A uniform rod of mass m and length L is hinged at one end and free to rotate in the horizontal plane. All the surface are smooth. A particle of same mass m collides with the rod perpendicular to the length of rod with a speed V0 . The coefficient of restitution for the collision is e=(1/2) . If hinge reaction during the collision is zero then the value of x is:

Question

A uniform rod of mass m and length L is hinged at one end and free to rotate in the horizontal plane. All the surface are smooth. A particle of same mass m collides with the rod perpendicular to the length of rod with a speed V0 . The coefficient of restitution for the collision is e=(1/2) . If hinge reaction during the collision is zero then the value of x is: <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-wx5nfxj0yub2.png" /> (A) No such value of x is possible (B) x=(L/2) (C) x=(2 L/3) (D) x=L

Tardigrade Admin · Accepted Answer

Angular momentum conservation about the hinge
Solution

m v_{0} x = m v_{1} x + \frac{m L ^{2}}{3} ω

v_{0} x = v_{1} x + \frac{ω L ^{2}}{3}

...(i)

e = \frac{ω x - v _{1}}{v _{0}} = \frac{1}{2}

...(ii)
Impulse of hinge

J = mω \frac{L}{2} + m v_{1} - m v_{0}

J = mω \frac{L}{2} - m \frac{ω L ^{2}}{3 x}

J = 0

x = \frac{2 L}{3}

It is independent of 'e'

No such value of x is possible

$x = \frac{L}{2}$

$x = \frac{2 L}{3}$

$x = L$

No such value of x is possible

x=2L​

x=32L​

x=L

Angular momentum conservation about the hinge mv0​x=mv1​x+3mL2​ω v0​x=v1​x+3ωL2​ ...(i) e=v0​ωx−v1​​=21​ ...(ii) Impulse of hinge J=mω2L​+mv1​−mv0​ J=mω2L​−m3xωL2​ J=0 x=32L​ It is independent of 'e'

$x = \frac{L}{2}$

$x = \frac{2 L}{3}$

$x = L$