Physics, 05.07.2019 04:20 morgaaaan3835

A4.4 kg mess kit sliding on a frictionless surface explodes into two 2.2 kg parts, one moving at 2.9 m/s, due north, and the other at 6.8 m/s, 35â° north of east. what is the original speed of the mess kit?

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Answer from: kaquino983

Original speed of the mess kit = 4.43 m/s at 50.67° north of east.

Explanation:

Let north represent positive y axis and east represent positive x axis.

Here momentum is conserved.

Let the initial velocity be v.

Initial momentum = 4.4 x v = 4.4v

Velocity of 2.2 kg moving at 2.9 m/s, due north = 2.9 j m/s

Velocity of 2.2 kg moving at 6.8 m/s, 35° north of east = 6.9 ( cos 35i + sin35 j ) = 5.62 i + 3.96 j m/s

Final momentum = 2.2 x 2.9 j + 2.2 x (5.62 i + 3.96 j) = 12.364 i + 15.092 j kgm/s

We have

Initial momentum = Final momentum

4.4v = 12.364 i + 15.092 j

v =2.81 i + 3.43 j

Magnitude

$v=\sqrt{2.81^2+3.43^2}=4.43m/s$

Direction

$\theta =tan^{-1}\left ( \frac{3.43}{2.81}\right )=50.67^0$

50.67° north of east.

Original speed of the mess kit = 4.43 m/s at 50.67° north of east.

Answer from: Quest

a.) travels faster in solids because the particles are closer together.

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A4.4 kg mess kit sliding on a frictionless surface explodes into two 2.2 kg parts, one moving at 2.9...

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