Question

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 1

Answer with Explanation:

We are given that

Mass of bullet,$m_1=0.03 kg$

$u_1=470 m/s$

$m_2=3 kg$

$\mu_k=0.2$

$m_3=30 kg$

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

$m_1u_1=(m_1+m_2)v$

$0.03(470)=(0.03+3)v$

$v=\frac{0.03(470)}{(0.03+3)}=4.65 m/s$

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

According to law of conservation of momentum

$(m_1+m_2)v=(m_1+m_2+m_3)V$

$(0.03+3)\times 4.65=(0.03+3+30)V$

$V=\frac{(0.03+3)\times 4.65}{(0.03+3+30)}$

$V=0.43 m/s$

Answer 2

Answer:

$v_B=4.65\ m.s^{-1}$ is  the velocity of the block and the bullet system after the first impact.

$v_c=0.43\ m.s^{-1}$ is the velocity of the carrier.

Explanation:

Given

mass of bullet, $m=0.03\ kg$

horizontal velocity of bullet, $v=470\ m.s^{-1}$

mass of block, $m_B=3\ kg$

mass of carrier, $m_c=30\ kg$

coefficient of kinetic friction between block and carrier, $\mu_k=0.2$

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

We have the conservation of linear momentum as:

$m.v=(m+m_B).v_B=(m+m_B+m_c).v_c$

$0.03\times 470=(0.03+3)\times v_B=(0.03+3+30)\times v_c$

$v_B=4.6535\ m.s^{-1}$ is  the velocity of the block and the bullet system after collision.

After the collision of block & carrier:

$v_c=0.4268\ m.s^{-1}$ is the velocity of the carrier.