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?

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.

connect the batteries in series and hope

1.195000

explanation:

total mass=0.0117+0.0137+0.0881+0.0111

total mass= 0.01172kg

total velocity = 0.283-0.743+0.871-0.799+0.155+0.637+0.301=0.102

momentum=mv=0.1172*0.102=1.195000//