A block is sliding down on an inclined plane of slope theta and at an instant t = 0 this block is given an upward momentum so that it starts moving up on the inclined surface with velocity u . The distance (S) travelled by the block before its velocity become zero, is ____. (g = gravitational acceleration)

v2 = u2 + 2as Since the surface is frictionless (implied as no friction coefficient is given), the only force along the incline opposing the upward motion is the component of gravity, mg theta. Acceleration a = -g theta. Using the third equation of motion: V2 = U2 + 2as Set final velocity V = 0: For upward motion on a smooth incline, deceleration is purely g theta. The stopping distance is always u2 / 2a. Class 11 Physics: Laws of Motion Class 11 Physics: Kinematics

A block is sliding down on an inclined plane of slope theta and at an instant t = 0 this block is given an upward momentum so that it starts moving up on the inclined surface with velocity u . The dis

v2 = u2 + 2as Since the surface is frictionless (implied as no friction coefficient is given), the only force along the incline opposing the upward motion is the component of gravity, mg theta. Acceleration a = -g theta. Using the third equation of motion: V2 = U2 + 2as Set final velocity V = 0: For upward motion on a smooth incline, deceleration is purely g theta. The stopping distance is always u2 / 2a. Class 11 Physics: Laws of Motion Class 11 Physics: Kinematics

A block is sliding down on an inclined plane of slope theta and at an instant t = 0 this block is given an upward momentum so that it starts moving up on the inclined surface with velocity u . The distance (S) travelled by the block before its velocity become zero, is ____. (g = gravitational acceleration)

v2 = u2 + 2as Since the surface is frictionless (implied as no friction coefficient is given), the only force along the incline opposing the upward motion is the component of gravity, mg theta. Acceleration a = -g theta. Using the third equation of motion: V2 = U2 + 2as Set final velocity V = 0: For upward motion on a smooth incline, deceleration is purely g theta. The stopping distance is always u2 / 2a. Class 11 Physics: Laws of Motion Class 11 Physics: Kinematics

A block is sliding down on an inclined plane of slope theta and at an instant t = 0 this block is given an upward momentum so that it starts moving up on the inclined surface with velocity u . The dis

v2 = u2 + 2as Since the surface is frictionless (implied as no friction coefficient is given), the only force along the incline opposing the upward motion is the component of gravity, mg theta. Acceleration a = -g theta. Using the third equation of motion: V2 = U2 + 2as Set final velocity V = 0: For upward motion on a smooth incline, deceleration is purely g theta. The stopping distance is always u2 / 2a. Class 11 Physics: Laws of Motion Class 11 Physics: Kinematics

A block is sliding down on an inclined plane of slope theta and at an instant t = 0 this block is given an upward…

Year	2026	2025	2024	Total
Questions	9	10	15	34

Q37 jee_main_2024_01_february_morning Circular Motion

A ball of mass

0.5~kg

is attached to a string of length

50~cm

. The ball is rotated on a horizontal circular path about its vertical axis. The maximum tension that the string can bear is

400~N

. The maximum possible value of angular velocity of the ball in

rad/s

is:

A. 1600
B. 40
C. 1000
D. 20

Solution

Related Formula

Centripetal force configuration for simplified horizontal rotation layout:

$T = mω² l$ $T = m\omega^2 l$

Core Logic

Given values: $m = 0.5~kg$ $m = 0.5\mathrm{~kg}$, $l = 50~cm = 0.5~m$ $l = 50\mathrm{~cm} = 0.5\mathrm{~m}$, $Tmax = 400~N$ $T_{\text{max}} = 400\mathrm{~N}$.

Equating max tension to centripetal requirement:

400 = 0.5 \times ω² \times 0.5

Step 1: Compute Angular Velocity

400 = 0.25 ω²

ω² = (400)/(0.25) = 1600

ω = \sqrt(1600) = 40~rad/s

Pattern Recognition

Ensure units are metric ( $50~cm \to 0.5~m$ $50\mathrm{~cm} \to 0.5\mathrm{~m}$). Direct mapping to horizontal string projection metrics.

Chapter Mix

Class 11 Physics: Laws of Motion

Q36 jee_main_2024_29_january_evening Circular Motion and Tension

A stone of mass

900 g

is tied to a string and moved in a vertical circle of radius

1 m

making

10 rpm

. The tension in the string, when the stone is at the lowest point is (if

π² = 9.8

and

g = 9.8 m/s²

):

A. $97 N$ $97\text{ N}$
B. $9.8 N$ $9.8\text{ N}$
C. $8.82 N$ $8.82\text{ N}$
D. $17.8 N$ $17.8\text{ N}$

Solution

Related Formula

At the lowest point of a vertical circle, the equation of motion for a mass $m$ $m$ is:

T - mg = m r ω²

Rearranging to solve for tension $T$ $T$:

T = mg + m r ω²

Core Logic

Given data:

Mass, $m = 900 g = 0.9 kg$ $m = 900\text{ g} = 0.9\text{ kg}$
Radius, $r = 1 m$ $r = 1\text{ m}$
Frequency, $N = 10 rpm = (10)/(60) rps = (1)/(6) rps$
Angular velocity, $ω = 2π N = 2π ((1)/(6)) = (π)/(3) rad/s$

Step 1: Calculate Force Values

Now we substitute our parameters into the tension equation:

T = (0.9)(9.8) + (0.9)(1)((π)/(3))²

T = 8.82 + 0.9 \times (π²)/(9)

Since $π² = 9.8$ $\pi^2 = 9.8$:

T = 8.82 + 0.1 \times 9.8

T = 8.82 + 0.98 = 9.80 N

Free-body diagram of stone at lowest point in vertical circle for Q36

Pattern Recognition

Always convert Mass to $kg$ $\text{kg}$ and rotational speed to $rad/s$ $\text{rad/s}$ first. Using the prompt constraint $π² = 9.8$ $\pi^2 = 9.8$ yields a perfect decimal addition match.

Chapter Mix

Class 11 Physics: Laws of Motion

Q59 jee_main_2024_29_january_evening Non-uniform Circular Motion

A particle is moving in a circle of radius

50 cm

in such a way that at any instant the normal and tangential components of its acceleration are equal. If its speed at

t = 0

is

4 m/s

, the time taken to complete the first revolution will be

(1)/(α)[ 1 - e-2π ] s

, where

α =

________.

Numerical Answer. Answer: 8 to 8

Solution

Related Formula

For a particle in circular motion:

Normal (centripetal) acceleration: $ac = (v²)/(r)$ $a_c = \frac{v^2}{r}$
Tangential acceleration: $aₜ = (dv)/(dt)$ $a_t = \frac{dv}{dt}$

Core Logic

Given $ac = aₜ$ $a_c = a_t$:

(v²)/(r) = (dv)/(dt)

\intv₀v (dv)/(v²) = \int₀t (dt)/(r)

[ -(1)/(v) ]v₀v = (t)/(r)

-(1)/(v) + (1)/(v₀) = (t)/(r) (1)/(v) = (1)/(v₀) - (t)/(r)

v = (v₀)/(1 - (v₀ t)/(r))

Step 1: Relate Velocity to Position and Integrate

Substitute the parameters $v₀ = 4 m/s$ $v_0 = 4\text{ m/s}$ and $r = 50 cm = 0.5 m$ $r = 50\text{ cm} = 0.5\text{ m}$:

v = (4)/(1 - 8t) = (ds)/(dt)

Integrating this to find the position $s(t)$ $s(t)$:

\int₀s ds = \int₀t (4)/(1 - 8t) dt

s = 4 [ (ln(1 - 8t))/(-8) ]₀^t = -(1)/(2) ln(1 - 8t)

Step 2: Solve for Time of First Revolution

To complete the first revolution, the distance covered is:

s = 2π r = 2π (0.5) = π m

Equating the distance:

π = -(1)/(2) ln(1 - 8t)

-2π = ln(1 - 8t)

1 - 8t = e-2π

8t = 1 - e-2π t = (1)/(8) [ 1 - e-2π ] s

Comparing this to $(1)/(α)[ 1 - e-2π ] s$ , we get:

$α = 8$ $\alpha = 8$

Pattern Recognition

The condition $aₜ = ac v (dv)/(ds) = (v²)/(r) (dv)/(v) = (ds)/(r)$ . Integrating directly gives $v = v₀ es/r$ $v = v_0 e^{s/r}$. Substituting this back into $v = ds/dt$ $v = ds/dt$ makes the final time integral much more intuitive.

Chapter Mix

Class 11 Physics: Laws of Motion

Q36 jee_main_2024_27_jan_morning Banking of Tracks

A train is moving with a speed of

12 m/s

on rails which are

1.5 m

apart. To negotiate a curve of radius

400 m

, the height by which the outer rail should be raised with respect to the inner rail is (Given,

g = 10 m/s²

):

A. $6.0 cm$ $6.0\text{ cm}$
B. $5.4 cm$ $5.4\text{ cm}$
C. $4.8 cm$ $4.8\text{ cm}$
D. $4.2 cm$ $4.2\text{ cm}$

Solution

Related Formula

θ = (v²)/(Rg)

For small angles, $θ \approx θ = (h)/(d)$ , where $h$ $h$ is the raised height and $d$ $d$ is the separation between the tracks.

Core Logic

Equating the two relationships:

(h)/(d) = (v²)/(Rg)

(h)/(1.5) = (12 \times 12)/(400 \times 10)

Step 1: Compute height value

h = 1.5 \times (144)/(4000) = 1.5 \times 0.036 = 0.054 m

Converting to centimeters:

h = 0.054 \times 100 = 5.4 cm

Pattern Recognition

Whenever $θ$ $\theta$ is small, geometry permits approximating $θ$ $\tan\theta$ with $hwidth$ $\frac{h}{\text{width}}$, vastly reducing computational transcendental overhead.

Chapter Mix

Class 11 Physics: Laws of Motion

Q45 jee_main_2024_27_jan_morning Conservation of Linear Momentum

A body of mass

1000 kg

is moving horizontally with a velocity

6 m/s

. If

200 kg

extra mass is added, the final velocity (in m/s) is:

A. $6$ $6$
B. $2$ $2$
C. $3$ $3$
D. $5$ $5$

Solution

Related Formula

m₁ v₁ = m₂ v₂

Core Logic

Since there is no external horizontal force acting on the body, linear momentum along the horizontal axis is conserved.

Initial mass $m₁ = 1000 kg$ $m_1 = 1000\text{ kg}$, initial velocity $v₁ = 6 m/s$ $v_1 = 6\text{ m/s}$. Final mass $m₂ = 1000 + 200 = 1200 kg$ $m_2 = 1000 + 200 = 1200\text{ kg}$.

Step 1: Balance conservation equation

1000 \times 6 = 1200 \times v₂

6000 = 1200 v₂ v₂ = (6000)/(1200) = 5 m/s

Pattern Recognition

Inelastic mass addition transitions are classic momentum-balance equations, scaling velocity inversely with total expanded mass profiles.

Chapter Mix

Class 11 Physics: Laws of Motion

Solution & Explanation

Related Formula

Core Logic

Step 1: Apply Kinematic Equation

Pattern Recognition

Chapter Mix

Reference Study Guides

More Laws of Motion Previous-Year Questions — Page 5

Solution

Related Formula

Core Logic

Step 1: Compute Angular Velocity

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Force Values

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Relate Velocity to Position and Integrate

Step 2: Solve for Time of First Revolution

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Compute height value

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Balance conservation equation

Pattern Recognition

Chapter Mix

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