Question

5. Someone fires a 0.04 kg bullet at a block of wood that has a mass of 0.5 kg. (The block of wood is sitting on a frictionless surface, so it moves freely when the bullet hits it). The wood block is initially at rest. The bullet is traveling 300 m/s when it hits the wood block and sticks inside it. Now the bullet and the wood block move together as one object. How fast are they traveling?

Answer 1

22.22 m/s

Step-by-step explanation:

Using the law of conservation of momentum:

Initial momentum of the system = final momentum of the system.

mu+m'u' = mv+m'v'

where u and u' are the initial velocities.

v and v' are the final velocities.

Then do the math= 0.04 kg × 300 m/s + 0 = 0.54 kg × v

the velocity is 22.22 m/s

Then when you do the math the bullet and the wood block move together as one object with the speed 22.2 m/s.

Answer 2

Answer: 22.22 m/s

Explanation:

Using the law of conservation of momentum:

Initial momentum of the system = final momentum of the system.

mu+m'u' = mv+m'v'

where u and u' are the initial velocities.

v and v' are the final velocities.

mass of bullet, m = 0.04 kg

mass of the block of wood, m' = 0.5 kg

initial velocity of bullet, u = 300 m/s

initial velocity of block, u' = 0

after collision, block and bullet would move together with velocity v

⇒0.04 kg × 300 m/s + 0 = 0.54 kg × v

⇒ v = 22.22 m/s

Thus, the bullet and the wood block move together as one object with the speed 22.2 m/s.