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a 0.015 kg bullet traveling at 500 m/s strikes a 1.0 kg block of wood that is balanced on a table edge 0.92 m above the ground as shown to the right. the bullet buries itself in the block. calculate the horizontal distance, dx where the block hits the floor.

Sagot :

onyebuchinnaji

The horizontal distance where the block hits the floor is 3.2 m.

The given parameters:

  • mass of the bullet, m₁ = 0.015 kg
  • speed of the bullet, u₁ = 500 m/s
  • mass of block wood, m₂ = 1.0 kg
  • height of the table, h = 0.92 m

The final velocity of the bullet-block system after the collision is calculated by applying the principle of conservation of linear momentum;

[tex]m_1u_1 + m_2u_2 = v(m_1 + m_2)\\\\0.015(500) + 1(0) =v(0.015 + 1)\\\\7.5 = 1.015v\\\\v = \frac{7.5}{1.015} \\\\v = 7.39 \ m/s[/tex]

The time taken for the bullet-block system to fall to the floor after collision is calculated as follows;

[tex]h = v_0_yt + \frac{1}{2} gt^2\\\\h = 0 + \frac{1}{2} gt^2\\\\h = \frac{1}{2} gt^2\\\\t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\times 0.92}{9.8} }\\\\t = 0.43 \ s[/tex]

The horizontal distance where the block hits the floor is calculated as follows;

[tex]X = v_x t\\\\X = 7.39 \times 0.43\\\\X = 3.2 \ m[/tex]

Learn more about conservation of linear momentum here:https://brainly.com/question/24424291

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