AIIMS Physics Motion In A Plane Class 11 Questions

An electron is moving in a circle of radius 2 m with speed 4 m/s. Find the acceleration of the electron.

The initial velocity of the water stream is given as, \(v = \sqrt{2g(H⁻h)}\) The time taken by stream to hit the surface is, \(h = rac{1}{2}gt^2\) \(t = \sqrt{rac{2h}{g}}\) The horizontal range can be calculated as, Horizontal range = \(vt = \sqrt{2g(H⁻h)} imes \sqrt{rac{2h}{g}} = 2\sqrt{(H⁻h)h} = 2\sqrt{(1-0.25)(0.25)} = 0.866 ext{ m}\)

The expression of force is given as, F = -du/dr = -[2a/r³ + b/r²] = 2a/r³ - b/r². Substitute the values.

A cricketer can throw a ball to a maximum horizontal distance of 100 m. The speed with which he throws the ball is (to the nearest integer)

The acceleration of particle is given by, a = dv/dt The velocity of particle is calculated by integrating above equation.

A body starts from rest with an acceleration a₁, after two seconds another body B starts from rest with an acceleration a₂. If they travel equal distance in fifth second, after the starts of A, the ratio a₁ : a₂ will be equal to

The formula for |𝐀⃗ + 𝐁⃗ |² is, |𝐀⃗ + 𝐁⃗ |² = |𝐀⃗ |² + |𝐁⃗ |² + 2𝐀⃗ ⋅𝐁⃗ = A + B + 2AB cosθ And The formula for |𝐀⃗ − 𝐁⃗ |² is, |𝐀⃗ − 𝐁⃗ |² = |𝐀⃗ |² + |𝐁⃗ |² − 2𝐀⃗ ⋅𝐁⃗ = A + B − 2AB cosθ It is given that,

The formula for the maximum height during projectile motion is, h = (u sinθ)² / 2g Now, Δh/h = 2Δu/u Δh/h = 2 × 2 = 4%

The maximum range of projectile motion is given as, R = u² sin 2θ / g. The time of flight of projectile motion is given as, T = 2u sin θ / g. The square of time of flight of projectile motion is, T² = (2u sin θ / g)² = 4u² sin² θ / g². The ratio of maximum range and square of time of flight in projectile motion is calculated as, R / T² = (u² sin 2θ / g) / (4u² sin² θ / g²) = g / 2 cot θ.

If F ⊥ V at all instants then motion will be circular.

Assertion: The equation of motion can be applied only if acceleration is along the direction of velocity and is constant. Reason: If the acceleration of the body is constant then its motion is known as uniform motion.

Assertion: The position-time graph of a uniform motion in one dimension of a body can have negative slope. Reason: When the speed of body decreases with time, the position-time graph of the moving body has negative slope.

Assertion: When a body is dropped or thrown upward from the same height, it would reach the ground at the same time. Reason: Direction of projection does not affect the time of flight.

Assertion: If $|\vec{A} + \vec{B}| = |\vec{A} - \vec{B}|$, then the angle between $\vec{A}$ and $\vec{B}$ is $90^\circ$. Reason: $\vec{A} + \vec{B} = \vec{B} + \vec{A}$

Assertion: Three projectiles are moving in different paths in the air. Vertical component of relative velocity between any of the pair does not change with time as long as they are in air. (Neglecting the effect of air friction) Reason: Relative acceleration between any of the pair of projection is zero.

A particle is projected with an angle of projection to the horizontal line passing through the points (P,Q) and (Q, P) referred to horizontal and vertical axes (can be treated as x-axis and y-axis respectively). The angle of projection can be given by

Consider the following diagram. Suppose particle falls down a distance y in t time, therefore, y = ½gt², t = √(2y/g). For x-axis, x = ½aₓt², x = ½(QE/m)(2y/g). Substitute the values.

Consider the figure below, vᵣ/vₘ = sin 30°, vᵣ = 0.25 m/s

A stone is moved in a horizontal circle of radius 4m by means of a string at the height of 20m above the ground. The string breaks and the particles flies off horizontally, striking the ground 10m away. The centripetal acceleration during circular motion is given by

The particle of mass m is moving in a circular path of constant radius r such that its centripetal acceleration aₚ is varying with time t as aₚ = k²rt², where k is a constant. The power delivered to particle by the force acting on it is

**Assertion:** For looping a vertical loop of radius $r$, the minimum velocity at the lowest point should be $\sqrt{5gr}$. **Reason:** In this even the velocity at the highest point will be zero.

A particle is projected with an angle of projection $\\theta$ to the horizontal line passing through the points $(P, Q)$ and $(Q, P)$ referred to horizontal and vertical axes (can be treated as x-axis and y-axis respectively).\n\nThe angle of projection can be given by

A particle is projected from the ground with an initial speed of 'v' at angle θ with horizontal. The average velocity of the particle between its point of projection and height point of trajectory is

If a vector 2î+3ĵ+8k̂ is perpendicular to the vector 4î−Âĵ+a k̂, then value of α is

Two projectiles of same mass have their maximum kinetic energies in ratio 4 : 1 and ratio of their maximum heights is also 4 : 1 then what is the ratio of their ranges?

Assertion : Two balls of different masses are thrown vertically upward with same speed. They will pass through their point of projection in the downward direction with the same speed. Reason : The maximum height and downward velocity attained at the point of projection are independent of the mass of the ball.

Assertion : In javelin throw, the athlete throws the projectile at an angle slightly more than 45°. Reason : The maximum range does not depend upon angle of projection.

A body is projected horizontally with a velocity of √(2gℓ) m/sec. The velocity of the body after 0.7 seconds will be nearly (Take g = 10 m/sec²)

Assertion : Generally the path of a projectile from the earth is parabolic but it is elliptical for projectiles going to a very great height. Reason : Up to ordinary height the projectile moves under a uniform gravitational force, but for great heights, projectile moves under a variable force.

At the uppermost point of a projectile, its velocity and acceleration are at an angle of