A particle of mass $m_1$ is moving with a velocity $v_1$ and another particle of mass $m_2$ is moving with a velocity $v_2$ . Both of them have the same momentum but their different kinetic energies are $E_1$ and $E_2$ respectively. If $m_1 > m_2$ then
$E_1 < E_2$
$\frac{{{E_1}}}{{{E_2}}}\, = \,\frac{{{m_1}}}{{{m_2}}}$
$E_1 > E_2$
$E_1 = E_2$
$A$ block of mass $m$ is hung vertically from an elastic thread of force constant $mg/a$. Initially the thread was at its natural length and the block is allowed to fall freely. The kinetic energy of the block when it passes through the equilibrium position will be :
A body of mass $10\,kg$ moving with speed of $3 \,ms ^{-1}$ collides with another stationary body of mass $5 \,kg$. As a result, the two bodies stick together. The $KE$ of composite mass will be .......... $J$
Equal force $F ( > mg)$ is applied to string in all the $3$ cases. Starting from rest, the point of application of force moves a distance of $2 m$ down in all cases. In which case the block has maximum kinetic energy?
If velocity of a body is twice of previous velocity, then kinetic energy will become
Write the definition of work and kinetic energy.