Let the mass of the wood ball to be M and the mass of the projectile to be m.

The maximum tension is then given as:

$\overline{){{\mathbf{T}}}_{\mathbf{m}\mathbf{a}\mathbf{x}}{\mathbf{=}}{\mathbf{(}}{\mathbf{M}}{\mathbf{+}}{\mathbf{m}}{\mathbf{)}}{\mathbf{g}}{\mathbf{+}}\frac{\mathbf{(}\mathbf{M}\mathbf{+}\mathbf{m}\mathbf{)}{\mathbf{v}}^{\mathbf{2}}}{\mathbf{l}}}$

Substituting:

$\begin{array}{rcl}\mathbf{400}& \mathbf{=}& \mathbf{(}\mathbf{20}\mathbf{.}\mathbf{0}\mathbf{+}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{)}\mathbf{(}\mathbf{9}\mathbf{.}\mathbf{8}\mathbf{)}\mathbf{+}\frac{\mathbf{(}\mathbf{20}\mathbf{.}\mathbf{0}\mathbf{+}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{)}{\mathbf{v}}^{\mathbf{2}}}{\mathbf{2}\mathbf{.}\mathbf{0}}\\ \mathbf{v}& \mathbf{=}& \sqrt{\mathbf{18}\mathbf{.}\mathbf{5}}\end{array}$

A 20.0 kg wood ball hangs from a 2.10 m-long wire. The maximum tension the wire can withstand without breaking is 400 N. A 0.900 kg projectile traveling horizontally hits and embeds itself in the wood ball.

What is the largest speed this projectile can have without causing the cable to break?

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