$A$ car comes to rest after traveling a distance $s$ under a constant retarding force $F$. If the mass of the car is increased by $50\%$,at what distance will the car come to rest,assuming the same retarding force $F$ and initial velocity $v$?

$A$ body of mass $m$ is travelling with a velocity $u$. When a constant retarding force $F$ is applied,it comes to rest after travelling a distance $s_{1}$. If the initial velocity is $2u$,with the same force $F$,the distance travelled before it comes to rest is $s_{2}$. Then,

$A$ body dropped from height $H$ reaches the ground with a speed of $1.1 \sqrt{gH}$. Calculate the work done by air friction in terms of $mgH$.

Two monkeys with the same mass stand on a branch at height $h$ above the horizontal jungle floor. Monkey $A$ steps off the branch holding the end of an inextensible rope of length $L$ whose other end is tied to another branch at height $H$,lets go at the bottom of the swing,and falls freely to the floor,as shown below. Monkey $B$ steps off and falls straight downward. Then,neglecting air resistance but not the tension in the rope,the total work $W$ done on each monkey and the speed $v$ with which each hits the floor are as follows:

$A$ simple pendulum is released from $A$ as shown in the figure. If $m$ and $l$ represent the mass of the bob and the length of the pendulum respectively,the gain in kinetic energy at $B$ is

$A$ ball is moving to and fro about the lowest point $A$ of a smooth hemispherical bowl. If it is able to rise up to a height of $20 \,cm$ on either side of $A$,its speed at $A$ must be .......... $m/s$ (Take $g = 10 \,m/s^2$,mass of the body = $5 \,g$).