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Vladimir [108]
3 years ago
10

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.)
Physics
2 answers:
Gala2k [10]3 years ago
3 0

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

Shalnov [3]3 years ago
3 0

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.

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