Particle $A$ moving with a velocity $v = 10 \ m/s$ experiences a head-on collision with a stationary particle $B$ of the same mass. As a result of the collision,the kinetic energy of the system decreases by $1 \%$. The speed of particle $A$ after the collision is: (in $m/s$)

$A$ ball of mass $m$ moving with velocity $v$ collides head-on with a second ball of mass $m$ at rest. If the coefficient of restitution is $e$,the velocity of the first ball after collision is $v_1$,and the velocity of the second ball after collision is $v_2$,then:

$A$ particle of mass $m$ moving with speed $2v$ collides with a mass $2m$ moving with speed $v$ in the same direction. After the collision,the first mass is stopped completely,while the second one splits into two particles each of mass $m$,which move at an angle of $45^o$ with respect to the original direction. The speed of each of the moving particles will be:

An electron collides with a free molecule initially in its ground state. The collision leaves the molecule in an excited state that is metastable and does not decay to the ground state by radiation. Let $K$ be the sum of the initial kinetic energies of the electron and the molecule,and $p$ be the sum of their initial momenta. Let $K^{\prime}$ and $p^{\prime}$ represent the same physical quantities after the collision. Then,

$A$ ball is dropped from a height $h$. It bounces repeatedly off the ground. What is the total distance covered by the ball before it comes to rest?

$A$ ball of mass $M$ falls from a height $h$ on a floor. If the coefficient of restitution is $e$,the height attained by the ball after two rebounds is