A battleship that is 6.00×107kg and is originally at rest fires a 1100-kg artillery shell horizontally with a velocity of 575 m/s. (a) If the shell is fired straight aft (toward the rear of the ship), there will be negligible friction opposing the ship’s recoil. Calculate its recoil velocity. (b) Calculate the increase in internal kinetic energy (that is, for the ship and the shell). This energy is less than the energy released by the gun powder—significant heat transfer occurs.

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aristocles aristocles · Answer 1 · 2019-08-27T02:31:10+02:00

Answer:

Part a)

v = 0.0105 m/s

Part b)

[tex]E = 1.82 \times 10^8 J[/tex]

Explanation:

As per momentum conservation we know that

[tex]P_1 = P_2[/tex]

[tex]P_1 = m_1 v_1[/tex]

[tex]P_2 = m_2v_2[/tex]

here we know

[tex]m_1 = 6.00 \times 10^7 kg[/tex]

[tex]v_1 = ?[/tex]

[tex]m_2 = 1100 kg[/tex]

[tex]v_2 = 575 m/s[/tex]

Part a)

now from above expression we can say

[tex]m_1v_1 = m_2v_2[/tex]

[tex](6.00 \times 10^7)(v) = (1100)(575)[/tex]

[tex]v = 0.0105 m/s[/tex]

Part b)

Now increase in the internal energy of the system

[tex]E = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2[/tex]

[tex]E = \frac{1}{2}(6.00 \times 10^7)(0.0105)^2 + \frac{1}{2}(1100)(575^2)[/tex]

[tex]E = 1.82 \times 10^8 J[/tex]