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
$\frac{1}{2}m{v^2} \times \frac{m}{{(m + M)}}$
$\frac{1}{2}m{v^2} \times \frac{M}{{(m + M)}}$
$\frac{1}{2}m{v^2} \times \frac{M+m}{{(M)}}$
$\frac{1}{2}m{v^2} \times \frac{2m}{{( M+m)}}$
The diagram to the right shows the velocity-time graph for two masses $R$ and $S$ that collided elastically. Which of the following statements is true?
$(I)$ $R$ and $S$ moved in the same direction after the collision.
$(II)$ Kinetic energy of the system $(R$ & $S)$ is minimum at $t = 2$ milli sec.
$(III)$ The mass of $R$ was greater than mass of $S.$
A ball of mass $m$ is dropped from a heigh $h$ on a platform fixed at the top of a vertical spring, as shown in figure. The platform is depressed by a distance $x.$ Then the spring constant is
$2$ particles of mass $1\,Kg$ and $5\,kg$ have same momentum, calculate ratio of their $K.E.$
Work equal to $25\,J$ is done on a mass of $2\,kg$ to set it in motion. If whole of it is used to increase the kinetic energy then velocity acquired by the mass is ............ $\mathrm{m}/ \mathrm{s}$
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 :