Question

A bullet loses 1/n if its velocity while penetrating a distance x into the target. The further distance travelled before coming to rest?

Answer 1

Answer on Question #72806-Physics-Mechanics-Relativity

A bullet loses $1/n$ if its velocity while penetrating a distance $x$ into the target. The further distance travelled before coming to rest?

Solution

$$Fx = \frac{mv^2}{2} - \frac{m}{2} \left( \left(1 - \frac{1}{n}\right) v \right)^2 = \frac{mv^2}{2} \left(1 - \left(1 - \frac{1}{n}\right)^2 \right) = \frac{mv^2}{2} \left(1 - 1 - \frac{1}{n^2} + \frac{2}{n} \right) = \frac{mv^2}{2} \left( \frac{2n - 1}{n^2} \right)$$

The initial kinetic energy is

$$\frac{mv^2}{2} = \frac{Fx}{\left( \frac{2n - 1}{n^2} \right)} = FD.$$

Where $D = x + d$

$$\frac{Fx}{\left( \frac{2n - 1}{n^2} \right)} = F(x + d)$$ $$(x + d) = \frac{x}{\left( \frac{2n - 1}{n^2} \right)} = \frac{n^2}{2n - 1} x$$

The further distance travelled before coming to rest is

$$d = \frac{n^2}{2n - 1} x - x = x \frac{n^2 - 2n + 1}{2n - 1} = x \frac{(n - 1)^2}{2n - 1}.$$

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