A 0.03-kg bullet is fired with a horizontal velocity of 470 m/s and becomes embedded in block B which has a mass of 3 kg. After the impact, block B slides on a 30-kg carrier C until it impacts the end of the carrier. Knowing the impact between B and C is perfectly plastic and the coefficient of kinetic friction between B and C is 0.2, determine the velocity of the bullet and block B after the first impact and the final velocity of the carrier. (Round the final answer answers to two decimal places.)

Answer with Explanation:

We are given that

Mass of bullet,

We have to find the velocity of the bullet and block B after the first impact and final velocity of the carrier.

According to law of conservation of momentum

Hence, the velocity of the bullet and block B after the first impact=4.65 m/s

According to law of conservation of momentum

is  the velocity of the block and the bullet system after the first impact.

is the velocity of the carrier.

Explanation:

Given

mass of bullet,

horizontal velocity of bullet,

mass of block,

mass of carrier,

coefficient of kinetic friction between block and carrier,

Since the collision between the bullet and the block is perfectly plastic:

We have the conservation of linear momentum as:

is  the velocity of the block and the bullet system after collision.

After the collision of block & carrier:

is the velocity of the carrier.

i think it'll be 7,000

probably not the "magnets"in the magnet, so to speak, are thought to be electrons orbiting in shells. vurrent loops two sides to the loop