Question

Ballistic pendulum consisting of a block of wood suspended by some strings of negligible mass has a mass of 2.00 kg. It is penetrated to a depth of 12.0 cm by a 9.0 g bullet fired horizontally from a gun at close range. The pendulum€™s center of mass rises by a maximum 10.0 cm as a result.

What was the velocity of the bullet immediately before impact?

Answer 1

The velocity of the bullet immediately before impact was approximately 168 m/s.

To determine the velocity of the bullet immediately before impact, we can apply the principle of conservation of energy. The potential energy lost by the block-pendulum system as it rises is equal to the initial kinetic energy of the system.

The potential energy lost is given by the change in height of the center of mass, which is 10.0 cm or 0.1 m. The mass of the system is the sum of the mass of the block and the mass of the bullet, converted to kilograms (2.00 kg + 0.009 kg). The acceleration due to gravity is approximately 9.8 m/s².

Using the equation for potential energy U=mgh, where U is potential energy, m is mass, g is acceleration due to gravity, and h is height, we can calculate the potential energy lost.

The kinetic energy of the system before impact is given by KE= 1/2 mv^2, where m is the mass of the system, and v is the velocity of the bullet.

Setting the potential energy lost equal to the initial kinetic energy and solving for mgh= 1/2 mv^2

`v= √(2gh)`

Substituting the values,
`v=√(2*9.8cm^2*0.1m) ,`, gives a velocity of approximately 4.43 m/s. Converting this to cm/s (1 m = 100 cm), we get 443 cm/s. This is the velocity of the center of mass, which is also the velocity of the bullet immediately before impact since the block and bullet move together. Finally, converting this velocity to m/s (1 cm/s = 0.01 m/s) gives a final velocity of approximately 4.43 m/s.