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astraxan [27]
3 years ago
9

An object, with mass 32 kg and speed 26 m/s relative to an observer, explodes into two pieces, one 5 times as massive as the oth

er; the explosion takes place in deep space. The less massive piece stops relative to the observer. How much kinetic energy is added to the system during the explosion, as measured in the observer's reference frame?
Physics
1 answer:
Brut [27]3 years ago
3 0

Answer:

\Delta K = 2164.053\,J

Explanation:

Let consider the observer as an inertial reference frame. The object is modelled after the Principle of Momentum Conservation:

(32\,kg)\cdot (26\,\frac{m}{s} ) = (5.333\,kg)\cdot (0\,\frac{m}{s} )+(26.665\,kg )\cdot v

The speed of the more massive piece is:

v = 31.202\,\frac{m}{s}

The kinetic energy added to the system is:

\Delta K = \frac{1}{2}\cdot [(5.333\,kg)\cdot (0\,\frac{m}{s} )^{2}+(26.665\,kg )\cdot (31.202\,\frac{m}{s} )^{2}]-\frac{1}{2}\cdot (32\,kg)\cdot (26\,\frac{m}{s} )^{2}

\Delta K = 2164.053\,J

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For both cases, + and -, after a certain time \delta t (\delta t >0), the displacement <em>y</em> of the wave will be determined by the kx\pm\omega (t+\delta t) term. For simplicity, if we imagine we are looking at the origin (x=0), this will be simply \pm \omega (t+\delta t).

To know which side, right or left of the origin, would go through the origin after a time \delta t (and thus know the direction of propagation) we have to see how we can achieve that same displacement <em>y</em> not by a time variation but by a space variation \delta x (we would be looking where in space is what we would have in the future in time). The term would be then k(x+\delta x)\pm\omega t, which at the origin is k \delta x \pm \omega t. This would mean that, when the original equation has kx+\omega t, we must have that \delta x>0 for k\delta x+\omega t to be equal to kx+\omega\delta t, and when the original equation has kx-\omega t, we must have that \delta x for k\delta x-\omega t to be equal to kx-\omega \delta t

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<em />

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3 years ago
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45 m / s North is a valid vector reading for an object.

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therefore two decays = 17.22/2= 8.61

8 0
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