A ball of mass $1\,\,kg.$  moving with a velocity of $4\,\,m/sec.$  collides with a stationary ball. The collision is oblique. After the collision the first moves at right angle to its, initial direction with a velocity of  $3\,\,m/s.$  The momentum of the second ball (in $kg.\,\,m/s.$ ) after collision would be nearly

A bullet when fired at a target with a velocity of $100\,\,m/sec$ penetrates one metre into it. If the bullet is fired at a similar target with a thickness $0.5\,\,metre,$  then it will emerge from it with a velocity of

$A$ flexible chain of length $2m$ and mass $1kg$ initially held in vertical position such that its lower end just touches a horizontal surface, is released from rest at time $t = 0$. Assuming that any part of chain which strikes the plane immediately comes to rest and that the portion of chain lying on horizontal surface does not from any heap, the height of its centre of mass above surface at any instant $t = 1/\sqrt 5$ befor it completely comes to rest) is ................ $\mathrm{m}$

In $a$ one-dimensional collision, $a$ particle of mass $2m$ collides with $a$ particle of mass $m$ at rest. If the particles stick together after the collision, what fraction of the initial kinetic energy is lost in the collision?

A mass $m_1$ moves with a great velocity. It strikes another mass $m _2$ at rest in a head on collision. It comes back along its path with low speed, after collision. Then

This question has statement $1$ and statement $2$ . Of the four choices given after the statements, choose the one that best describes the two statements. 
Statement $- 1$: A point particle of mass m moving with speed $u$ collides with stationary point particle of mass $M$. If the maximum energy loss possible is given as $f$ $\left( {\frac{1}{2}m{v^2}} \right)$ then $ f = \left( {\frac{m}{{M + m}}} \right)$ 

Statement $-2$: Maximum energy loss occurs when the particles get stuck together as a result of the collision.