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3 years ago

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a 59kg physics student jumps off the back of her laser sailboat (42kg). after she jumps the laser is found to be travelling at 1

.5m/s. what is the speed of the physics student?

1 answer:

evablogger [386]3 years ago

5 0

From the conservation of linear momentum of closed system,

Initial momentum = final momentum

Mass of the student, M = 59 kg

Mass of the laser boat, m = 42 kg

Initial speed of student + laser boat, u =0

Final speed of laser boat, v = 1.5 m/s

Final speed of the student = V

(M+m) u =M V +m v

0 = (59 kg) V + (42 kg) (1.5m/s)

V = - 1.06 m/s

Thus, the speed of the student is 1.06 m/s in the opposite direction of the motion of boat.

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a. $\Delta s ^2 = 8.0888 \ 10^{17} m^2$

b. $\Delta s ^2 = 3.0234 \ 10^{16} m^2$

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Explanation:

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$\Delta s ^2 = - \Delta (c t) ^ 2 + \Delta \vec{x}^2$ .

Lets work with the signature ( + - - - ), and, if needed in the other signature, we can multiply our interval by -1.

<h3>a.</h3>

$\Delta \vec{x}^2 = (7.5 \ 10 \ m)^2$

$\Delta \vec{x}^2 = 5,625 m^2$

$\Delta (c t) ^ 2 = (299,792,458 \frac{m}{s} \ 3 \ s)^2$

$\Delta (c t) ^ 2 = (899,377,374 \ m)^2$

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<h3>b.</h3>

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$\Delta \vec{x}^2 = 2,500 m^2$

$\Delta (c t) ^ 2 = (299,792,458 \frac{m}{s} \ 0.58 \ s)^2$

$\Delta (c t) ^ 2 = (173,879,625.6 \ m)^2$

$\Delta (c t) ^ 2 = 3.0234 \ 10^{16} m^2$

so

$\Delta s ^2 = 3.0234 \ 10^{16} m^2 - 2,500 m^2$

$\Delta s ^2 = 3.0234 \ 10^{16} m^2$

<h3>c.</h3>

$\Delta \vec{x}^2 = (5 \ 10 \ m)^2$

$\Delta \vec{x}^2 = 2,500 m^2$

$\Delta (c t) ^ 2 = (299,792,458 \frac{m}{s} \ 58 \ s)^2$

$\Delta (c t) ^ 2 = (1.73879 \ 10^{10} \ m)^2$

$\Delta (c t) ^ 2 = 3.0234 \ 10^{20} m^2$

so

$\Delta s ^2 = 3.0234 \ 10^{20} m^2 - 2,500 m^2$

$\Delta s ^2 = 3.0234 \ 10^{20} m^2$

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