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Pendulum hit by bullet

  • Thread starter joex444
  • Start date
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...
 

kp

53
0
the speed you have of 21.7m/s is the speed of the pendulum/bullet system. Energy is conserved.

you still have one more step to find the initial speed of the bullet, where energy is not conserved.
 
I think I got it. Momentum during the collision is conserved, so Velocity of the block+bullet = m(bullet)*v(bullet)/(m(bullet)+M(block)). Then I use that V in the KE and GPE conservation equation, and I did get 308m/s.
 

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