$A$ bullet of mass $0.01\,kg$ and travelling at a speed of $500\,m/s$ strikes a block of mass $2\,kg$ which is suspended by a string of length $5\,m$. The centre of gravity of the block is found to rise a vertical distance of $0.1\,m$. What is the speed of the bullet after it emerges from the block?

The linear momentum of a body of mass $8 \ kg$ is $24 \ kg \ m \ s^{-1}$. If a constant force of $24 \ N$ acts on the body in the direction of motion of the body for a time of $3 \ s$,then the increase in the kinetic energy of the body is (in $J$)

$A$ 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.)

$A$ force exerts an impulse $I$ on a particle changing its speed from $u$ to $2u$. The applied force and the initial velocity are oppositely directed along the same line. The work done by the force is

Which of the following graphs correctly represents the relationship between kinetic energy $(E)$,potential energy $(U)$,and height $(h)$ from the ground for a particle moving vertically?

$A$ wooden block of mass $M$ is suspended by a cord and is at rest. $A$ bullet of mass $m$,moving with a velocity $v$,passes through the block and emerges with a velocity $v/2$ in the same direction. If the block rises to a height $h$,calculate the value of $h$.