In a perfectly inelastic collision, two spheres made of the same material with masses 15 kg and 25 kg, moving in…

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More Work, Energy and Power Previous-Year Questions — Page 2

Q33 jee_main_2026_23_january_evening Motion on Curved Path

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

Motion on Curved Path diagram for Q33 - JEE Main 2026 Evening — Diagram showing bead P on a curved path and bead Q on a straight horizontal path.

A. $tP > tQ$ $t_P > t_Q$
B. $tₚ < tQ$ $t_p < t_Q$
C. $tₚ > 1.25tQ$ $t_p > 1.25t_Q$
D. $tₚ = tQ$ $t_p = t_Q$

Solution

Related Formula

t = SVhorizontal

Core Logic

For bead Q on the straight horizontal wire, its horizontal velocity remains constant at $v$ $v$. Total distance is $XQ$ $X_Q$. 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.

Step 1: Compare Horizontal Displacement and Time

Horizontal displacement required to reach B from their respective starting states: $XQ > XP$ $X_Q > X_P$.

Time taken by bead Q:

tQ = (XQ)/(v)

Time taken by bead P:

tP = XPvhorizontal, avg

Since $XQ > XP$ $X_Q > X_P$ and $vhorizontal, P > v$ $v_{\text{horizontal, P}} > v$ for a major part of the journey, $tP < tQ$ $t_P < t_Q$.

Pattern Recognition

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.

Chapter Mix

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

Q30 jee_main_2026_24_january_evening Vertical Circular Motion

In case of vertical circular motion of a particle by a thread of length r if the tension in the thread is zero at an angle

30°

shown in figure, the velocity at the bottom point (A) of the circular path is (g = gravitational acceleration)

Vertical Circular Motion diagram for Q30 - JEE Main 2026 Evening — A particle on a circular track of radius r with tension becoming zero at 30 degrees to the horizontal.

A. $5gr$ $\sqrt{5\mathrm{gr}}$
B. $(7)/(2) gr$ $\sqrt{\frac{7}{2} \mathrm{gr}}$
C. $4gr$ $\sqrt{4\mathrm{gr}}$
D. $(5)/(2) gr$ $\sqrt{\frac{5}{2} \mathrm{gr}}$

Solution

Related Formula

T + mg θ = (mV²)/(r)

$Einitial = Efinal$ $E_{\text{initial}} = E_{\text{final}}$ (Mechanical Energy Conservation)

Core Logic

At the point where the string makes an angle of $60°$ $60^{\circ}$ with the upward vertical (which corresponds to $30°$ $30^{\circ}$ above the horizontal diameter):

T + mg 60° = (mV²)/( )

Given that $T = 0$ $T = 0$:

V² = (g )/(2)

Here, $V$ $V$ is the speed at the topmost point.

Step 1: Energy Conservation

Applying Mechanical Energy Conservation (M.E.C.) between the bottom point and the point of zero tension:

(1)/(2) m u² = mg( + 60°) + (1)/(2) m V²

Step 2: Solving for u

(1)/(2) m u² = mg ( + ( )/(2)) + (1)/(2) m ((g )/(2))

u² = 3g + (g )/(2)

u = \sqrt((7g )/(2))

(Note: The problem uses $r$ $r$ as the length, so substituting $r$ $r$ for $\ell$ yields $\sqrt((7)/(2) gr)$ $\sqrt{\frac{7}{2} gr}$).

Pattern Recognition

When tension is zero at an angle $θ$ $\theta$ above the horizontal, the required bottom velocity is derived by finding the critical velocity $(mv²)/(r) = mg θ$ $\frac{mv^2}{r} = mg\sin\theta$ and then equating energy.

Chapter Mix

Class 11 Physics: Work, Energy and Power Class 11 Physics: Laws of Motion

Q10 jee_main_2025_03_april_evening Collisions and Spring Potential Energy

Consider two blocks A and B of masses

m₁=10~kg

and

m₂=5~kg

that are placed on a frictionless table. The block A moves with a constant speed

v=3~m/s

towards the block B kept at rest. A spring with spring constant

k=3000~N/m

is attached with the block B as shown in the figure.

Spring block collision diagram for Q10 - JEE Main 2025 Evening — Diagram showing Block A of mass m1 moving with velocity v towards Block B of mass m2 with a spring attached on a frictionless table.

After the collision, suppose that the blocks A and B, along with the spring in constant compression state, move together, then the compression in the spring is, (Neglect the mass of the spring)

A. 0.2 m
B. 0.4 m
C. 0.1 m
D. 0.3 m

Solution

Related Formula

By conservation of linear momentum, the common velocity $vcm$ $v_{\text{cm}}$ of the combined mass system is:

m₁ v₁ + m₂ v₂ = (m₁ + m₂) vcm

By conservation of energy, the loss in Kinetic Energy during maximum compression converted to the potential energy of the spring:

Δ K = (1)/(2) k x² \Rightarrow (1)/(2) m₁ v² - (1)/(2) (m₁ + m₂) vcm² = (1)/(2) k x²

Core Logic

Given parameters:

$m₁ = 10~kg$ $m_1 = 10\mathrm{~kg}$, $m₂ = 5~kg$ $m_2 = 5\mathrm{~kg}$
Initial velocity of A: $v = 3~m/s$ $v = 3\mathrm{~m/s}$
Initial velocity of B: $v₂ = 0$ $v_2 = 0$
Spring constant $k = 3000~N/m$ $k = 3000\mathrm{~N/m}$

Step 1: Calculate the common center of mass velocity ( $vcm$ $v_{\text{cm}}$)

vcm = (10 \times 3 + 5 \times 0)/(10 + 5) = (30)/(15) = 2~m/s

Step 2: Apply Energy Conservation to find spring compression ($ Step 2: Apply Energy Conservation to find spring compression ($x) $) $

(1)/(2) k x² = Kᵢ - Kf

(1)/(2) (3000) x² = [ (1)/(2) (10) (3²) ] - [ (1)/(2) (10 + 5) (2²) ]

1500 x² = (1)/(2)(90) - (1)/(2)(15)(4)

1500 x² = 45 - 30 = 15

x² = (15)/(1500) = (1)/(100)

x = (1)/(10)~m = 0.1~m

Pattern Recognition For maximum compression in block-spring-block collisions, the relative kinetic energy gets fully transformed into spring potential energy.

(1)/(2) μ vrel² = (1)/(2) k x²

where$ where $\mu = \frac{m_1 m_2}{m_1 + m_2}is the reduced mass. Here:

μ = (10 \times 5)/(15) = (10)/(3)~kg

(1)/(2) ((10)/(3)) (3)² = (1)/(2) (3000) x² \Rightarrow 15 = 1500 x² \Rightarrow x = 0.1~m

Using reduced mass simplifies center-of-mass collision problems instantly.

Chapter Mix

Class 11 Physics: Work, Energy and Power

Q18 jee_main_2025_03_april_evening Power and Efficiency of Motor

A motor operating on

100~V

draws a current of 1 A. If the efficiency of the motor is

91.6%

, then the loss of power in units of cal/s is :

A. 4
B. 8.4
C. 2
D. 6.2

Solution

Related Formula

The electrical input power $Pᵢₙ$ $P_{\text{in}}$ delivered to the motor is:

Pᵢₙ = V \cdot I

The fraction of power lost is determined by the efficiency $η$ $\eta$:

Plost = Pᵢₙ - Pout = Pᵢₙ (1 - η)

To convert watts ( $J/s$ $\mathrm{J/s}$) to calories per second:

Plost (cal/s) = Plost (W)4.2

Core Logic Given parameters: Potential$ Core Logic Given parameters: Potential $V = 100\mathrm{~V} Current$ Current $I = 1\mathrm{~A} Efficiency$ Efficiency $\eta = 91.6\% = 0.916 Step 1: Calculate Input Power ($ Step 1: Calculate Input Power ($P_{\text{in}})

Pᵢₙ = 100 \times 1 = 100~W

Step 2: Calculate Power Loss in Watts

Plost = 100 \times (1 - 0.916) = 100 \times 0.084 = 8.4~W

Step 3: Convert Power Loss to cal/s

Plost = (8.4)/(4.2) = 2~cal/s

Pattern Recognition Efficiency calculations often feature clean conversion rates. Recognizing that $ Pattern Recognition Efficiency calculations often feature clean conversion rates. Recognizing that $8.4 is exactly twice $ is exactly twice $ 4.2$ (the conversion factor for Joules to calories) simplifies the final step. Chapter Mix Class 11 Physics: Work, Energy and Power

Solution & Explanation

Related Formula

Step 1: Loss of Kinetic Energy

Step 2: Heat Conversion

Step 3: Calculate Temperature Rise

Pattern Recognition

Chapter Mix

Reference Study Guides

More Work, Energy and Power Previous-Year Questions — Page 2

Solution

Related Formula

Core Logic

Step 1: Compare Horizontal Displacement and Time

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Energy Conservation

Step 2: Solving for u

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate the common center of mass velocity (vcm$v_{\text{cm}}$)

Step 2: Apply Energy Conservation to find spring compression ($ Step 2: Apply Energy Conservation to find spring compression ($x)$)

Step 2: Apply Energy Conservation to find spring compression ($x)

Pattern Recognition

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Core Logic

Step 1: Calculate Input Power ($

Step 1: Calculate Input Power ($P_{\text{in}})

Step 2: Calculate Power Loss in Watts

Step 2: Calculate Power Loss in Watts

Step 3: Convert Power Loss to cal/s

Step 3: Convert Power Loss to cal/s

Pattern Recognition

Pattern Recognition

Chapter Mix

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Step 1: Calculate the common center of mass velocity ( $vcm$ $v_{\text{cm}}$)

Step 2: Apply Energy Conservation to find spring compression ($)

Step 2: Apply Energy Conservation to find spring compression ($x $)$

Step 1: Calculate Input Power ($P_{\text{in}} $)$