A small block of mass m slides down from the top of a frictionless inclined surface, while the inclined plane is moving towards left with constant acceleration a₀$a_{0}$ . The angle between the inclined plane and ground is θ$\theta$ and its base length is L. Assuming that initially the small block is at the top of the inclined plane, the time it takes to reach the lowest point of the inclined plane is ____.
A mass m on an inclined plane of base L, accelerating to the left with a0.
s = ut + (1)/(2)at²$$s = ut + \frac{1}{2}at^2$$Fpseudo = m a₀$$F_{\text{pseudo}} = m a_0$$
Core Logic
A mass m on an inclined plane of base L, accelerating to the left with a0.
We solve the problem from the non-inertial frame of reference of the inclined plane. A pseudo force m a₀$m a_0$ acts on the block towards the right.
Forces acting parallel to the incline (downward positive):
Component of gravity: mg θ$mg \sin \theta$
Component of pseudo force (pointing up the incline): -ma₀ θ$-ma_0 \cos \theta$
Step 1: Calculate Effective Acceleration
Net force down the incline:
Fₙₑₜ = mg θ - m a₀ θ$$F_{\text{net}} = mg \sin \theta - m a_0 \cos \theta$$
Thus, the acceleration relative to the incline is:
a = g θ - a₀ θ$$a = g \sin \theta - a_0 \cos \theta$$
Step 2: Distance Travelled
The base length of the incline is L$L$. Thus, the total length of the inclined surface is:
Whenever an inclined plane accelerates horizontally, resolve the pseudo force ma₀$ma_0$ parallel and perpendicular to the incline. Multiply by 2 inside the root to simplify into double-angle formats if options demand it.
Chapter Mix
Class 11 Physics: Laws of Motion
Class 11 Physics: Kinematics
Keywords:#slides down from the top of a frictionless inclined surface#JEE Main 2026 Evening Q40#Laws of Motion JEE Main 2026#Pseudo Force and Inclined Plane JEE Main 2026#Inclined plane#acceleration a0#mass m#angle theta#base length L
More Laws of Motion Previous-Year Questions — Page 5
A ball of mass 0.5~kg$0.5\mathrm{~kg}$ is attached to a string of length 50~cm$50\mathrm{~cm}$. The ball is rotated on a horizontal circular path about its vertical axis. The maximum tension that the string can bear is 400~N$400\mathrm{~N}$. The maximum possible value of angular velocity of the ball in rad/s$\mathrm{rad/s}$ 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$T_{\text{max}} = 400\mathrm{~N}$.
Ensure units are metric (50~cm → 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
Q36jee_main_2024_29_january_eveningCircular Motion and Tension
A stone of mass 900 g$900\text{ g}$ is tied to a string and moved in a vertical circle of radius 1 m$1\text{ m}$ making 10 rpm$10\text{ rpm}$. The tension in the string, when the stone is at the lowest point is (if π² = 9.8$\pi^2 = 9.8$ and g = 9.8 m/s²$g = 9.8\text{ m/s}^2$):
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$$T - mg = m r \omega^2$$
Rearranging to solve for tension T$T$:
T = mg + m r ω²$$T = mg + m r \omega^2$$
Core Logic
Given data:
Mass, m = 900 g = 0.9 kg$m = 900\text{ g} = 0.9\text{ kg}$
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.
A particle is moving in a circle of radius 50 cm$50\text{ cm}$ in such a way that at any instant the normal and tangential components of its acceleration are equal. If its speed at t = 0$t = 0$ is 4 m/s$4\text{ m/s}$, the time taken to complete the first revolution will be (1)/(α)[ 1 - e-2π ] s$\frac{1}{\alpha}\left[ 1 - e^{-2\pi} \right]\text{ s}$, where α =$\alpha =$ ________.
Numerical Answer.Answer: 8 to 8
Solution
Related Formula
For a particle in circular motion:
Normal (centripetal) acceleration: ac = (v²)/(r)$a_c = \frac{v^2}{r}$
Comparing this to (1)/(α)[ 1 - e-2π ] s$\frac{1}{\alpha}\left[ 1 - e^{-2\pi} \right]\text{ s}$, we get:
α = 8$\alpha = 8$
Pattern Recognition
The condition aₜ = ac v (dv)/(ds) = (v²)/(r) (dv)/(v) = (ds)/(r)$a_t = a_c \implies v \frac{dv}{ds} = \frac{v^2}{r} \implies \frac{dv}{v} = \frac{ds}{r}$. 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
Q36jee_main_2024_27_jan_morningBanking of Tracks
A train is moving with a speed of 12 m/s$12\text{ m/s}$ on rails which are 1.5 m$1.5\text{ m}$ apart. To negotiate a curve of radius 400 m$400\text{ m}$, the height by which the outer rail should be raised with respect to the inner rail is (Given, g = 10 m/s²$g = 10\text{ m/s}^{2}$):
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)$$\tan\theta = \frac{v^2}{Rg}$$
For small angles, θ ≈ θ = (h)/(d)$\tan\theta \approx \sin\theta = \frac{h}{d}$, where h$h$ is the raised height and d$d$ is the separation between the tracks.
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
Q45jee_main_2024_27_jan_morningConservation of Linear Momentum
A body of mass 1000 kg$1000\text{ kg}$ is moving horizontally with a velocity 6 m/s$6\text{ m/s}$. If 200 kg$200\text{ kg}$ 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$$m_1 v_1 = m_2 v_2$$
Core Logic
Since there is no external horizontal force acting on the body, linear momentum along the horizontal axis is conserved.
Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.
