Conservation of Momentum
Momentum
In a closed system, the total momentum of a group of objects before an interaction equals the total momentum after the interaction. Conserv. of Momentum – Fill in the Blanks Notes
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Warm-Up
A teacher pushed ice cubes of different masses with different amounts of force. Which diagram shows the cube that will show the biggest velocity change when pushed with the force represented by the arrow?
A) A B) B C) C D) D
E) To find the velocity change, you need the force's contact time.
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What is the momentum of the big fish after it swallows the little fish?
A) Collision Problems
A 3.0 kg object traveling 6.0 m/s East has a perfectly elastic collision with a 4.0 kg object traveling 8.0 m/s West. After the collision, the 3.0 kg object will travel 10. m/s West.
a) What was the total momentum before the collision?
| Before | After |
| m1 = 3.0 kg | |
| v1 = 6.0 m/s (East) | |
| m2 = 4.0 kg | |
| v2 = -8.0 m/s (West) |
Total momentum before:
= m1v1 + m2v2
= 3.0 kg(6.0 m/s) + 4.0 kg(-8.0 m/s)
= 18. kg·m/s + -32. kg·m/s
Total Momentum Before Interaction = Total Momentum After Interaction
b) What is the total momentum of these objects after this collision?
c) What velocity will the 4.0 kg object have after the collision?
| Before | After |
| m1 = 3.0 kg | |
| v1′ = -10. m/s | |
| m2 = 4.0 kg | |
| v2′ = ? |
Total Momenta Before Interaction = Total Momenta After Interaction
-14. kg·m/s = m1v1′ + m2v2′
-14. kg·m/s = 3.0 kg(-10. m/s) + 4.0 kg(v2′)
-14. kg·m/s = (-30. kg·m/s) + 4.0 kg(v2′)
16. kg·m/s = 4.0 kg(v2′)
What is the momentum of the attached carts?
| p = 15 kg·m/s | p = -30 kg·m/s |
A 10. kg Block A moves with a velocity of 2.0 m/s to the right and collides with a 10. kg Block B which is at rest. After the collision Block A stops moving and Block B moves to the right.
a) Find the total momentum after the collision.
| Before | After |
| mA = 10. kg | mA = 10. kg |
| vA = 2.0 m/s (East) | vA′ = 0 m/s |
| mB = 10. kg | mB = 10. kg |
| vB = 0 m/s | vB′ = ? |
Total mom. before = Total mom. after:
= mAvA + mBvB
= 10. kg(2.0 m/s) + 10. kg(0 m/s)
Find the velocity of Block B after the collision.
20. kg·m/s = mAVA + mBVB
20. kg·m/s = 10. kg(0 m/s) + 10. kg·VB
A 10. kg cart moving with a velocity of 10. m/s East collides and attaches itself to a 10. kg cart moving at a velocity of 50. m/s West.
a) Find the total momentum before the collision.
| Before | After |
| mA = 10. kg | |
| vA = 10. m/s (East) | |
| mB = 10. kg | |
| vB = -50. m/s |
Total momentum = m1v1 + m2v2
Total momentum = 100. kg·m/s + -500. kg·m/s
Impulse
Hi-speed video of a golf ball compressed by driver.
Recorded at 10,000 fps with the Photron ultima APX slow motion video camera, the golf ball can be seen to compress as the golf club comes into contact with it at high speed. Recorded with the APX at 10K fps, with a 10 microsecond shutter and 512 (H) x 256 (V) at 10-bit pixel depth.
b) Find the total momentum after the collision.
Total mom. before = Total mom. after:
c) What is the velocity of the attached carts after the collision?
| m = 20. kg (masses combine) |
| v = ? |
Total mom. before = Total mom. after:
-400. kg·m/s = (20. kg)V
What is the magnitude of the total momentum of these carts?
Total momentum = m1v1 + m2v2
Total mom. = (4.0 kg)(3 m/s) + 6.0 kg(-3.0 m/s)
= 12. kg·m/s + -18. kg·m/s
"Left train momentum = 30 kg·m/s. What is the momentum of the right train?"
What's the total momentum before the collision? After the collision?
