$A$ truck of mass $M$ and a car of mass $\frac{M}{10}$ moving with the same momentum are brought to a halt by the application of the same braking force. The ratio of the distances travelled by the truck and the car before they come to a stop is

$A$ 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.)

$A$ body moving with velocity $v$ has momentum and kinetic energy numerically equal. What is the value of $v$ in $m/s$?

$A$ lift of mass $2000 \ kg$ starts from rest in the basement and moves to the fourth floor,which is at a height of $25 \ m$. As it passes the fourth floor,its speed is $3 \ ms^{-1}$. There is a constant frictional force of $500 \ N$ acting on it. Calculate the work done by the lift's motor in $kJ$.

$A$ block moving horizontally on a smooth surface with a speed of $40\, m/s$ splits into two parts with masses in the ratio of $1:2$. If the smaller part moves at $60\, m/s$ in the same direction,then the fractional change in kinetic energy is:

$A$ 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 $,m$)?