A pendulum with a mass of 1kg. on a light string 1 meter long is dropped from a

Question

A pendulum with a mass of 1kg. on a light string 1 meter long is dropped from a position with the string held level falling until it as its lowest point. At that instant it is struck by a bullet moving in precisely the opposite direction. The bullet has a mass of 50 grams, and it is noted that the two objects which stick together continue in the direction of the pendulum's motion but rise only half as high as the starting point. What is the velocity of the bullet how much energy is lost (converted to non mechanical form) during the collision.

Explanation / Answer

Initial State: PE = mgy PE = (1)(9.8)(1) PE = 9.8 J KE = 0.5mv^2 KE = 0.5(1)(0^2) KE = 0 ME = PE + KE = 9.8 J Final State: ME = 9.8 J (no frictional losses) PE = mgy PE = mg(0) PE = 0 therefore, all PE is converted to KE KE = 9.8 J 0.5mvp^2 = 9.8 vp^2 = 19.6 vp = 4.43 m/s Now, the collision takes place and momentum must be conserved in this perfectly inelastic collision: (mp)(vp) + (mb)(vb) = (mb + mp)v (1)(-4.43) + (.05)(vb) = (1 + .05)v Now, the pendulum and bullet combination rise back up to a point that is only half as high as before. Again KE will be converted to PE: KE = 0.5(1+0.05)(v^2) = PE 0.5(1.05)(v^2) = (1+0.05)(9.8)(0.5) v = 3.13 m/s Substitute into the conservation of momentum equation: -4.43 + 0.05vb = 1.05(3.13) vb = 154.33 m/s Energy before the collision: KE = 9.8 + 0.5(.05)(154.33)^2 KE = 605.24 Energy after the collision: KE = 0.5(1+0.05)(3.13)^2 KE = 5.14 J KE lost = 605.24 - 5.14 KE lost = 600.1 J BOL (note, the work written has some mistakes, namely in them calculating the speed of the bullet)

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