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19. A 0.0340 kg bullet traveling at 120 m/s embeds itself in a 1.24 kg wooden block which
is at rest on a smooth surface. The block then slides toward a spring and collides with it.
The block compresses the spring (k = 99.0 N/m). Calculate how far the block bullet
compresses the spring.
I

Sagot :

Onyebuchinnajiidn

The distance the block-bullet compress the spring is 0.36 m.

The given parameters:

  • Mass of the bullet, m₁ = 0.034 kg
  • Velocity of the bullet, u₁ = 120 m/s
  • Mass of the wooden block, m₂ = 1.24 kg
  • Spring constant, k = 99 N/m

The final velocity of the bullet- block system is calculated as follows;

[tex]m_1 u_1 + m_2 u_2 = v(m_1 + m_2)\\\\0.034(120) + 1.24(0) = v(0.034+ 1.24)\\\\4.08 = 1.274v\\\\v = \frac{4.08}{1.274} \\\\v = 3.2 \ m/s[/tex]

The compression of the spring is calculated by applying principle of conservation of energy;

[tex]\frac{1}{2} kx^2 = \frac{1}{2} mv^2\\\\kx^2 = mv^2\\\\x^2 = \frac{mv^2}{k} \\\\x = \sqrt{\frac{mv^2}{k} } \\\\x = \sqrt{\frac{(1.274 \times 3.2^2)}{99} }\\\\x = 0.36 \ m[/tex]

Thus, the distance the block-bullet compress the spring is 0.36 m.

Learn more about conservation of energy here: https://brainly.com/question/166559

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