A 500g bullet is fired horizontally into a 1.20kg wooden block resting on a hori

Question

A 500g bullet is fired horizontally into a 1.20kg wooden block resting on a horizontal surface. The coefficient of kinetic friction between block and surface is 0.20. The bullet remains embedded in the block, which is observed to slide .230m along the surface before stopping. a) What is the initial speed of the bullet. b) What kind of collision took place between bullet and the block? Explain your anweer comparing the values of the kinetic energy before and after the collision c) Calculate the impulse of the block from the moment just after the impact to this final position. Analyze the result

Explanation / Answer

Answer --> Original velocity is 228.95 m/s How... Given: m1 = 0.005 kg (SI unit) m2 = 1.2 kg u2 = 0.0 m/s µ = 0.2 d = 0.23 m g = 9.81 m/s^2 Solve for the post-impact initial system velocity v2 = SQRT [2µgd] v2 = SQRT [ 2 * (0.2) * (9.81 m/s^2) * (0.23 m) ] v2 = SQRT [ 0.90252 m^2/s^2 ] v2 = 0.95 m/s Now find original u1 = { [m1+m2] * v2 } / [m1] u1 = { [ (0.005 kg) + (1.2 kg) ] * (0.95 m/s) } / [ 0.005 kg ] u1 = { [ 1.205 kg ] * (0.95 m/s) } / [ 0.005 kg ] u1 = { 1.14475 kg-m/s } / [ 0.005 kg ] u1 = 228.95 m/s

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