A bullet of mass m is fired with a velocity of 50 m s - 1 at an angle θ with the horizontal. At the highest point of its trajectory, it collides head on with a bob connected to massless string of length l and gets embedded in the bob of mass 3 m . After the collision, the string moves to an angle of 120 ° . What is the angle θ ?

Question

A bullet of mass m is fired with a velocity of 50 m s - 1 at an angle θ with the horizontal. At the highest point of its trajectory, it collides head on with a bob connected to massless string of length l and gets embedded in the bob of mass 3 m . After the collision, the string moves to an angle of 120 ° . What is the angle θ ?, cos - 1 ⁡ 4 5, cos - 1 ⁡ 5 4, sin - 1 ⁡ 4 5, sin - 1 ⁡ 5 4

MARKS App · Accepted Answer

Velocity of bullet at highest point of its trajectory = 50 cos ⁡ θ in horizontal direction. As bullet of mass m collides with pendulum bob of mass 3 m and two stick together, their common velocity v ′ = 50 m cos ⁡ θ m + 3 m = 25 2 cos ⁡ θ m s - 1 As now under this velocity v ′ pendulum bob goes up to an angle 120 ° , hence v ′ 2 2 g = h = l 1 - cos ⁡ 120 ° = 10 3 1 - f - 1 2 = 5 ⇒ v ′ 2 = 2 × 10 × 5 = 100 or v ′ = 10 Comparing two answer of v ′ , we get 25 2 cos ⁡ θ = 10 ⇒ cos ⁡ θ = 4 5 or θ = cos - 1 ⁡ 4 5

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