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A block of wood mass 0.60kg is balanced on top of a vertical port 2.0m high. A 10gm bullet is fired horizontally into the block and the embedded bullet land at a 4.0m from the base of the port. Find the initial velocity of the bullet.

Respuesta :

Marinette09

Answer:

Mass of bullet is m=0.01kg

Mass of the block is M=4kg

Coefficient=0.25,distance=20m

So, let the speed of the block just after the bullet embedded in it be V and v be the speed of bullet before striking the block,

By applying conservation of momentum,

mv=(m+M)V

V=

M+m

mv

Explanation:

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onyebuchinnaji

The initial velocity of the bullet is 382 m/s

The given parameters;

  • mass of the wood, m₁ = 0.6 kg
  • height of the port, h = 2
  • mass of the bullet, m₂ = 10 g = 0.01 kg
  • horizontal distance traveled by the bullet, x = 4 m

Apply the principle of conservation of mechanical energy;

The maximum potential energy of the bullet-wood system at the top of the port  = maximum kinetic energy of the system at the base of the port.

[tex]K.E_{max} = P.E_{max}\\\\\frac{1}{2} mv_{max}^2 = mgh_{max}\\\\v^2_{max} = 2gh_{max}\\\\v_{max} = \sqrt{2gh_{max}} \\\\v_{max} = \sqrt{2\times 9.8\times 2} \\\\v_{max} = 6.26 \ m/s[/tex]

Apply the principle of conservation of linear momentum to determine the initial velocity of the bullet;

  • let the initial velocity of the bullet = u₂

[tex]m_1u_1 + m_2u_2 = v(m_1 + m_2)\\\\0.6(0) + 0.01(u_2) = 6.26((0.6 + 0.01)\\\\0.01u_2 = 3.819\\\\u_2 = \frac{3.819}{0.01} \\\\u_2 = 381.9 \ \approx 382 \ m/s[/tex]

Thus, the initial velocity of the bullet is 382 m/s

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