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JEE Main Physics question

PhysicsSystems of Particles and Rotational MotionJEE Main

A golf ball of mass 0.05 kg, placed on a tee, is struck by a golf club. The speed of the golf ball as it leaves the tee is 100 ms⁻¹, and the time of contact between them is 0.002 s. If the force decreases to zero linearly with time, then the force at the beginning of the contact is

Answer: (A)

Solution

Impulse = mv = 0.05×100 = 5 N·s. Since the force decreases linearly from F₀ to zero over the contact time t, the impulse equals the area of this triangular F–t profile: Impulse = ½F₀t ⟹ F₀ = 2(mv)/t = 2(5)/0.002 = 5000 N. (The contact time is

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Question ID VK-153098