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# 2D Momentum question

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A 5.0 kg bomb at rest explodes into three pieces, each of which travels parallel to the ground. The first piece, with a mass of 1.2 kg travels at 5.5 m/s at an angle of 20 degrees south of east. The second piece has a mass of 2.5 kg and travels 4.1 m/s at an angle of 25 degrees north of east. Determine the velocity of the third piece.

Guest Mar 16, 2017
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#1
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A 5.0 kg bomb at rest explodes into three pieces, each of which travels parallel to the ground. The first piece, with a mass of 1.2 kg travels at 5.5 m/s at an angle of 20 degrees south of east. The second piece has a mass of 2.5 kg and travels 4.1 m/s at an angle of 25 degrees north of east. Determine the velocity of the third piece.

$$Cosine\ in \ the\ vector \ triangle$$

$$c^2=a^2+b^2-2ac\cdot cos\gamma$$

$$2P_a=1.2kg\cdot 5.5^2\frac{m^2}{s^2}$$

$$2P_b=2.5kg\cdot 4.1^2\frac{m^2}{s^2}$$

$$\gamma=(180-20-25)°=135°$$

$$4P_c^2=(1.2^2\cdot 5.5^4+2.5^2\cdot 4.1^4-2\cdot 1.2\cdot 5.5^2\cdot 2.5\cdot 4.1^2\cdot cos135°)kg^2\cdot \frac{m^4}{s^4}$$

$$2P_c=\sqrt{5241.184021\ kg^2\cdot \frac{m^4}{s^4}}$$

$$2P_c=72.3960221352 kg\cdot\frac{m^2}{s^2}$$

$$2P_c=m_c\cdot v_c^2$$

$$\large v_c=\sqrt{\frac{2P_c}{m_c}}=\sqrt{\frac{72.3960221352 kg\cdot\frac{m^2}{s^2}}{(5-1.2-2.5)kg}}$$

$$v_c=7.463\ m/s$$

$$The\ third \ piece \ of \ the \ bomb \ is \ removed \ at \ a \ rate \ of \ 7.463 \ m/s.$$

!

asinus  Mar 17, 2017
#2
+7159
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sorry!

$$Cosine\ law,\ vector\ triangle:$$  $$c^2=a^2+b^2-2ab\ cos\gamma$$

asinus :- (

asinus  Mar 17, 2017

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