Answer to Question #117873 in Mechanics | Relativity for Thandeka

Question #117873
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.
1
Expert's answer
2020-05-25T11:00:42-0400

System momentum at first time:

"p=(m+M)V=mv, \\newline\nwhere\\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\nh_{1}+r(1-\\cos(\\theta))=V\\cos(\\theta)t-\\dfrac{gt^2}{2}\\newline\nh_{1}+r(1-\\cos(\\theta))=V\\cos(\\theta)\\dfrac{d_2}{V\\sin(\\theta)}-\\dfrac{gd^2_{2}}{2V^2\\sin^2(\\theta)}\\newline\nh_{1}+r(1-\\cos(\\theta))=d_2\\cot(\\theta)-\\dfrac{gd^2_{2}}{2V^2\\sin^2(\\theta)}=>\\newline\nh_{1}+r(1-\\cos(\\theta))-d_2\\cot(\\theta)=-\\dfrac{gd^2_{2}}{2V^2\\sin^2(\\theta)}=>\\newline\nV=\\dfrac{d_2\\sqrt{g}}{\\sin(\\theta)\\sqrt{2(-h_{1}-r(1-\\cos(\\theta))+d_2\\cot(\\theta)\n)}}\\newline\nV=\\dfrac{d_2}{\\sin(\\theta)}\\sqrt{\\dfrac{g}{2(-h_{1}-r(1-\\cos(\\theta))+d_2\\cot(\\theta))}}\\newline\nV=\\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\nV\\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"


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Comments

Thandeka
30.05.20, 03:01

Thank you so much. Such great help

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