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 ? $(g = 10\,m/s^2)$
$1\,m$
$\sqrt 2\,m$
$\sqrt 3\,m$
$4\,m$
A ball of mass $2 \,m$ and a system of two balls with equal masses $m$ connected by a massless spring, are placed on a smooth horizontal surface (see figure below). Initially, the ball of mass $2 \,m$ moves along the line passing through the centres of all the balls and the spring, whereas the system of two balls is at rest. Assuming that the collision between the individual balls is perfectly elastic, the ratio of vibrational energy stored in the system of two connected balls to the initial kinetic energy of the ball of mass $2 \,m$ is
A block is simply released from the top of an inclined plane as shown in the figure above. The maximum compression in the spring when the block hits the spring is :
Two similar springs $P$ and $Q$ have spring constants $K_P$ and $K_Q$, such that $K_P > K_Q .$ They are stretched first by the same amount (case $a$), then by the same force (case $b$). The work done by the springs $W_P$ and $W_Q$ are related as, in case $(a)$ and case $(b)$ respectively
Two springs $A$ and $B$ having spring constant $K_{A}$ and $K_{B}\left(K_{A}=2 K_{B}\right)$ are stretched by applying force of equal magnitude. If energy stored in spring $A$ is $E_{A}$ then energy stored in $B$ will be
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?