+0

# 2D Momentum question

0
126
2

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
Sort:

#1
+6709
0

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
+6709
0

sorry!

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

asinus :- (

asinus  Mar 17, 2017

### 2 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details