A block of mass m = 1 kg slides with velocity v = 6 m s - 1 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

Step-by-step solution

Angular momentum mv l = m l 2 + 2 m l 2 3 ω mv l = 5 3 m l 2 ω ω = 3 v 5 l 1 2 Iω 2 = 2 mg l 2 1 - cos   θ + mg l 1 - cos   θ 1 2 5 3 m l 2 9 v 2 25 l 2 = 2 mg l   1 - cos   θ 3 5 × 2 mv 2 = 2 mg l 1 - cosθ 3 10 × 36 2 × 10 = 1 - cos   θ 1 - 27 50 = cos   θ cos   θ = 23 50 = 0 . 46 θ = cos - 1 0 . 46 = 63 °