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II. Conservation of Momentum

 

 

 

 

Collision/3 Laws

 

 

 

In a closed system:

 When two objects interact,…….

 

....the SUM of their momenta BEFORE the interaction ...

 

...equals the SUM of their momenta AFTER the interaction.

 

 

Total Momentum
Before Interaction

=

Total Momentum
After Interaction

 

m1v1 + m2v2 = m1v1′ + m2v2′

 

 

 

 

Momentum is only conserved
in a totally elastic collision

Totally Elastic Collision?

 

 

 

What is the momentum

of the big fish after it

 swallows the little fish?

 

 

 

 

 

 

-4 kgm/s

Total momentum before =
 Total momentum after

 

A) Collision Problems

 

Ex 1) 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 W. 

After the collision, the 3.0 kg object will travel 10. m/s west. 

a) Total momentum before the collision? 

 

 

Ex 1) 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 kg(-8 m/s)

 

= 18. kg(m/s) + -32. kgm/s

 

= - 14. kg(m/s)

 

OR

 

14. kg(m/s) West

 

 

b) What is the total momentum of these objects after this collision?

 

 

= -14. kg(m/s)

OR 14. kg(m/s) West

 

Total Momentum Before Interaction = Total Momentum After Interaction

 

Ex 1) ..... After the collision, the 3.0 kg object will travel 10. m/s west. 

c) What velocity will the 4.0 kg object have after the collision?

 

m1 = 3.0 kg
v1′ = -10. m/s
m2 = 4.0 kg   

v2′ = ?

 

Total Momenta Before Interaction = Total Momenta After Interaction

Before  After 
 

-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. kgm/s)+4.0 kg(v2′)

 

16. kg(m/s) = 4.0 kg(v2′)

 

v2′ = 4.0 m/s

OR

v2′ = 4.0 m/s East

 

 

 

 

 

 

What is the momentum

of the attached carts?



p = 15 kgm/s		 p = -30 kgm/s

 

 

 

 

 

 

 

Answer: - 15 kgm/s

Total momentum before = total momentum after

 

 

 

 

 

 

Ex 2) 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 total momentum after the collision.

 

Ex 2) 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(2.0m/s)+10. kg(0 m/s)

 

= 20 kgm/s

 

 

 

 

 

Ex 2) 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. 

Find the velocity of Block B 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:

 

 

 

20. kgm/s = mAVA + mBVB

20. kgm/s =

 

10. kg(0m/s)+10. kgV2

 

 

VB = 2.0 m/s

 

 


Ex 3) 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

 

 

Ex 3) 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. kgm/s + -500. kgm/s

 

= -4.0 x 102 kgm/s

 

 

 

 

 

Ex 3) 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.

 

b) Find the total momentum after the collision

 

Total mom. before
= Total mom. after:

 

= -4.0 x 102 kgm/s

 

 

Ex 3) 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.

 

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. kgm/s  = (20. kg)V

 

V = -20. m/s

 

V = 20. m/s W

 

 

Ex 4) What is the magnitude of the total momentum of these carts?

Total momentum = m1v1 + m2v2

Total mom. =

(4.0 kg)(3 m/s)+6.0kg(-3.0m/s)

 

= 12. kgm/s + -18. kgm/s 

 

= -6.0 kgm/s

 

 

 

 

 

"Left train momentum

= 30 kgm/s.

What is the momentum

 of the right train?"

What's the total momentum before the collision?

After the collision?

 

 

 

 

 

 

 

Answer: -30 kgm/s

 

 

 

 

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