Work-energy-power practice · Mixed difficulty

The correct answer is A because a collision between two carom coins is an example of an oblique collision, where the objects do not move along the same straight line after the collision.

Option C is correct because momentum is always conserved in collisions due to the absence of external forces, but kinetic energy is only conserved in elastic collisions.

Option A (The light body rebounds after collision) is correct because, in an elastic collision with a much more massive object, the lighter object reverses its velocity.

Option A (9.5 W) is correct because it is the power calculated using the dot product of force and displacement divided by time.

Option C (-k/2r) is correct because the total energy is the sum of kinetic and potential energies, where kinetic energy is k/2r and potential energy is -k/r.

Option C (Only A is wrong) is correct because a particle moving with constant speed can have changing velocity (e.g., circular motion), and zero velocity doesn't imply zero acceleration.

Statement I and Statement II are both true, and Statement II correctly explains Statement I. The tendency to overturn increases with the square of the velocity, as shown by the equation $\tan \theta = \frac{{v^2}}{{Rg}}$.

Option D ($8.1 \mathrm{~kW}$) is correct because the power generated by the turbine is 90% of the input power, which is calculated from the potential energy of the water falling from a height of 60 m.

Option D ($\frac{1}{{\sqrt t }}$) is correct because the force is inversely proportional to the square root of time when kinetic energy increases uniformly with time.

Option D is correct because in circular motion with constant speed, the kinetic energy remains constant, and hence the work done is zero.

Option A ($69 \mathrm{~J}$) is correct because it is the dot product of the force and displacement vectors.

Option A (7.46 J) is correct because it is the work done by the motor in one rotation, calculated considering the power, speed, and efficiency.

Option D ($\frac{V^2}{T^2} t^2$) is correct because work done is proportional to the square of velocity, which is proportional to the square of time.

Option B is correct because the force is found to be increasing with time.

Option D ($2\Omega$) is correct because the resistance is calculated by dividing the voltage for full-scale deflection by the corresponding current.

Option A is correct because the work done by a conservative force, like the spring force, in a cyclic process is zero, and the reason correctly explains why.

Option B is correct because a body at rest can possess potential energy, which is a form of mechanical energy.

Option A ($m_1=m_2$) is correct because maximum energy transfer occurs when masses are equal in an elastic collision.

Option C is correct because in uniform circular motion, the magnitude of velocity (speed) remains constant, so the change in magnitude of velocity is zero.

Option C (240 J) is correct because it is the difference between the initial and final kinetic energies, calculated using the coefficient of restitution to find the final velocities.

Option D ($\frac{T}{8}$) is correct because the potential energy is half the total energy when $\sin^2(\omega t) = \frac{1}{2}$, which occurs at $t = \frac{T}{8}$.

Option B ($\sqrt {\frac{{2Fd + mv_1^2}}{m}}$) is correct because it is derived from the work-energy theorem, relating work done to the change in kinetic energy.

Option B ($\frac{\sqrt{34}}{3} {~m} / {s}$) is correct because it is calculated by finding the minimum potential energy and then using conservation of energy to find the maximum kinetic energy and thus the maximum speed.

Option B ($400 \mathrm{~m}^3$) is correct because it's the volume of water required per second to generate the specified power, considering the efficiency of conversion.

Option A (1.414) is correct because the ratio of velocities at 60 and 120 degrees is calculated using the conservation of energy in a vertical circle.

Option B (The acceleration vector is tangent to the circle) is correct because in uniform circular motion, the acceleration vector always points towards the center of the circle (centripetal acceleration), not tangent to it.

Option A (61) is correct because the instantaneous power is the dot product of the force and velocity vectors.

Option A (Stable equilibrium) is correct because the second derivative of the potential energy at x=2 is positive, indicating a minimum.

Option C (The height by which it is raised) is correct because the work done against gravity depends only on the change in potential energy, which is determined by the height.

Option A ($6 \mathrm{~W}$) is correct because power is the dot product of force and velocity.