Definition of Force Inertia and First Law: 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
Short Solution (Best for NEET):
To solve this problem, we apply the core principles of Definition of Force Inertia and First Law. Understanding the underlying formula is key to arriving at the correct answer below:
The velocity of the ball just before striking the ground is \(u = \sqrt{2gh} = \sqrt{2 \times 10 \times 10} = 14.14\text{ m/s}\). Since it rebounds to the same height, its velocity just after is \(v = 14.14\text{ m/s}\) upwards. The change in momentum is \(Delta p = m(v - (-u)) = 2mu = 2 \times 0.15 \times 14.14 \approx 4.24\text{ kg m/s}\).
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