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Nookie1986 [14]
3 years ago
6

A 4.4 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 ot

her at 6.8 m/s, 35° north of east. What is the original speed of the mess kit?
Physics
1 answer:
sp2606 [1]3 years ago
6 0

Answer:

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.

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2 / F2 = (Gm1m2 / 4) × (16 / Gm1m2)

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<u>Given:</u>

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