A 25.0 g bullet strikes a 0.600 kg block attached to a fixed horizontal spring w

Question

A 25.0 g bullet strikes a 0.600 kg block attached to a fixed horizontal spring whose spring stiffness constant is 7.70*10^3 N/m. The block is set into vibration with an amplitude of 21.5 cm. What was the speed of the bullet before impact if the bullet and block move together after impact? Part B: Determine the time for the bullet-block combination to first reach the spring maximum amplitude position immediately after impact.

Explanation / Answer

As the bullet and the block move together after the impact, the impact is inelastic and hence their kinetic energies are not conserved. However, their linear momentum will be conserved. Total mechanical energy of the bullet-block system = (1/2) kA^2 = (1/2) * 7.7 x 10^3 * (0.215)^2 J = 177.96625 J If v = their combined velocity after the impact (1/2) mv^2 = 177.96625 => (1/2) (0.600 + 0.021) v^2 = 177.96625 => v = v[(2 x 177.96625) / (0.621)] = 23.94 m/s If u = velocity of the bullet, then by the law of conservation of linear momentum, 0.021 u + 0 = (0.600 + 0.021) * (23.94) => u = (0.621 * 23.94) / (0.021) m/s = 707.96 m/s.

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