Question

A 23 g bullet traveling at 230 m/s penetrates a 2.0 kg block of wood and emerges cleanly at 170 m/s. If the block is stationary on a surface with a coefficient of kinetic friction of 0.15 when hit, how far does it move after the bullet emerges?

Answer 1

The distance traveled by the wood after the bullet emerges is 0.16 m.

The given parameters;

• mass of the bullet, m = 23 g = 0.023 g
• speed of the bullet, u = 230 m/s
• mass of the wood, m = 2 kg
• final speed of the bullet, v = 170 m/s
• coefficient of friction, μ = 0.15

The final velocity of the wood after the bullet hits is calculated as follows;

$m_1u_1 + m_2 u_2 = m_1v_1 + m_2v_2\\\\0.023(230) + 2(0) = 0.023(170) + 2v_2\\\\5.29 = 3.91 + 2v_2\\\\2v_2 = 1.38\\\\v_2 = \frac{1.38}{2} = 0.69 \ m/s$

The acceleration of the wood is calculated as follows;

$\mu = \frac{a}{g} \\\\a = \mu g\\\\a = 0.15 \times 9.8\\\\a = 1.47 \ m/s^2$

The distance traveled by the wood after the bullet emerges is calculated as follows;

$v^2 = v_0^2 + 2as\\\\v^2 = 0 + 2as\\\\v^2 = 2as\\\\s = \frac{v^2}{2a} \\\\s = \frac{(0.69)^2}{2(1.47)} \\\\s = 0.16 \ m$

Thus, the distance traveled by the wood after the bullet emerges is 0.16 m.

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