In an elastic collision of two particles the following quantity is conserved
Momentum of each particle
Speed of each particle
Kinetic energy of each particle
Total kinetic energy of both the particles
If the potential energy of a gas molecule is
$U = \frac{M}{{{r^6}}} - \frac{N}{{{r^{12}}}}$,
$M$ and $N$ being positive constants, then the potential energy at equilibrium must be
The force $F$ acting on a body moving in a circle of radius $r$ is always perpendicular to the instantaneous velocity $v$. The work done by the force on the body in one complete rotation is :
State if each of the following statements is true or false. Give reasons for your answer.
$(a)$ In an elastic collision of two bodies, the momentum and energy of each body is conserved.
$(b)$ Total energy of a system is always conserved, no matter what internal and external forces on the body are present.
$(c)$ Work done in the motion of a body over a closed loop is zero for every force in nature.
$(d)$ In an inelastic collision, the final kinetic energy is always less than the initial kinetic energy of the system.
A block of mass $0.50\, kg$ is moving with a speed of $2.00\, ms^{-1}$ on a smooth surface. It strikes another mass of $1.00\, kg$ and then they move together as a single body. The energy loss during the collision is .............. $\mathrm{J}$
A body of mass $m$ is moving in a circle of radius $r$ with a constant speed $v$. The force on the body is $\frac{mv^2}{r}$ and is directed towards the centre. What is the work done by this force in moving the body over half the circumference of the circle