Two identical particles are moving with same velocity $v$ as shown in figure. If the collision is completely inelastic then
The velocity of separation is zero
The velocity of approach is $2\,v \sin\, (\theta /2)$
The common velocity after collision is $v \cos\,(\theta /2)$
All of the above
Pulley and spring are massless and the friction is absent everwhere. $5\,kg$ block is released from rest. The speed of $5\,kg$ block when $2\,kg$ block leaves the contact with ground is (take force constant of the spring $K = 40\,N/m$ and $g = 10\,m/s^2$ )
The force acting on a body moving along $x-$ axis varies with the position of the particle as shown in the figure. The body is in stable equilibrium at
Two blocks $A$ and $B$ of masses $1\,\,kg$ and $2\,\,kg$ are connected together by a spring and are resting on a horizontal surface. The blocks are pulled apart so as to stretch the spring and then released. The ratio of $K.E.s$ of both the blocks is
A ball $P$ collides with another identical ball $Q$ at rest. For what value of coefficient of restitution $e$, the velocity of ball $Q$ become two times that of ball $P$ after collision
A body of mass $2\, kg$ slides down a curved track which is quadrant of a circle of radius $1$ $meter$ as shown in figure. All the surfaces are frictionless. If the body starts from rest, its speed at the bottom of the track is ............. $\mathrm{m}/ \mathrm{s}$