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joex444
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A bullet, 10g, hits a pendulum, 2kg. The center of mass of the pendulum rises a vertical distance of 12cm. Assuming the bullet embeds itself in the pendulum, calculate the bullet's initial speed.
What I did was assume that at 12cm vertical distance the pendulum and bullet are momentarily at 0m/s velocity, and all their energy is gravitational potential, so mgh = 2*9.8*0.12 = 2.36J. The bullet's inital speed could be found by assuming the pendulum was at rest before being shot, and its GPE was 0. So, 2.36J came from the bullet, and 1/2mv^2 shows that v^2=472m^2/s^2. Taking the square root, v=21.7m/s.
Now, my professor says the answer is 308m/s. 21.7m/s is quite slow for a bullet...
What I did was assume that at 12cm vertical distance the pendulum and bullet are momentarily at 0m/s velocity, and all their energy is gravitational potential, so mgh = 2*9.8*0.12 = 2.36J. The bullet's inital speed could be found by assuming the pendulum was at rest before being shot, and its GPE was 0. So, 2.36J came from the bullet, and 1/2mv^2 shows that v^2=472m^2/s^2. Taking the square root, v=21.7m/s.
Now, my professor says the answer is 308m/s. 21.7m/s is quite slow for a bullet...