Momentum Question: Asteroid Collision

[student34]

Homework Statement

Asteroid A is traveling at 40.0m/s and collides with asteroid B which is at rest. Both asteroids have the same mass, but asteroid A deflects off of its path by 30.0° above the horizontal, and asteroid B begins to travel 45.0° below the horizontal. What is the speed of asteroid A?

Homework Equations

PA1X = PA2X + PB2X

PA1X = 40.0m/s*mA

PA2X = mA*VA2*cos30°

PB2X = mB*VB2*cos45°

PA2Y = mA*VA2*sin30°

PB2Y = mB*VB2*sin45°

The Attempt at a Solution

Equation #1 mB*VB2*sin45° = mA*VA2*sin30°
Equation #2 mB*VB2*cos45° + mA*VA2*cos30° = 40.0m/s*mA

I isolate mB*VB2 from Equation #1 and get mB*VB2 = (mA*VA2*sin30°)/(sin45).

Now I substitute mB*VB2 into Equation #2, and get (cos45°*VA2*sin30°)/(sin45°) + VA2*cos30° = 40.0m/s, I canceled out mA.

Next I isolate VA2*(sin30°)/(tan45°) + cos30° = 40.0m/s

Finally VA2 = 78.2679m/s which is impossible because it shouldn't go faster after the collision. Also, my textbook's answer is 23.3m/s.

 

[gneill]

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Now I substitute mB*VB2 into Equation #2, and get (cos45°*VA2*sin30°)/(sin45°) + VA2*cos30° = 40.0m/s, I canceled out mA.

Next I isolate VA2*(sin30°)/(tan45°) + cos30° = 40.0m/s

Careful... didn't VA2 multiply both terms on the left?

Finally VA2 = 78.2679m/s which is impossible because it shouldn't go faster after the collision. Also, my textbook's answer is 23.3m/s.

Yup, 78 m/s is definitely too large. It's also possible that your textbook's answer is not correct

 

[student34]

Careful... didn't VA2 multiply both terms on the left?

You're right - thank-you so much!