A horizontal spring with stiffness 0.5 N/m has a relaxed length of 19 cm (0.19 m). A mass of 22 grams (0.022 kg) is attached and
1 answer:
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Answer:
=(5.7 x
i - 2.24 x
j) kgm/s
Explanation:
Due to earths gravity, force on asteroid is given by:
r^
Plugging in the values, we have
F= [(6.67x)(1500)(5.97 x
)(8x
i + 9x
j)] / ((8x
)² + (9x
)²
F= 2736 i^ + 3078 j^
In order find the final momentum of the Asteroid, apply impulse momentum theorem
=
+ FΔt
= 1500(3.5 x
i - 1.8x
j) + (2736i + 3078j)(1.5x
)
=(5.7 x
i- 2.24 x
j)kgm/s