The mass of a person is $60 \, kg$. If he gains $10^5 \, \text{calories}$ of heat energy from food and the efficiency of his body is $28 \%$,then up to what height can he climb? (approximately) ($g = 9.8 \, m/s^2$,$J = 4.2 \, J/\text{cal}$)

Two bodies of different masses $m_1$ and $m_2$ have equal momenta. Their kinetic energies $E_1$ and $E_2$ are in the ratio:

In the figure shown,two identical balls of mass $M$ and radius $R$ each are placed in contact with each other on a frictionless horizontal surface. $A$ third ball of mass $M$ and radius $R$ is coming down vertically with a velocity $v_0$. It simultaneously hits the two balls. The smaller ball (top ball) does not stop after the collision but continues to move downwards with a speed $v_0/2$. Find the speed of each bigger ball after the collision.

$A$ block of mass $m$ starts from rest and slides down a frictionless semi-circular track from a height $h$ as shown. When it reaches the lowest point of the track,it collides with a stationary piece of putty also having mass $m$. If the block and the putty stick together and continue to slide,the maximum height that the block-putty system could reach is:

Two wooden blocks of mass $M_1$ and $M_2$ rest on a frictionless table. $A$ bullet of mass $m$ is fired at $M_1$ with speed $v$,which gets embedded in it,and the two together finally collide with $M_2$. Find the velocity of $M_2$ after the collision. (Assume the collision between the $(M_1+m)$ system and $M_2$ is elastic and treat the problem as one-dimensional.)

$A$ 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?