A particle of mass $m$ moving with a speed $v$ hits elastically another stationary particle of mass $2\ m$ in a fixed smooth horizontal circular tube of radius $R$. Find the time when the next collision will take place?
$\frac{{\pi R}}{v}$
$\frac{{2\pi R}}{v}$
$\frac{{4\pi R}}{v}$
$\frac{{6\pi R}}{v}$
Four particles $A, B, C$ and $D$ of equal mass are placed at four corners of a square. They move with equal uniform speed $v$ towards the intersection of the diagonals. After collision, $A$ comes to rest, $B$ traces its path back with same speed and $C$ and $D$ move with equal speeds. What is the velocity of $C$ after collision
A ball of mass $m$ moving with speed $u$ collides with a smooth horizontal surface at angle $\theta$ with it as shown in figure. The magnitude of impulse imparted to surface by ball is [Coefficient of restitution of collision is $e$]
Two particles of equal mass $\mathrm{m}$ have respective initial velocities $u\hat{i}$ and $u\left(\frac{\hat{\mathrm{i}}+ \hat{\mathrm{j}}}{2}\right) .$ They collide completely inelastically. The energy lost in the process is
A body of mass $m$ is at rest. Another body of same mass moving with velocity $ V $ makes head on elastic collision with the first body. After collision the first body starts to move with velocity
Two masses $m_1$ and $m_2$ are connected by a string of length $l$. They are held in a horizontal plane at a height $H$ above two heavy plates $A$ and $B$ made of different material placed on the floor. Initially distance between two masses is $a < l$. When the masses are released under gravity they make collision with $A$ and $B$ with coefficient of restitution $0.8$ and $0.4$ respectively. The time after the collision when the string becomes tight is :- (Assume $H>>l$)