Answer to Question #117873 in Mechanics | Relativity for Thandeka

Question

A bullet with speed 511 m/s and mass m = 5.00 g is shot into a stationary block of wood hanging a distance h1 = 1.00 m above the ground, as shown in Figure 1. The block is hanging by a string of length r = 0.30 m. The bullet is lodged in the block, and the block-bullet system swings so that the string makes an angle of θ = 60.0° when the string breaks, as shown in Figure 2. The block-bullet system then falls to the ground a horizontal distance d2 = 2.00 m from its position when the string breaks. Ignoring the effects of air resistance, determine the mass, M of the block of wood.

Answer 1

System momentum at first time:

$p=(m+M)V=mv, \newline where\space v \space initial \space speed\space of \space bullet.$

Time while system fall:

$t=\dfrac{d_2}{V_x}=\dfrac{d_2}{V\sin(\theta)}$

Also we know:

$h_{1}+(r-r\cos(\theta))=V_{0y}t-\dfrac{gt^2}{2}\newline h_{1}+r(1-\cos(\theta))=V\cos(\theta)t-\dfrac{gt^2}{2}\newline h_{1}+r(1-\cos(\theta))=V\cos(\theta)\dfrac{d_2}{V\sin(\theta)}-\dfrac{gd^2_{2}}{2V^2\sin^2(\theta)}\newline h_{1}+r(1-\cos(\theta))=d_2\cot(\theta)-\dfrac{gd^2_{2}}{2V^2\sin^2(\theta)}=>\newline h_{1}+r(1-\cos(\theta))-d_2\cot(\theta)=-\dfrac{gd^2_{2}}{2V^2\sin^2(\theta)}=>\newline V=\dfrac{d_2\sqrt{g}}{\sin(\theta)\sqrt{2(-h_{1}-r(1-\cos(\theta))+d_2\cot(\theta) )}}\newline V=\dfrac{d_2}{\sin(\theta)}\sqrt{\dfrac{g}{2(-h_{1}-r(1-\cos(\theta))+d_2\cot(\theta))}}\newline V=\dfrac{2}{\sqrt{3}/2}\sqrt{\dfrac{10}{2(-1-0.3(1-1/2)+2/\sqrt3)}}\approx\dfrac{2}{\sqrt{3}/2}\sqrt{\dfrac{10}{0.0094}}\newline V\approx\dfrac{4}{1.73}\times32.62\approx75.42m/s.$

From the first equation:

$M=\dfrac{mv}{V}-m=\dfrac{5\times 511}{75.42}-5\approx28.88 g$