Question

g A 24-gg bullet strikes and becomes embedded in a 1.50-kgkg block of wood placed on a horizontal surface just in front of the gun. Part A If the coefficient of kinetic friction between the block and the surface is 0.23, and the impact drives the block a distance of 9.5 mm before it comes to rest, what was the muzzle speed of the bullet

Answer 1

The muzzle speed of the bullet before the collision is 415.3 m/s.

The given parameters:

• Mass of the bullet, m₁ = 24 g
• Mass of the wood, m₂ = 1.5 kg
• Coefficient of kinetic friction, μk = 0.23
• Distance traveled by the block before stopping, d = 9.5 m

Apply the principle of work-energy theorem to determine the final velocity of the block-bullet system;

$F_f \times d = \frac{1}{2} mv^2\\\\\mu_k F_n \times d = \frac{1}{2} mv^2\\\\\mu_ k (m_1 + m_2)g \times d = \frac{1}{2} (m_1 + m_2)v^2\\\\\mu_k g \times d= \frac{1}{2} v^2\\\\2\mu _k gd = v^2\\\\v= \sqrt{2\mu _k gd } \\\\v = \sqrt{2 \times 0.23 \times 9.8 \times 9.5} \\\\v = 6.54 \ m/s$

Apply the principle of conservation of linear momentum to determine the muzzle speed of the bullet;

$m_1 u_1 \ + \ m_2u_2 = v(m_1 + m_2)\\\\0.024(u_1) \ + \ 1.5(0) = 6.54(0.024 + 1.5)\\\\0.024u_1 = 9.967\\\\u_1 = \frac{9.967}{0.024} \\\\u_1 = 415.3 \ m/s$

Thus, the muzzle speed of the bullet before the collision is 415.3 m/s.

Learn more about conservation of linear momentum here: brainly.com/question/7538238