$A$ block of mass $1.9\, kg$ is at rest at the edge of a table of height $1\, m$. $A$ bullet of mass $0.1\, kg$ collides with the block and sticks to it. If the velocity of the bullet is $20\, m/s$ in the horizontal direction just before the collision,then the kinetic energy just before the combined system strikes the floor is $....J$. (Take $g = 10\, m/s^2$. Assume there is no rotational motion and loss of energy after the collision is negligible.)

$A$ block of mass $m$ starts at rest at height $h$ on a frictionless inclined plane. The block slides down the plane,travels across a rough horizontal surface with coefficient of kinetic friction $\mu$,and compresses a spring with force constant $k$ a distance $x$ before momentarily coming to rest. Then the spring extends and the block travels back across the rough surface,sliding up the plane. The block travels a total distance $d$ on the rough horizontal surface. The correct expression for the maximum height $h'$ that the block reaches on its return is

The quantities remaining constant in a general collision are

$A$ body of mass $2 \, \text{kg}$ moving with a speed of $4 \, \text{m/s}$ makes an elastic collision with another body at rest and continues to move in the original direction but with one-fourth of its initial speed. The speed of the two-body centre of mass is $\frac{x}{10} \, \text{m/s}$. Then the value of $x$ is ..... .

$A$ tennis ball dropped from a height of $2 \,m$ rebounds only $1.5 \,m$ after hitting the ground. What fraction of its energy is lost in the impact?

Two identical blocks $A$ and $B$,each of mass $m$,resting on a smooth floor,are connected by a light spring of natural length $L$ and spring constant $k$. The spring is at its natural length. $A$ third identical block $C$ (mass $m$) moving with a speed $v$ along the line joining $A$ and $B$ collides with $A$. The maximum compression in the spring is proportional to