A bullet of mass $m$ moving with velocity $v$ strikes a suspended wooden block of mass $M$. If the block rises to a height $h$, the initial velocity of the bullet will be
$\sqrt {2gh} $
$\frac{{(M + m)}}{m}\sqrt {2gh} $
$\frac{m}{{(M + m)}}\sqrt {2gh} $
$\frac{{(M - m)}}{m}\sqrt {2gh} $
A body of mass $2\, kg$ moving with a velocity of $3\, m/sec$ collides head on with a body of mass $1\, kg$ moving in opposite direction with a velocity of $4\, m/sec$. After collision, two bodies stick together and move with a common velocity which in $m/sec$ is equal to
$A$ particle of mass $m$ is released from $a$ height $H$ on $a$ smooth curved surface which ends into a vertical loop of radius $R$, as shown The minimum value of $H$ required so that the particle makes a complete vertical circle is given by
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
Two identical balls $A$ and $B$ are released from the positions shown in figure. They collide elastically on horizontal portion $MN$. The ratio of the heights attained by $A$ and $B$ after collision will be (neglect friction)
The total work done on a particle is equal to the change in its kinetic energy. This is applicable