JEE Main · 2020 · 03 Sep 2020 · Shift 1 · Rotational Mechanics

A block of mass

m = 1

kg slides with velocity

v = 6

m/s on a frictionless horizontal surface and collides with a uniform vertical rod and sticks to it as shown. The rod is pivoted about O and swings as a result of the collision making angle

θ

before momentarily coming to rest. If the rod has mass

M = 2

kg, and length

ℓ = 1

m, the value of

θ

is approximately: (take

g = 10

m/s

^{2}

)

(A) $4 9^{\circ}$
(B) $5 5^{\circ}$
(C) $6 9^{\circ}$
(D) $6 3^{\circ}$ ✓ Correct

Answer: (D)

Solution

During the collision, angular momentum about the pivot

O

is conserved as the block sticks to the rod:

m v ℓ = (\frac{M ℓ ^{2}}{3} + m ℓ^{2}) ω_{0}

. After that, energy is conserved as the system swings up, the rotational kinetic energy raising the block through

ℓ (1 - cos θ)

and the rod's centre through

\frac{ℓ}{2} (1 - cos θ)

. Substituting

m = 1

kg,

M = 2

kg,

ℓ = 1

v = 6

m s

^{- 1}

and

g = 10

m s

^{- 2}

gives

1 - cos θ \approx 0.54

, so

cos θ \approx 0.46

and

θ \approx 6 3^{\circ}

JEE Main · 2020 · 03 Sep 2020 · Shift 1 · Single correct option

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JEE Main · 2020 · 03 Sep 2020 · Shift 1 · Rotational Mechanics

Solution

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