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3 years ago

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A 2.2-kg model rocket is launched vertically and reaches an altitude of 70 m with a speed of 30 m/s at the end of powered flight

, time t = 0. As the rocket approaches its maximum altitude it explodes into two parts of masses mA = 0.77 kg and mB = 1.43 kg. Part A is observed to strike the ground 80 m west of the launch point at t = 6 s. Determine the position of part B at that time.

1 answer:

kirill115 [55]3 years ago

8 0

Answer:

$r_b= (30.8\hat{i} + 69.96 \hat{j}) m$

Explanation:

given,

mass = 2.2 kg

altitude(r₀) = (70 j) m

speed = 30 m/s

m_a = 0.77 kg

m_b =1.43 kg

part A strike ground (r_a)= (80 i) m

t = 6 s

$r = r_0 + v_ot-\dfrac{1}{2}gt^2$

$r = 60\hat{j} + (30\hat{j})\times 6-\dfrac{1}{2}\times 9.8 \times 6^2$

r = 63.6 j m

by conservation of energy

$mr = m_ar_a+m_br_b$

$2.2\times 63.6\hat{j} = 0.77\times (-80 \hat{i})+2\times r_b$

$r_b= (30.8\hat{i} + 69.96 \hat{j}) m$

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$\Rightarrow n_p=13$

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$C_c= C_mT_c+\frac{C_e}{n_p}$

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$C_e=$ $\$8/6=\$1.33/edge$

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$V_{opt}= \frac {650}{\left[ \left(1+\frac{1.33}{0.667}\right)\left(\frac{1}{0.3}-1\right)\right]^{0.3}}=363m/min$ [from(i)]

(b) Tool life,

$T=\left[ \left(1+\frac{1.33}{0.667}\right)\left(\frac{1}{0.3}-1\right)\right]=7min$ [from(ii)]

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$T_c=T_h+T_m+\frac{T_t}{n_p}$

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So, the cycle time

$T_c=2.5+0.53+\frac{1}{13}=3.11 min/pc$

(d) Cost per production unit:

$C_c= C_mT_c+\frac{C_e}{n_p}$

$\Rightarrow C_c=0.667\times3.11+\frac{1.33}{13}=\$2.18/pc$

(e) Total time to complete the batch $=2\times60+ {50\times 3.11}{50}=275.5 min=4.59 hr$ .

(f) The proportion of time spent actually cutting metal

$=\frac{50\times0.53}{275.5}=0.0962=9.62\%$

Similarly, for the ceramic tool:

$C_e=$ $\$10/6=\$1.67/edge$

$T_t-1min$

$C= 3500 m/min$

$n=0.6$

(a) Cutting speed:

$V_{opt}= \frac {3500}{\left[ \left(1+\frac{1.67}{0.667}\right)\left(\frac{1}{0.6}-1\right)\right]^{0.6}}$

$\Rightarrow V_{opt}=2105 m/min$

(b) Tool life,

$T=\left[ \left(1+\frac{1.67}{0.667}\right)\left(\frac{1}{0.6}-1\right)\right]=2.33 min$

(c) Cycle time:

$T_c=T_h+T_m+\frac{T_t}{n_p}$

$\Rightarrow T_m=\frac{\pi \times 73 \times 250\times 10^{-6}}{0.3\times 10^{-3}\times 2105}=0.091 min/pc$

$n_p=\frac {2.33}{0.091}=25.6$

$\Rightarrow n_p=25 pc/tool\; life$

So,

$T_c=2.5+0.091+\frac{1}{25}=2.63 min/pc$

(d) Cost per production unit:

$C_c= C_mT_c+\frac{C_e}{n_p}$

$\Rightarrow C_c=0.667\times2.63+\frac{1.67}{25}=$1.82/pc$

(e) Total time to complete the batch

$=2\times60+ {50\times 2.63}=251.5 min=4.19 hr$ .

(f) The proportion of time spent actually cutting metal

$=\frac{50\times0.091}{251.5}=0.0181=1.81\%$

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3 years ago