A bullet of mass $m$ moving with velocity $v$ strikes a block of mass $M$ at rest and gets embedded into it. The kinetic energy of the composite block will be
$\frac{1}{2}m{v^2} \times \frac{m}{{(m + M)}}$
$\frac{1}{2}m{v^2} \times \frac{M}{{(m + M)}}$
$\frac{1}{2}m{v^2} \times \frac{M+m}{{(M)}}$
$\frac{1}{2}m{v^2} \times \frac{2m}{{( M+m)}}$
Work done in time $t$ on a body of mass $m$ which is accelerated from rest to a speed $v$ in time $t_1$ as a function of time $t$ is given by
A force $F = - K(yi + xj)$ (where K is a positive constant) acts on a particle moving in the xy-plane. Starting from the origin, the particle is taken along the positive x-axis to the point $(a, 0)$ and then parallel to the y-axis to the point $(a, a)$. The total work done by the force F on the particles is
A body of mass $m$ moving with velocity $v$ collides head on with another body of mass $2\, m$ which is initially at rest. The ratio of $K.E.$ of the colliding body before and after collision will be
Work done in time $t$ on a body of mass $m$ which is accelerated from rest to a speed $v$ in time $t_1$ as a function of time $t$ is given by
When the momentum of a body increases by $100\%$, its $KE$ increases by .............. $\%$