Answered

Get the answers you need at Westonci.ca, where our expert community is always ready to help with accurate information. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.

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

Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Thank you for choosing Westonci.ca as your information source. We look forward to your next visit.