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A dry cleaner throws a 22 kg bag of laundry onto a stationary 9 kg cart. The cart and bag begin moving at 3 m/s to the right. Find the velocity of the bag before the collision. (4.2 m/s)

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Sagot :

rafaleo84

Answer:

Vbag = 4.22 [m/s]

Explanation:

In order to solve this problem we must use the principle of conservation of momentum. Which tells us that the momentum of a system is preserved before and after a collision.

Let's take both movements (bag and cart) as positive, since they both move in the same direction. Now everything that happens before the collision will be taken to the left of the equation and after the collision to the right of the equal sign.

[tex](m_{bag}*v_{bag})=(m_{bag}+m_{cart})*v_{final}[/tex]

where:

mbag = mass of the bag = 22 [kg]

vbag = velocity of the bag [m/s]

mcart = mass of the cart = 9 [kg]

vfinal = mass of both bodies moving together = 3 [m/s]

Now replacing:

[tex](22*v_{bag})=(22+9)*3\\22*v_{bag}=93\\v_{bag}=4.22[m/s][/tex]

Animedefender223

Answer:

Vbag = 4.22 [m/s]

Explanation:

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