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A 2
A 2.50 g bullet, traveling at a speed of 425 m/s, strikes the wooden block of a ballistic pendulum. The block has a mass of 215 g. (a)What is the speed of the bullet/block combination immediately after the collision. (b)How high does the combination rise above its initial position?
Expert Solution
Please see the attached file for detailed solution.
(a) Using the conservation of momentum, the initial velocity of the bullet v0=425m/s, the block is 0.
After the collision, the speed of the combination is vf.
(b) After the collision, the energy is conserved for the combination system.
At the moment just after the collision, the total energy is only the kinetic energy. At the highest point, the speed is 0, that is, the kinetic energy is 0. The total energy is equal to the potential energy.
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