$A$ man of mass $60 \ kg$ gets $10^5 \ calories$ of energy from food. If his efficiency is $28\%$,up to what height in $m$ can he climb?

$A$ ball of mass $m$ falls vertically to the ground from a height $h_1$ and rebounds to a height $h_2$. The change in momentum of the ball on striking the ground is

$A$ block of mass $5 \ kg$ is lifted to a height of $5 \ m$ by a force of $60 \ N$. Find:
$(1)$ Work done by the force.
$(2)$ Potential energy of the block at $5 \ m$.
$(3)$ Kinetic energy of the block at $5 \ m$.
$(4)$ Velocity of the block at $5 \ m$. (Take $g = 9.8 \ m/s^2$)

$A$ ball of mass $0.2 \ kg$ rests on a vertical post of height $5 \ m$. $A$ bullet of mass $0.01 \ kg$,traveling with a velocity $V \ m/s$ in a horizontal direction,hits the centre of the ball. After the collision,the ball and bullet travel independently. The ball hits the ground at a distance of $20 \ m$ and the bullet at a distance of $100 \ m$ from the foot of the post. The initial velocity $V$ of the bullet is

$A$ and $B$ are blocks of same mass $m$ exactly equivalent to each other. Both are placed on a frictionless surface connected by a spring. The natural length of the spring is $L$ and the force constant is $K$. Initially,the spring is at its natural length. Another equivalent block $C$ of mass $m$ travelling at speed $v$ along the line joining $A$ and $B$ collides with $A$. In an ideal condition,the maximum compression of the spring is:

$A$ particle of mass $m$ moving eastward with a speed $v$ collides with another particle of same mass moving northward with same speed $v$. The two particles coalesce after collision. The new particle of mass $2m$ will move in north-east direction with a speed (in $m/s$) of: