A bead P sliding on a frictionless semi-circular string (ACB) and it is at point S at t = 0 and at this instant the horizontal component of its velocity is v. Another bead Q of the same mass as P is ejected from point A at t = 0 along the horizontal string AB, with the speed v, friction between the beads and the respective strings may be neglected in both cases. Let tp and tQ be the respective times taken by beads P and Q to reach the point B, then the relation between tp and tQ is

For bead Q on the straight horizontal wire, its horizontal velocity remains constant at v. Total distance is XQ. For bead P on the semi-circular wire, as it moves down the curve, gravity increases its total speed, thus increasing its horizontal velocity component significantly compared to its initial value. Over the path, horizontal displacement of Q is greater than the direct horizontal projection of P's remaining path. Horizontal displacement required to reach B from their respective starting states: XQ > XP. Time taken by bead Q: Time taken by bead P: Since XQ > XP and vhorizontal, P > v for a major part of the journey, tP < tQ. A particle on a frictionless downward curve gains kinetic energy, resulting in a higher average horizontal velocity than a particle moving strictly horizontally at the initial horizontal velocity. It covers less or equal horizontal distance but at a higher average speed. Class 11 Physics: Work, Energy and Power Class 11 Physics: Kinematics

A bead P sliding on a frictionless semi-circular string (ACB) and it is at point S at t = 0 and at this instant the horizontal component of its velocity is v. Another bead Q of the same mass as P is e

For bead Q on the straight horizontal wire, its horizontal velocity remains constant at v. Total distance is XQ. For bead P on the semi-circular wire, as it moves down the curve, gravity increases its total speed, thus increasing its horizontal velocity component significantly compared to its initial value. Over the path, horizontal displacement of Q is greater than the direct horizontal projection of P's remaining path. Horizontal displacement required to reach B from their respective starting

A bead P sliding on a frictionless semi-circular string (ACB) and it is at point S at t = 0 and at this instant the horizontal component of its velocity is v. Another bead Q of the same mass as P is ejected from point A at t = 0 along the horizontal string AB, with the speed v, friction between the beads and the respective strings may be neglected in both cases. Let tp and tQ be the respective times taken by beads P and Q to reach the point B, then the relation between tp and tQ is

For bead Q on the straight horizontal wire, its horizontal velocity remains constant at v. Total distance is XQ. For bead P on the semi-circular wire, as it moves down the curve, gravity increases its total speed, thus increasing its horizontal velocity component significantly compared to its initial value. Over the path, horizontal displacement of Q is greater than the direct horizontal projection of P's remaining path. Horizontal displacement required to reach B from their respective starting states: XQ > XP. Time taken by bead Q: Time taken by bead P: Since XQ > XP and vhorizontal, P > v for a major part of the journey, tP < tQ. A particle on a frictionless downward curve gains kinetic energy, resulting in a higher average horizontal velocity than a particle moving strictly horizontally at the initial horizontal velocity. It covers less or equal horizontal distance but at a higher average speed. Class 11 Physics: Work, Energy and Power Class 11 Physics: Kinematics

A bead P sliding on a frictionless semi-circular string (ACB) and it is at point S at t = 0 and at this instant the horizontal component of its velocity is v. Another bead Q of the same mass as P is e

For bead Q on the straight horizontal wire, its horizontal velocity remains constant at v. Total distance is XQ. For bead P on the semi-circular wire, as it moves down the curve, gravity increases its total speed, thus increasing its horizontal velocity component significantly compared to its initial value. Over the path, horizontal displacement of Q is greater than the direct horizontal projection of P's remaining path. Horizontal displacement required to reach B from their respective starting

A bead P sliding on a frictionless semi-circular string (ACB) and it is at point S at t = 0 and at this instant the…

Year	2026	2025	2024	Total
Questions	8	15	8	31

Q19 jee_main_2025_03_april_evening Work-Energy Theorem with Variable Force

A block of mass 1 kg, moving along x with speed

vᵢ=10~m/s

enters a rough region ranging from

x=0.1~m

to

x=1.9~m

. The retarding force acting on the block in this range is

Fᵣ=-kx~N

with

k=10~N/m

. Then the final speed of the block as it crosses rough region is :

A. $10~m/s$ $10\mathrm{~m/s}$
B. $4~m/s$ $4\mathrm{~m/s}$
C. $6~m/s$ $6\mathrm{~m/s}$
D. $8~m/s$ $8\mathrm{~m/s}$

Solution

Related Formula

By the Work-Energy Theorem, the work done by the retarding force equals the change in kinetic energy:

W = Δ K = Kf - Kᵢ

W = \intxᵢxf Fᵣ(x) dx = (1)/(2) m vf² - (1)/(2) m vᵢ²

Core Logic

Given parameters:

Mass $m = 1~kg$ $m = 1\mathrm{~kg}$
Initial velocity $vᵢ = 10~m/s$ $v_i = 10\mathrm{~m/s}$
Region bounds: $xᵢ = 0.1~m$ $x_i = 0.1\mathrm{~m}$, $xf = 1.9~m$ $x_f = 1.9\mathrm{~m}$
Retarding force $Fᵣ = -kx = -10x~N$ $F_r = -kx = -10x\mathrm{~N}$

Step 1: Calculate Work Done by the Retarding Force ( $W$ $W$)

W = \int0.11.9 (-10x) dx = -10 [ (x²)/(2) ]0.11.9 = -5 [ (1.9)² - (0.1)² ]

Using the algebraic identity $a² - b² = (a-b)(a+b)$ $a^2 - b^2 = (a-b)(a+b)$:

(1.9)² - (0.1)² = (1.9 - 0.1)(1.9 + 0.1) = (1.8)(2.0) = 3.6

W = -5 \times 3.6 = -18~J

Step 2: Solve for final velocity ( $vf$ $v_f$)

Apply the Work-Energy Theorem:

-18 = (1)/(2) (1) vf² - (1)/(2) (1) (10²)

-18 = 0.5 vf² - 50

0.5 vf² = 50 - 18 = 32

vf² = 64 \Rightarrow vf = 8~m/s

Pattern Recognition Notice that integrating a linear force$ Pattern Recognition Notice that integrating a linear force $F = -kxyields a potential-energy-like term$ yields a potential-energy-like term $\frac{1}{2}k(x_f^2 - x_i^2). Combining this with the Work-Energy theorem gives$. Combining this with the Work-Energy theorem gives $\frac{1}{2} m v_f^2 + \frac{1}{2} k x_f^2 = \frac{1}{2} m v_i^2 + \frac{1}{2} k x_i^2$, which is identical to conservation of mechanical energy. Chapter Mix Class 11 Physics: Work, Energy and Power

Q18 jee_main_2025_07_april_morning Power

An object of mass

1000 g

experiences a time dependent force

F = (2t i + 3t² j)N

. The power generated by the force at time

t

is:

A. $(2t² + 3t³)W$ $(2t^{2} + 3t^{3})\mathrm{W}$
B. $(2t² + 18t³)W$
C. $(3t³ + 5t⁵)W$ $(3t^{3} + 5t^{5})\mathrm{W}$
D. $(2t³ + 3t⁵)W$ $(2t^{3} + 3t^{5})\mathrm{W}$

Solution

Related Formula

Instantaneous power $P$ $P$ generated by a force is given by:

P = F \cdot v

Newton's second law:

a = Fm = d vdt v = \int a dt

Core Logic

Convert mass to SI units:

m = 1000 ~g = 1 ~kg

Calculate acceleration:

a = 2t i + 3t² j1 = 2t i + 3t² j

Step 1: Determine Velocity Vector

Assuming the object starts from rest at $t = 0$ $t = 0$:

v = \int₀t (2t i + 3t² j) dt = t² i + t³ j

Step 2: Calculate Power

Compute the dot product of force and velocity:

P = F \cdot v = (2t i + 3t² j) \cdot (t² i + t³ j)

P = (2t)(t²) + (3t²)(t³) = 2t³ + 3t⁵ ~W

Pattern Recognition Sees: Time-dependent force$ Pattern Recognition Sees: Time-dependent force $\vec{F} \propto t^non a 1 kg mass. Shortcut: For$ on a 1 kg mass. Shortcut: For $m=1kg, velocity is the integral of the force components. Power is the dot product of the force vector and its integral. Since$ kg, velocity is the integral of the force components. Power is the dot product of the force vector and its integral. Since $\int at^n \mathrm{d}t = \frac{a}{n+1} t^{n+1}, power component becomes$, power component becomes $\frac{a^2}{n+1} t^{2n+1}. Here,$. Here, $2^2/2 t^3 + 3^2/3 t^5 = 2t^3 + 3t^5 \mathrm{~W}$. Chapter Mix Class 11 Physics: Work, Energy and Power

Q jee_main_2025_08_april_evening Conservation of Mechanical Energy

A block of mass

2~kg

is attached to one end of a massless spring whose other end is fixed at a wall. The spring-mass system moves on a frictionless horizontal table. The spring's natural length is

2~m

and spring constant is

200~N/m

. The block is pushed such that the length of the spring becomes

1~m

and then released. At distance

x~m

(

x < 2

) from the wall, the speed of the block will be:

A. $10[1 - (2 - x)]3/2~m/s$ $10[1 - (2 - x)]^{3/2}\mathrm{~m/s}$
B. $10[1 - (2 - x)² ]1/2~m/s$ $10[1 - (2 - x)^2 ]^{1/2}\mathrm{~m/s}$
C. $10[1 - (2 - x)²]~m/s$ $10[1 - (2 - x)^{2}]\mathrm{~m/s}$
D. $10[1 - (2 - x)²]²~m/s$ $10[1 - (2 - x)^{2}]^{2}\mathrm{~m/s}$

Solution

Related Formula

E = Kᵢ + Uᵢ = Kf + Uf

U = (1)/(2) k y²

where, $E$ $E$ = total mechanical energy $K = (1)/(2) m v²$ $K = \frac{1}{2} m v^2$ = kinetic energy $U$ $U$ = potential energy of spring with deformation $y$ $y$

Core Logic

Given parameters:

Mass, $m = 2~kg$ $m = 2\mathrm{~kg}$
Natural length of spring, $L₀ = 2~m$ $L_0 = 2\mathrm{~m}$
Spring constant, $k = 200~N/m$ $k = 200\mathrm{~N/m}$

Initial State (when block is pushed):

Length of spring is $1~m$ $1\mathrm{~m}$.
Deformation (compression), $yᵢ = L₀ - 1 = 2 - 1 = 1~m$ $y_i = L_0 - 1 = 2 - 1 = 1\mathrm{~m}$.
Released from rest: $vᵢ = 0 Kᵢ = 0$ $v_i = 0 \implies K_i = 0$.

x

Since the spring is attached to the wall, its length is $x~m$ $x\mathrm{~m}$.
Deformation (compression) at this position, $yf = L₀ - x = (2 - x)~m$ $y_f = L_0 - x = (2 - x)\mathrm{~m}$.
Kinetic energy $Kf = (1)/(2) m v² = (1)/(2) (2) v² = v²$ .

Step 1: Conservation of Energy Equation

Equate initial and final energies:

Kᵢ + Uᵢ = Kf + Uf

0 + (1)/(2) k yᵢ² = (1)/(2) m v² + (1)/(2) k yf²

Substitute the parameters:

(1)/(2) (200) (1)² = v² + (1)/(2) (200) (2 - x)²

100 = v² + 100 (2 - x)²

v² = 100 [ 1 - (2 - x)² ]

v = 10 [ 1 - (2 - x)² ]1/2~m/s

Pattern Recognition

Sees: Horizontal spring-mass energy conservation. Trap: The deformation is not $x$ $x$; it is the difference from natural length, i.e., $(L₀ - x) = (2 - x)$ $(L_0 - x) = (2 - x)$. Shortcut: Writing out energy conservation directly allows mass to cancel beautifully, simplifying the algebra immediately. ✓

Chapter Mix

Class 11 Physics: Work, Energy and Power Class 11 Physics: Oscillations

Q11 jee_main_2025_29_jan_evening Elastic Collisions in One Dimension

Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): Three identical spheres of same mass undergo one dimensional motion as shown in figure with initial velocities

vA = 5~m/s, vB = 2~m/s, vC = 4~m/s

. If we wait sufficiently long for elastic collision to happen, then

vA = 4~m/s, vB = 2~m/s, vC = 5~m/s

will be the final velocities. Reason (R): In an elastic collision between identical masses, two objects exchange their velocities. In the light of the above statements, choose the correct answer from the options given below:

A. $Both (A) and (R) are true but (R) is NOT the correct explanation of (A)$
B. $(A) is true but (R) is false$ $\text{(A) is true but (R) is false}$
C. $Both (A) and (R) are true and (R) is the correct explanation of (A)$
D. $(A) is false but (R) is true$ $\text{(A) is false but (R) is true}$

Solution

Related Formula

v₁' = v₂ and v₂' = v₁

for perfectly elastic collision ( $e=1$ $e=1$) when masses are identical ( $m₁ = m₂$ $m_1 = m_2$).

Core Logic

Reason (R) states that identical masses exchange their velocities during elastic collision, which is mathematically correct.

Let us trace the sequence of collisions chronologically:

Since $vA = 5~m/s$ $v_A = 5\mathrm{~m/s}$ and $vB = 2~m/s$ $v_B = 2\mathrm{~m/s}$, sphere A collides with sphere B. After this collision, they swap velocities:

vA' = 2~m/s, vB' = 5~m/s

Now sphere B has velocity $vB' = 5~m/s$ $v_B' = 5\mathrm{~m/s}$ and sphere C has $vC = 4~m/s$ $v_C = 4\mathrm{~m/s}$. Sphere B will collide with C. After swapping:

vB'' = 4~m/s, vC' = 5~m/s

Looking at the values now: $vA' = 2~m/s$ $v_A' = 2\mathrm{~m/s}$ and $vB'' = 4~m/s$ $v_B'' = 4\mathrm{~m/s}$. No more collisions occur.

Therefore, the final velocities are $vA = 2~m/s, vB = 4~m/s, vC = 5~m/s$ . The values given in Assertion (A) are wrong. Hence, (A) is false but (R) is true.

Elastic Collisions Step 1 diagram for Q11 - JEE Main 2025 Evening

Pattern Recognition

Velocity exchange happens pairwise in sequential order. Do not try to solve simultaneous conservation laws across all three blocks at once; handle each collision step-by-step from left to right.

Chapter Mix

Class 11 Physics: Work, Energy and Power

Q jee_main_2025_28_jan_morning Conservation of Mechanical Energy

A bead of mass

m

slides without friction on the wall of a vertical circular hoop of radius

R

as shown in figure. The bead moves under the combined action of gravity and a massless spring (k) attached to the bottom of the hoop. The equilibrium length of the spring is

R

. If the bead is released from top of the hoop with (negligible) zero initial speed, velocity of bead, when the length of spring becomes

R

, would be (spring constant is

k

, g is acceleration due to gravity)

Conservation of Mechanical Energy diagram for Q10 - JEE Main 2025 Morning — A bead sliding along a vertical circular ring constrained by a tracking baseline spring system.

A. $2 gR + kR²m$
B. $2 Rg + 4 kR²m$
C. $\sqrt(2Rg + (kR²)/(m))$ $\sqrt{2Rg + \frac{kR^2}{m}}$
D. $\sqrt(3Rg + (kR²)/(m))$ $\sqrt{3Rg + \frac{kR^2}{m}}$

Solution

Core Logic

Let's apply the comprehensive Work-Energy theorem framework across key layout tracking nodes:

A bead sliding along a vertical circular ring constrained by a tracking baseline spring system.

Wall = Δ K

M g (R + R 60°) + (1)/(2) k (R² - 0²) = (1)/(2) m v²

Simplifying the gravitational shift and potential expansions:

M g 3R2 + kR²2 = (1)/(2) m v²

Step 1: Final Kinematic Value

v = 3gR + kR²m

Matches parameters specified by option (4).

Pattern Recognition

Isolate spring metrics at node points: Initial extension equals $2R - R = R$ $2\mathrm{R} - \mathrm{R} = \mathrm{R}$. Final extension is $0$ $0$ since spring length matching $R$ $\mathrm{R}$ satisfies unextended conditions.

Chapter Mix

Class 11 Physics: Work, Energy and Power

Solution & Explanation

Related Formula

Core Logic

Step 1: Compare Horizontal Displacement and Time

Pattern Recognition

Chapter Mix

Reference Study Guides

More Work, Energy and Power Previous-Year Questions — Page 3

Solution

Related Formula

Core Logic

Step 1: Calculate Work Done by the Retarding Force (W$W$)

Step 2: Solve for final velocity (vf$v_f$)

Pattern Recognition

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Determine Velocity Vector

Step 2: Calculate Power

Pattern Recognition

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Conservation of Energy Equation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Final Kinematic Value

Pattern Recognition

Chapter Mix

More Work, Energy and Power Questions — jee_main_2026_23_january_evening

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Step 1: Calculate Work Done by the Retarding Force ( $W$ $W$)

Step 2: Solve for final velocity ( $vf$ $v_f$)