$A$ ball is released from a height of $10 \, m$. If after the impact there is a loss of $40 \%$ in its energy,the ball shall rise up to ................. $m$.

$A$ small bob $A$ of mass $m$ is attached to a massless rigid rod of length $1 \ m$ pivoted at point $P$ and kept at an angle of $60^{\circ}$ with the vertical as shown in the figure. At a distance of $1 \ m$ below point $P$, an identical bob $B$ is kept at rest on a smooth horizontal surface that extends to a circular track of radius $R$ as shown in the figure. If bob $B$ just manages to complete the circular path of radius $R$ up to a point $Q$ after being hit elastically by bob $A$, then the radius $R$ is . . . . . . $m$.

If a lighter body (mass $M_1$ and velocity $V_1$) and a heavier body (mass $M_2$ and velocity $V_2$) have the same kinetic energy,then

Two bodies of masses $m_{1}$ and $m_{2}$ are acted upon by a constant force $F$ for a time $t$. They start from rest and acquire kinetic energies,$E_{1}$ and $E_{2}$ respectively. Then $\frac{E_{1}}{E_{2}}$ is

$A$ block of mass $m$ starts from rest and slides down a frictionless semi-circular track from a height $h$ as shown. When it reaches the lowest point of the track,it collides with a stationary piece of putty also having mass $m$. If the block and the putty stick together and continue to slide,the maximum height that the block-putty system could reach is:

$A$ bullet of mass $4\,g$ is fired horizontally with a speed of $300\,m/s$ into a $0.8\,kg$ block of wood at rest on a table. If the coefficient of friction between the block and the table is $0.3$,how far will the block slide approximately (in $,m$)?