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A 500-g lump of clay is dropped onto a 1-kg cart moving at 60 cm/s. The clay is moving downward at 30 cm/s just before l dith t Aft th lli i thd f th tlanding on the cart. After the collision, the speed ofthe cartand clay is closest to:

Respuesta :

Answer:

The speed of the cart and clay after the collision is 50 cm/s .

Explanation:

Given :

Mass of lump , m = 500 g = 0.5 kg .

Velocity of lump , v = 30 cm/s .

Mass of cart , M = 1 kg .

Velocity of cart , V = 60 cm/s .

We know by conservation of momentum :

[tex]mv+MV=(m+M)v'[/tex]

Here , [tex]v'[/tex] is the speed of the cart and clay after the collision .

Putting all value in above equation .

We get :

[tex]0.5\times 30+1\times 60=(0.5+1)\times v'\\\\v'=\dfrac{15+60}{1.5}\\\\v'=50\ cm/s[/tex]

Hence , this is the required solution .

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