$A$ pendulum of length $2\,m$ is released from point $P$ as shown in the figure. When it reaches point $Q$,it loses $10\%$ of its total energy due to air resistance. The velocity at $Q$ is .... $m/s$.

An object of mass $M$ is at rest on a smooth horizontal surface. Objects of different masses collide head-on elastically with the object of mass $M$. All colliding objects have the same fixed kinetic energy $E$,and in each case,mass $M$ is initially at rest. The kinetic energy transferred to the stationary mass $M$ depends on the linear momentum $P$ of the incoming colliding mass. How does the energy transferred to $M$ vary with the linear momentum $P$?

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 $m$)

$A$ body of mass $2 \; kg$ initially at rest moves under the action of an applied horizontal force of $7 \; N$ on a table with coefficient of kinetic friction $= 0.1$. Compute the
$(a)$ work done by the applied force in $10 \; s$,
$(b)$ work done by friction in $10 \; s$,
$(c)$ work done by the net force on the body in $10 \; s$,
$(d)$ change in kinetic energy of the body in $10 \; s$,
and interpret your results.

$A$ body of mass $4 \,kg$ is moving with momentum of $8 \,kg \,m/s$. $A$ force of $0.2 \,N$ acts on it in the direction of motion of the body for $10 \,s$. The increase in kinetic energy in joules is

In the case of circular motion of a body,if a tangential force also acts on the body in addition to a centripetal force,then the work done: