Answered

Discover the answers you need at Westonci.ca, where experts provide clear and concise information on various topics. Get immediate and reliable solutions to your questions from a knowledgeable community of professionals on our platform. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.

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

We hope our answers were helpful. Return anytime for more information and answers to any other questions you may have. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Stay curious and keep coming back to Westonci.ca for answers to all your burning questions.