Rotational MotionHard
Question
A rod of mass ′m′ is suspended vertically from a hinge A and is free to rotate along a horizontal axis passing through A as shown. A ball of mass m moving horizontally with speed u hits the bottom of the vertical rod elastically at t = 0 sec. Then the linear momentum of the ball and rod system after the collision
Options
A.is same as before the collision
B.becomes lesser than before collision
C.becomes more than before the collision
D.no inference can be drawn
Solution
After collision end of the rod which is at hinge A try to move left so hinge provide force to the rod towards right.
Proof:
Conservation of angular momentum: let U′ is final velocity of block.
mul = m u′l + Iω
Conservation of energy
1/2 mu2 = 1/2 mu′2 + 1/2 Iω2
solving these equation:
We get Lω = 3/2 V
Vcom = 3/4V
V′ = V/2
∴ net linear momentum of the system will be :
3/4MV + MV/2 = 5MV/4
Thus linear momentum will increase
Create a free account to view solution
View Solution FreeMore Rotational Motion Questions
Two disc one of density 7.2 g/cm3 and the other of density 8.9 g/cm3 are of same mass and thickness. Their moments of in...For rotational motion, the Newton′s second law of motion is indicated by :-...A round uniform body of radius R, mass M and moment of inertia ′I′, rolls down (without slipping) an incline...Two discs of same moment of inertia rotating about their regular axis passing through centre and perpendicular to the pl...A circular disk of moment of inertia It is roatating in a horizontal plane, about the symmetry axis, another disc of mom...