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A 62.0 kg child is sitting in a wagon full of bricks. The wagon and all of the bricks together have mass of 150. kg. In order to move the wagon east without touching the ground, the child throws two bricks, each of mass 3.00 kg, westward at 2.00 m/s. What is the new velocity of the child, wagon, and remaining bricks

physics
Guest Jan 25, 2015

Best Answer 

 #2
avatar+26973 
+5

 

Anonymous is correct, but just to flesh it out a little:

 

Momentum is conserved (assuming no friction etc.) so:

 

Initial momentum = 0

 

Final momentum of wagon + child after bricks are thrown = (62 + 150 - 6)*velocity

Final momentum of bricks = -6*2  (assuming velocity is negative to the west and positive to the east)

 

(62+150-6)*velocity - 6*2 = 0

 

velocity = 6*2/(62+150-6) m/s

 

$${\mathtt{velocity}} = {\frac{{\mathtt{6}}{\mathtt{\,\times\,}}{\mathtt{2}}}{\left({\mathtt{62}}{\mathtt{\,\small\textbf+\,}}{\mathtt{150}}{\mathtt{\,-\,}}{\mathtt{6}}\right)}} \Rightarrow {\mathtt{velocity}} = {\mathtt{0.058\: \!252\: \!427\: \!184\: \!466}}$$ 

 

or velocity ≈ 0.058 m/s

.

Alan  Jan 26, 2015
 #1
avatar
+5

$$\left({\mathtt{3}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}\right){\mathtt{\,\times\,}}{\mathtt{2}} = {\mathtt{12}}$$

$${\mathtt{150}}{\mathtt{\,-\,}}{\mathtt{3}}{\mathtt{\,-\,}}{\mathtt{3}}{\mathtt{\,\small\textbf+\,}}{\mathtt{62}} = {\mathtt{206}}$$

$${\mathtt{206}}{V} = {\mathtt{12}}$$

$${\mathtt{V}} = {\frac{{\mathtt{12}}}{{\mathtt{206}}}}$$ m.s-1

Guest Jan 26, 2015
 #2
avatar+26973 
+5
Best Answer

 

Anonymous is correct, but just to flesh it out a little:

 

Momentum is conserved (assuming no friction etc.) so:

 

Initial momentum = 0

 

Final momentum of wagon + child after bricks are thrown = (62 + 150 - 6)*velocity

Final momentum of bricks = -6*2  (assuming velocity is negative to the west and positive to the east)

 

(62+150-6)*velocity - 6*2 = 0

 

velocity = 6*2/(62+150-6) m/s

 

$${\mathtt{velocity}} = {\frac{{\mathtt{6}}{\mathtt{\,\times\,}}{\mathtt{2}}}{\left({\mathtt{62}}{\mathtt{\,\small\textbf+\,}}{\mathtt{150}}{\mathtt{\,-\,}}{\mathtt{6}}\right)}} \Rightarrow {\mathtt{velocity}} = {\mathtt{0.058\: \!252\: \!427\: \!184\: \!466}}$$ 

 

or velocity ≈ 0.058 m/s

.

Alan  Jan 26, 2015

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