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 ? $(g = 10\,m/s^2)$
$1\,m$
$\sqrt 2\,m$
$\sqrt 3\,m$
$4\,m$
The system of the wedge and the block connected by a massless spring as shown in the figure is released with the spring in its natural length. Friction is absent. maximum elongation in the spring will be
A ball is dropped from a height of $80\,m$ on a surface which is at rest. Find the height attainded by ball after $2^{nd}$ collision if coefficient of restitution $e = 0.5$ ............ $\mathrm{m}$
Two springs have their force constant as $k_1$ and $k_2 (k_1 > k_2)$. when they are stretched by the same force
$A$ 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
According to observer $B$, the potential energy of the spring increases
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