A particle P strikes the rod R perpendicularly as shown. The rod is suspended vertically with upper end hinged. Arrangement For the collision (A) x = \frac{ℓ}{2}, elastic collision (P) Linear momentum of P, R system increases (B) x = ℓ, elastic collision (Q) Linear momentum of P, R system decreases (C) x = \frac{ℓ}{2}, P sticks to R (R) KE of the P decreases (D) x = ℓ, P sticks to R (S) Angular momentum of the P, R system is conserved about hinge.

Engineering

Physics

Conservation of Angular Momentum

Question

A particle P strikes the rod R perpendicularly as shown. The rod is suspended vertically with upper end hinged.

Arrangement	For the collision
(A) x = $\frac{ℓ}{2}$ , elastic collision	(P) Linear momentum of P, R system increases
(B) x = ℓ, elastic collision	(Q) Linear momentum of P, R system decreases
(C) x = $\frac{ℓ}{2}$ , P sticks to R	(R) KE of the P decreases
(D) x = ℓ, P sticks to R	(S) Angular momentum of the P, R system is conserved about hinge.

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(A) Angular momentum conservation about A
mv₀ $\frac{ℓ}{2}$ = Iw + mv₁ $\frac{ℓ}{2}$ ...(1)

   e = $\frac{\frac{ωℓ}{2} - v_{1}}{v_{0}}$ = 1  ⟹ $\frac{ωℓ}{2}$ – v₁ = v₀   ⟹ v₁ = $\frac{ωℓ}{2}$ – v₀       ...(2)
   Solving (1) & (2)
           v₁ = $\frac{v_{0} [3 m - 4 M]}{[3 m + 4 M]}$ < v₀ ∴ KE of P decreases
           ω = $\frac{12 {mv}_{0}}{l [3 m + 4 M]}$
   On solving linear momentum of P₁R system decreases.
(B)    A.M about A
   mv₀ℓ = Iω + mv₁ℓ   ...(1)
   I = $\frac{ω l - v_{1}}{v_{0}}$     ...(2)
Solving (1) & (2)
ω = $\frac{6 {mv}_{0}}{l [3 m + M]}$ , v₁ = $\frac{v_{0} [3 m - M]}{[3 m + M]}$ < v₀   ∴ KE decreases

On solving linear momentum increases
(C)   A.M. about A
mv₀ $\frac{ℓ}{2}$ = Iω + mv₁ $\frac{ℓ}{2}$        ...(1)    $[I = \frac{M l^{2}}{3}]$
       $\frac{ω l}{2}$ – v₁ = 0           ...(2)
   Solving,
     ω = $\frac{6 {mv}_{0}}{l [3 m + 4 M]}$ ,   v₁ = $\frac{3 {mv}_{0}}{[3 m + 4 M]}$ < v₀   ∴ KE decrease
∴  Linear momentum of P₁R system decreases
(D)  A.M. about A
mv₀ℓ = Iω + mv₁ℓ       ...(1)   $[I = \frac{M l^{2}}{3}]$
ωℓ = v₁               ...(2)
Solving,
ω = $\frac{3 {mv}_{0}}{[M + 3 m] l}$ , v₁ = $\frac{3 {mv}_{0}}{[M + 3 m]}$ < v₀     ∴ KE decreases
∴ Linear momentum increases

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