To simulate car accidents,auto manufacturers study the collisions of moving cars with mounted springs of different spring constants. Consider a typical simulation with a car of mass $1000 \; kg$ moving with a speed $18.0 \; km/h$ on a rough road having $\mu = 0.5$ and colliding with a horizontally mounted spring of spring constant $6.25 \times 10^{3} \; N m^{-1}$. What is the maximum compression of the spring in $m$?

Consider two blocks $A$ and $B$ of masses $m_1 = 10 \ kg$ and $m_2 = 5 \ kg$ that are placed on a frictionless table. The block $A$ moves with a constant speed $v = 3 \ m/s$ towards the block $B$ kept at rest. $A$ spring with spring constant $k = 3000 \ N/m$ is attached to the block $B$ as shown in the figure. After the collision,suppose that the blocks $A$ and $B$,along with the spring in a constant compression state,move together. Then the compression in the spring is (neglect the mass of the spring). (in $m$)

Four particles $A, B, C$ and $D$ of equal mass $m$ are placed at four corners of a square. They move with equal uniform speed $v$ towards the intersection of the diagonals. After collision,$A$ comes to rest,$B$ traces its path back with the same speed $v$,and $C$ and $D$ move with equal speeds $v'$. What is the velocity of $C$ after collision?

$A$ spring-block system is resting on a frictionless floor as shown in the figure. The spring constant is $2.0 \ N \ m^{-1}$ and the mass of the block is $2.0 \ kg$. Ignore the mass of the spring. Initially,the spring is in an unstretched condition. Another block of mass $1.0 \ kg$ moving with a speed of $2.0 \ m \ s^{-1}$ collides elastically with the first block. The collision is such that the $2.0 \ kg$ block does not hit the wall. The distance,in metres,between the two blocks when the spring returns to its unstretched position for the first time after the collision is . . . . .

$A$ pendulum of length $1 \,m$ and having a bob of mass $1 \,g$ is pulled aside through an angle $60^{\circ}$ with the vertical and then released. The power delivered by all the forces acting on the bob when the pendulum makes $30^{\circ}$ with the vertical is . . . . . . $\left(g=10 \,ms^{-2}\right)$ (in $\,mW$)

$A$ 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 $,m$)?