Uncategorized: Practice Problem & Solution
A ball of mass \(0.15 text{ kg}\) is dropped from a height \(10 text{ m}\), strikes the ground and rebounds to the same height. The magnitude of impulse imparted to the ball is (\(g = 10 text{ m/s}^2\)) nearly: (2021)
Solution Explained:
To solve this problem, we apply the core principles of Uncategorized. Understanding the underlying formula is key to arriving at the correct answer below:
Speed before impact \(v = sqrt{2gh} = sqrt{2 times 10 times 10} = 10sqrt{2} text{ m/s}\).Since it rebounds to the same height, speed after impact is also \(v = 10sqrt{2} text{ m/s}\).Impulse \(J = Delta P = m(v_{final} - v_{initial})\). Considering upward as positive, \(J = m(v - (-v)) = 2mv\).\(J = 2 times 0.15 times 10sqrt{2} = 3sqrt{2} approx 4.24 text{ Ns}\).
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