Two particles of masses $m_1, m_2$ move with initial velocities $u_1$ and $u_2.$ On collision, one of the particles gets excited to a higher level, after absorbing energy $\varepsilon.$ If final velocities of the particles be $v_1$ and $v_2$ then we must have:

A block of mass $10$ kg, moving in $x$ direction with a constant speed of $10$ ms$^{-1},$ is subjected to a retarding force $F = 0.1 \times J$ m$^{-1}$ during its travel from $x = 20$ to $30$ m. Its final K.E. will be

Two spherical bodies of mass $M$ and $5M$ and radii $R$ and $2R$ are released in free space with initial separation between their centres equal to $12R.$ If they attract each other due to gravitational force only, then the distance covered by the smaller body before collision is

Two similar springs P and Q have spring constants $K_P$ and $K_Q,$ such that $K_P > K_Q.$ They are stretched, first by the same amount (case a,) then by the same force (case b). The work done by the springs $W_P$ and $W_Q$ are related as, in case (a) and case (b), respectively

A particle of mass $m$ is driven by a machine that delivers a constant power $k$ watt. If the particle starts from rest, the force on the particle at time $t$ is

If the magnitude of the sum of two vectors is equal to the magnitude of the difference of the two vectors, the angle between these vectors is

A spring of force constant $k$ is cut into lengths of ratio $1:2:3.$ They are connected in series and the new force constant is $k'.$ Then, they are connected in parallel and force constant is $k''.$ Then the ratio $k': k''$ is

Consider a drop of rainwater having mass 1 g falling from a height of 1 km. It hits the ground with a speed of 50 m/s. Take $g$ constant with a value 10 m/s$^2.$ The work done by the (i) gravitational force and the (ii) resistive force of air is

A moving block having mass $m,$ collides with another stationary block having mass $4m.$ The lighter block comes to rest after collision. When the initial velocity of the lighter block is $v,$ then the value of coefficient of restitution $(e)$ will be

A force $F = 20 + 10y$ acts on a particle in the y-direction where $F$ is in newton and $y$ in meter. Work done by this force to move the particle from $y = 0$ to $y = 1$ m is:

Body $A$ of mass $4m$ moving with speed $u$ collides with another body $B$ of mass $2m$, at rest. The collision is head-on and elastic in nature. After the collision the fraction of energy lost by the colliding body $A$ is:

Water falls from a height of $60$ m at the rate of $15$ kg/s to operate a turbine. The losses due to frictional force are $10\%$ of the input energy. How much power is generated by the turbine?$\newline (g=10$ m/s$^2)$

An electric lift with a maximum load of $2000$ kg (lift + passengers) is moving up with a constant speed of $1.5$ m/s. The frictional force opposing the motion is $3000$ N. The minimum power delivered by the motor to the lift in watts is: ($g = 10$ m/s$^2$)

Two bodies A and B of same mass undergo completely inelastic one-dimensional collision. The body A moves with velocity $v_1$ while body B is at rest before collision. The velocity of the system after collision is $v_2.$ The ratio $v_1: v_2$ is:

At any instant of time $t,$ the displacement of any particle is given by $2t - 1$ (SI unit) under the influence of force of 5N. The value of instantaneous power is (in SI unit):