A bullet of mass $m$ strikes a block of mass $M$ connected to a light spring of stiffness $k,$ with a speed $v_0.$ If the bullet gets embedded in the block then, the maximum compression in the spring is
${\left( {\frac{{{m^2}v_0^2}}{{(M + m)k}}} \right)^{1/2}}$
${\left( {\frac{{Mmv_0^2}}{{2(M + m)k}}} \right)^{1/2}}$
${\left( {\frac{{mv_0^2}}{{2(M + m)k}}} \right)^{1/2}}$
${\left( {\frac{{Mv^2}}{{(M + m)k}}} \right)^{1/2}}$
As shown in figure there is a spring block system. Block of mass $500\,g$ is pressed against a horizontal spring fixed at one end to compress the spring through $5.0\,cm$ . The spring constant is $500\,N/m$ . When released, calculate the distance where it will hit the ground $4\,m$ below the spring ? $(g = 10\,m/s^2)$
A block is attached to a spring as shown and very-very gradually lowered so that finally spring expands by $"d"$. If same block is attached to spring & released suddenly then maximum expansion in spring will be-
A block of mass $m$, lying on a smooth horizontal surface, is attached to a spring (of negligible mass) of spring constant $k$. The other end of the spring is fixed, as shown in the figure. The block is initially at rest in a equilibrium position. If now the block is pulled with a constant force $F$, the maximum speed of the block is
A block $C$ of mass $m$ is moving with velocity $v_0$ and collides elastically with block $A$ of mass $m$ which connected to another block $B$ of mass $2\,m$ through a spring of spring constant $k$. What is $k$ if $x_0$ is the compression of spring when velocity of $A$ and $B$ is same?
$A$ block of mass $m$ moving with a velocity $v_0$ on a smooth horizontal surface strikes and compresses a spring of stiffness $k$ till mass comes to rest as shown in the figure. This phenomenon is observed by two observers:
$A$: standing on the horizontal surface
$B$: standing on the block To an observer
$A$, the work done by the normal reaction $N$ between the block and the spring on the block is