Question

A bullet of mass 0.02kg is moving with a speed 10m/s. It can penetrate 10 cm if a given target before coming to rest. If the same target were only 6 cm thick, what will be the speed and kinetic energy of the bullet, when it comes out.

Answer 1

Question #64399, Physics / Mechanics | Relativity

A bullet of mass $0.02\mathrm{kg}$ is moving with a speed $10\mathrm{m/s}$. It can penetrate $10\mathrm{cm}$ if a given target before coming to rest. If the same target were only $6\mathrm{cm}$ thick, what will be the speed and kinetic energy of the bullet, when it comes out.

Solution

The kinetic energy of the bullet right before it hits the target:

$$E_k = \frac{m v^2}{2};$$ $$E_{k\_i} = \frac{0.02 \times 10^2}{2} = 1\mathrm{J}$$

The change in kinetic energy is equal to the work done by resistive forces.

$$E_{k\_i} = W = F d;$$ $$F = \frac{E_{k\_i}}{d};$$ $$F = \frac{1}{0.1} = 10\mathrm{N}$$

The kinetic energy on the bullet when it comes out:

$$E_{k\_f} = E_{k\_i} - F d;$$ $$E_k = 1 - 10 \times 0.06 = 0.4\mathrm{J}$$

The final speed of the bullet:

$$v = \sqrt{\frac{2 E_k}{m}};$$ $$v_f = \sqrt{\frac{2 \times 0.4}{0.02}} = 6.32\mathrm{m/s}.$$

Answer: $6.32\mathrm{m/s}$; $0.4\mathrm{J}$.

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