Substitute the values into the angular momentum formula:

L = m v d

L = 5text kg times (3sqrt2text ms^-1) times (2sqrt2text m)

L = 5 times 3 times 4 = 60text kg m^2texts^-1

Thus, the angular momentum of the particle about the origin is

60text kg m^2texts^-1

. ### Pattern Recognition Instead of complicated vector cross products, find the perpendicular distance of the straight line from the origin using standard coordinate geometry.

L = mvd

is extremely fast and reliable. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: System of Particles and Rotational Motion

Q56 jee_main_2024_27_jan_morning Moment of Inertia

Four particles each of mass

1text kg

are placed at four corners of a square of side

2text m

. The moment of inertia of the system about an axis perpendicular to its plane and passing through one of its vertices is ______

textkgcdottextm^2

.

Numerical Answer. Answer: 16 to 16

Solution

### Related Formula

I = sum m_i r_i^2

### Core Logic Let the axis pass through vertex 1. Evaluate distances (

r

) for each corner particle: - Particle at vertex 1:

r_1 = 0

- Particle at adjacent vertex 2:

r_2 = a

- Particle at adjacent vertex 4:

r_4 = a

- Particle at diagonally opposite vertex 3:

r_3 = sqrt2a

### Step 1: Set up substitution formula

I = m(0)^2 + m(a)^2 + m(a)^2 + m(sqrt2a)^2

I = ma^2 + ma^2 + 2ma^2 = 4ma^2

### Step 2: Numeric Evaluation Substitute

m = 1text kg

and side length

a = 2text m

:

I = 4 times 1 times (2)^2 = 4 times 4 = 16text kgcdottextm^2

### Pattern Recognition For a standard planar configuration system, total orthogonal moment components map predictably via basic summation configurations matching

4ma^2

exactly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: System of Particles and Rotational Motion

Q54 jee_main_2024_29_jan_morning Rolling Motion

A cylinder is rolling down on an inclined plane of inclination

60^circ

. It's acceleration during rolling down will be

fracxsqrt3 mathrm~m / s^2

, where

x =

________ (use

g = 10 mathrm~m/s^2

).

Numerical Answer. Answer: 10 to 10

Solution

### Related Formula The linear acceleration (

a

) of a symmetric body performing pure rolling down an inclined plane of angle

theta

is given by:

a = fracg sin theta1 + fracI_textcmM R^2

### Core Logic For a solid cylinder, the moment of inertia about its central longitudinal axis is:

I_textcm = frac12 M R^2 implies fracI_textcmM R^2 = frac12

Given inclination angle,

theta = 60^circ

, and

g = 10 mathrm~m/s^2

.

Free body diagram of a rolling cylinder on an incline for Q54

### Step 1: Calculate Linear Acceleration Substituting the values into the acceleration template:

a = frac10 times sin 60^circ1 + frac12 = frac10 times fracsqrt32frac32

a = frac10 sqrt33 = frac10sqrt3 mathrm~m/s^2

### Step 2: Solve for x Comparing this evaluated value with the expression

\frac{x}{\sqrt{3}}

:

frac10sqrt3 = fracxsqrt3 implies x = 10

Therefore, the value of

x

is

10

. ### Pattern Recognition Pure rolling problems reduce down to tracking the shape factor fraction

\beta = 1 + \frac{I}{MR^2}

. For solid cylinders it is

1.5

, for solid spheres it is

1.4

, and for hoops it is

2.0$. This value acts as an effective inertial scaling factor for gravity. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: System of Particles and Rotational Motion

Q53 jee_main_2024_30_january_evening Loss in Kinetic Energy on Coupling

Two discs of moment of inertia

mathrmI_1 = 4 mathrm~kg mathrm~m^2

and

mathrmI_2 = 2 mathrm~kg mathrm~m^2

about their central axes & normal to their planes, rotating with angular speeds

10 mathrm~rad/s

&

4 mathrm~rad/s

respectively are brought into contact face to face with their axe of rotation coincident. The loss in kinetic energy of the system in the process is ________

mathrmJ

.

Numerical Answer. Answer: 24 to 24

Solution

### Related Formula

textC.O.A.M: I_1 omega_1 + I_2 omega_2 = (I_1 + I_2) omega_textfinal

Delta K.E. = frac12 fracI_1 I_2I_1 + I_2 (omega_1 - omega_2)^2

### Core Logic When two rotating discs are brought into contact, they exert friction on each other until they reach a common angular velocity. Angular momentum is conserved about the central axis. We can use the conservation of angular momentum to find the final angular velocity, or use the direct formula for loss in kinetic energy. ### Step 1: Calculate Final Angular Velocity (Alternative Method)

I_1 omega_1 + I_2 omega_2 = (I_1 + I_2)omega_0

4(10) + 2(4) = (4 + 2)omega_0

40 + 8 = 6omega_0 implies omega_0 = 8 mathrm~rad/s

### Step 2: Calculate Kinetic Energy Loss Initial Energy

mathrmE_1 = frac12 I_1 omega_1^2 + frac12 I_2 omega_2^2

mathrmE_1 = frac12(4)(100) + frac12(2)(16) = 200 + 16 = 216 mathrm~J

Final Energy

mathrmE_2 = frac12 (I_1 + I_2) omega_0^2

mathrmE_2 = frac12 (6) (8^2) = 3 times 64 = 192 mathrm~J

Loss

Delta E = mathrmE_1 - mathrmE_2 = 216 - 192 = 24 mathrm~J

### Pattern Recognition The loss formula

\Delta K = \frac{1}{2} \frac{I_1 I_2}{I_1 + I_2} (\omega_1 - \omega_2)^2

is incredibly fast:

\frac{1}{2} \times \frac{4 \times 2}{6} \times (10 - 4)^2 = \frac{1}{2} \times \frac{8}{6} \times 36 = \frac{4}{6} \times 36 = 24 \mathrm{~J}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: System of Particles and Rotational Motion

Q47 jee_main_2024_30_jan_morning Impulse and Momentum

A spherical body of mass

100 mathrm~g

is dropped from a height of

10 mathrm~m

from the ground. After hitting the ground, the body rebounds to a height of

5 mathrm~m

. The impulse of force imparted by the ground to the body is given by: (given

g = 9.8 mathrm~m / s^2

)

A. $4.32 mathrm~kg\,ms^-1$ $4.32 \mathrm{~kg\,ms^{-1}}$
B. $43.2 mathrm~kg\,ms^-1$ $43.2 \mathrm{~kg\,ms^{-1}}$
C. $23.9 mathrm~kg\,ms^-1$ $23.9 \mathrm{~kg\,ms^{-1}}$
D. $2.39 mathrm~kg\,ms^-1$ $2.39 \mathrm{~kg\,ms^{-1}}$

Solution

### Related Formula

v = sqrt2gh

vecI = Delta vecP = m(vecv_f - vecv_i)

### Core Logic Impulse delivered by the ground equals the total change in momentum of the body during the collision. We must compute the velocity immediately before impact and immediately after rebound, respecting their opposite vector directions. ### Step 1: Calculate Velocities Velocity just before hitting the ground (

v_i

):

v_i = sqrt2 times 9.8 times 10 = sqrt196 = -14 mathrm~m/s quad (textdownwards)

Velocity just after rebounding (

v_f

):

v_f = sqrt2 times 9.8 times 5 = sqrt98 = +7sqrt2 mathrm~m/s quad (textupwards)

### Step 2: Calculate Impulse Mass

M = 100 mathrm~g = 0.1 mathrm~kg

vecI = Delta P = m(v_f - v_i)

vecI = 0.1 times [7sqrt2 - (-14)]

vecI = 0.1(14 + 7sqrt2)

Since

sqrt2 approx 1.414

:

vecI = 0.1(14 + 7(1.414)) = 0.1(14 + 9.898) = 0.1(23.898)

vecI approx 2.39 mathrm~kg\,ms^-1

### Pattern Recognition When dealing with rebounds,

Delta v

is the sum of magnitudes

|v_1| + |v_2|

because the direction reverses. A common trap is to subtract the magnitudes instead of adding. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: System of Particles and Rotational Motion Class 11 Physics: Kinematics

Solution & Explanation

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