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3 years ago

7

3.00 kg block moving 2.09 m/s right hits a 2.22 kg block moving 3.92 m/s left. afterwards, the 3.00 kg block moves 1.71 m/s left

. find the velocity of the 2.22 kg block afterwards

1 answer:

Leviafan [203]3 years ago

7 0

Momentum is conserved, so the total momentum before collision is equal to the total momentum after collision. Take the right direction to be positive. Then

(3.00 kg) (2.09 m/s) + (2.22 kg) (-3.92 m/s) = (3.00 kg) (-1.71 m/s) + (2.22 kg) <em>v</em>

where <em>v</em> is the velocity of the 2.22 kg block after collision. Solve for <em>v</em> :

6.27 kg•m/s - 8.70 kg•m/s = -5.13 kg•m/s + (2.22 kg) <em>v</em>

(2.22 kg) <em>v</em> = 2.70 kg•m/s

<em>v</em> = (2.70 kg•m/s) / (2.22 kg)

<em>v</em> ≈ 1.22 m/s

i.e. a velocity of about 1.22 m/s to the right.

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