A body of mass 2~kg$2\mathrm{~kg}$ moving with velocity of vᵢₙ = 3 i +4 j~m/s$\vec{v}_{\mathrm{in}} = 3\hat{i} +4\hat{j}\mathrm{~m/s}$ enters into a constant force field of 6~N$6\mathrm{~N}$ directed along positive z-axis. If the body remains in the field for a period of (5)/(3)$\frac{5}{3}$ seconds, then velocity of the body when it emerges from force field is:
F = m a$$\vec{F} = m \vec{a}$$v = u + at$$\vec{v} = \vec{u} + \vec{a}t$$
where,
F$\vec{F}$ = constant force vector
m$m$ = mass of body
a$\vec{a}$ = acceleration vector
u$\vec{u}$ = initial velocity vector
v$\vec{v}$ = final velocity vector
Core Logic
Given parameters:
Mass, m = 2~kg$m = 2\mathrm{~kg}$
Initial velocity, u = 3 i + 4 j~m/s$\vec{u} = 3\hat{i} + 4\hat{j}\mathrm{~m/s}$
Force, F = 6 k~N$\vec{F} = 6\hat{k}\mathrm{~N}$ (directed along positive z-axis)
Time interval, t = (5)/(3)~s$t = \frac{5}{3}\mathrm{~s}$
Calculate the acceleration vector a$\vec{a}$:
a = Fm = 6 k2 = 3 k~m/s²$$\vec{a} = \frac{\vec{F}}{m} = \frac{6\hat{k}}{2} = 3\hat{k}\mathrm{~m/s}^{2}$$
Thus, the emerging velocity of the body is 3 i + 4 j + 5 k~m/s$3\hat{i} + 4\hat{j} + 5\hat{k}\mathrm{~m/s}$.
Pattern Recognition
Sees: Orthogonal initial velocity and force field direction.
Shortcut: Since the force acts entirely along the z-axis, the x and y components of the velocity remain unchanged (3 i + 4 j$3\hat{i} + 4\hat{j}$). Simply compute the z-component change: vz = az t = ((6)/(2)) ((5)/(3)) = 5$v_z = a_z t = \left(\frac{6}{2}\right) \left(\frac{5}{3}\right) = 5$. Result: 3 i + 4 j + 5 k$3\hat{i} + 4\hat{j} + 5\hat{k}$. ✓
Chapter Mix
Class 11 Physics: Laws of Motion
Class 11 Physics: Kinematics
Keywords:#Force field constant acceleration velocity vector#JEE Main 2025 Evening Q15#Newton's laws kinematics vectors#Laws of motion three dimensional trajectory
More Laws of Motion Previous-Year Questions — Page 7
Q31jee_main_2024_31_jan_eveningPulley Systems and Tension
A light string passing over a smooth light fixed pulley connects two blocks of masses m₁$m_1$ and m₂$m_2$. If the acceleration of the system is g/8$g/8$, then the ratio of masses is
The image displays two masses suspended over a single fixed smooth pulley.
For standard Atwood machines, a = g × Difference in massSum of mass$a = g \times \frac{\text{Difference in mass}}{\text{Sum of mass}}$. If a/g = 1/8$a/g = 1/8$, then (m₁-m₂)/(m₁+m₂) = 1/8$(m_1-m_2)/(m_1+m_2) = 1/8$, which can be solved using componendo and dividendo: m₁/m₂ = (8+1)/(8-1) = 9/7$m_1/m_2 = (8+1)/(8-1) = 9/7$.
Chapter Mix
Class 11 Physics: Laws of Motion
Q42jee_main_2024_31_jan_eveningFriction on an Inclined Plane
A block of mass 5 kg$5 \text{ kg}$ is placed on a rough inclined surface as shown in the figure.
The image shows a 5 kg block on a rough plane inclined at 30 degrees, with coefficient of friction mu = 0.1.
If F₁$\vec{F}_1$ is the force required to just move the block up the inclined plane and F₂$\vec{F}_2$ is the force required to just prevent the block from sliding down, then the value of | F₁| - | F₂|$|\vec{F}_1| - |\vec{F}_2|$ is: [Use g = 10 m/s²$g = 10 \text{ m/s}^2$]
To move the block up, the applied force F₁$F_1$ must overcome both the downward gravitational component and the downward frictional force.
To prevent it from sliding down, the applied force F₂$F_2$ acts upwards and is aided by friction which acts upwards to oppose impending downward slip.
The image shows a 5 kg block on a rough plane inclined at 30 degrees, with coefficient of friction mu = 0.1.
The image shows a 5 kg block on a rough plane inclined at 30 degrees, with coefficient of friction mu = 0.1.
Note: The official options had an anomaly where 5√(3) N$5\sqrt{3} \text{ N}$ was missing or evaluated as a bonus. Option 2 was listed as 50√(3)$50\sqrt{3}$ in the primary text. We track the closest logic path indicating Bonus.
Pattern Recognition
The difference between 'push up' and 'hold from sliding' forces on an incline is always precisely 2 fk$2 f_k$ (2 μ mg θ$2 \mu mg \cos \theta$). Bypass calculating the mg θ$mg \sin \theta$ terms entirely.
Chapter Mix
Class 11 Physics: Laws of Motion
Qjee_main_2024_31_jan_morningPulley And Incline Friction
In the given arrangement of a doubly inclined plane two blocks of masses M$M$ and m$m$ are placed. The blocks are connected by a light string passing over an ideal pulley as shown. The coefficient of friction between the surface of the plane and the blocks is 0.25$0.25$. The value of m$m$, for which M = 10 kg$M = 10\mathrm{\ kg}$ will move down with an acceleration of 2 m/s²$2\mathrm{\ m/s^2}$ is : (take g = 10 m/s²$g = 10\mathrm{\ m/s^2}$ and 37° = 3 / 4$\tan 37^{\circ} = 3 / 4$)
Two blocks M and m on opposite sides of a double inclined plane linked by a rope over a top pulley. M is on the 53-degree slope and moving downwards, m is on the 37-degree slope.
A.9 kg$9\mathrm{\ kg}$
B.4.5 kg$4.5\mathrm{\ kg}$
C.6.5 kg$6.5\mathrm{\ kg}$
D.2.25 kg$2.25\mathrm{\ kg}$
Solution
Related Formula
Σ F = ma$$\sum F = ma$$fk = μk N = μk mg θ$$f_k = \mu_k N = \mu_k mg \cos\theta$$
Core Logic
Two blocks M and m on opposite sides of a double inclined plane linked by a rope over a top pulley. M is on the 53-degree slope and moving downwards, m is on the 37-degree slope.
Since block M$M$ moves down the incline, kinetic friction opposes its motion (acts upwards).
Block m$m$ is pulled up the incline, so kinetic friction opposes its motion (acts downwards).
A coin is placed on a disc. The coefficient of friction between the coin and the disc is μ$\mu$. If the distance of the coin from the center of the disc is r$r$, the maximum angular velocity which can be given to the disc, so that the coin does not slip away, is:
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.
