$A$ block of mass $M$ moving on a frictionless horizontal surface collides with a spring of spring constant $K$ and compresses it by a length $L$. The maximum momentum of the block during the collision process is

$A$ spring of force constant $K$ is first stretched by distance $a$ from its natural length and then further by distance $b$. The work done in stretching the part $b$ is .............

Two springs of spring constants $K$ and $2K$ are stretched by the same force. If $W_{1}$ and $W_{2}$ are the energies stored in them respectively,then:

When a spring is stretched by $2 \ cm$,the potential energy stored is $U$. If it is stretched by $10 \ cm$,the potential energy stored will be:

Show that the law of conservation of mechanical energy is obeyed by pulling or compressing the block tied at the end of a spring.

$A$ spring has a natural length $l$ with one end fixed to the ceiling. The other end is fitted with a smooth ring which can slide on a horizontal rod fixed at distance $l$ below the ceiling. Initially, the spring makes an angle of $60^{\circ}$ with the vertical, when the system is released from rest. Find the angle of the spring with the vertical, when the velocity of the ring reaches half of the maximum velocity, which the ring can attain during the motion.