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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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b) Potential difference: -26,800 V

c) Work: $4.3\cdot 10^{-15} J$

Explanation:

a)

The electric potential at a distance r from a single-point charge is given by:

$V(r)=\frac{kq}{r}$

where

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

q is the charge

r is the distance from the charge

In this problem, we have a system of two charges, so the total potential at a certain point will be given by the algebraic sum of the two potentials.

Charge 1 is

$q_1=+2.00\mu C=+2.00\cdot 10^{-6}C$

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$q_2=-3.00 \mu C=-3.00\cdot 10^{-6}C$

and is located at (x=0, y = 0.40 m)

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$r_{1A}=0.40 m$

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$V_1=\frac{(8.99\cdot 10^9)(+2.00\cdot 10^{-6})}{0.40}=+4.50\cdot 10^4 V$

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$q=-1.6\cdot 10^{-19}C$ is the charge of the electron

$\Delta V=-26,800 V$ is the potential difference

Therefore, the work required on the electron is

$W=(-1.6\cdot 10^{-19})(-26,800)=4.3\cdot 10^{-15} J$

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