A ballistic pendulum was used to measured the speeds of bullets before electroni

Question

A ballistic pendulum was used to measured the speeds of bullets before electronic timing devices were developed. A large wooden block (m=5.40kg) is hanging from a rod forming a pendulum. A bullet (m=19.5g) is fired into the block and quickly stops within the block. The block + bullet then swing upward and their center of mass rises a vertical distance h=6.30cm before the pendulum swings back down. What is the speed of the bullet just prior to the collision?

A: This is a two-stage problem. First there is a collection between blovk and bullet. What is conserved in a collision?

B: Then, the block+bullet swing upward. What is conserved during the upward swing?

C: To solve the problem, first consider the block+bullet. At what speed were they traveling in order for the pendulum to rise distance h?

D. The answer to part C is the soeed of the block + bullet after the collision. now calculate the speed of the bullet before the collision

Explanation / Answer

Just after the collision, the bullet + block have speed V. Applying the conservation of linear momentum to the collisions, we have

mv = (M + m)V

Since the bullet and block stick together, the collisions is completely inelastic, and kinetic energy is not conserved during it. However, after the collision, mechanical energy is conserved, because no force then acts to dissipate that energy. So the kinetic energy of the system when the block is at the bottom of its arc must equal the potential energy of the system when the block is at the top:

1/2(M + m)V 2 = (M + m)gh

Eliminating V between these two equations leads to

v = [(M + m)/m]*sqrt(2gh)

v = [(5.4 + 0.0195)/0.0195]*sqrt(2*9.8*0.063) = 308.83 m/s

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