# Momentum and Impulse.

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Momentum and Impulse

Momentum Momentum can be defined as "mass in motion." All objects have mass; so if an object is moving, then it has momentum Momentum depends upon the variables mass and velocity Momentum = (mass) (velocity) p = (m)(v) where m = mass and v=velocity

Momentum Momentum = (mass) (velocity) p = mv p m v

Momentum is a vector quantity
To fully describe the momentum of a 5-kg bowling ball moving westward at 2 m/s, you must include information about both the magnitude and the direction of the bowling ball p = (m)(v) p = (5 kg)(2 m/s west) p = 10 kgm / s west Givens: m = 5kg v = 2 m/s west

Elastic and Inelastic Collisions
When a Ball hits the ground and sticks, the collision would be totally inelastic When a Ball hits the ground and bounces to the same height, the collision is elastic All other collisions are partially elastic collision

Determine the momentum of a ... 60-kg halfback moving eastward at 9 m/s. p = mv = 60 kg ( 9 m/s ) 540 kgm /s east 1000-kg car moving northward at 20 m/s. p = mv = 1000 kg ( 20 m/s ) 20,000 kgm /s north Given: m = 60Kg v= 9 m/s Find : momentum (p) Given: m = 1000Kg v= 20 m/s

Momentum and Impulse Connection
To stop an object, it is necessary to apply a force against its motion for a given period of time J = F (t) = m D v J Impulse = Change in momentum F t

Long Time Period: When momentum is changed over a long time period, less force is needed:

Short Time Period: When momentum is changed over a short time period, a larger force is needed. This can produce some drastic results.

Notice how the normally rigid golf ball is temporarily deformed from the large force applied over the short time interval.

Bungee jumping with a non stretch rope would NOT be a good idea.
The bungee cord spreads the change in momentum over a longer time so that the force on you is less.

This increases the force, thus breaking the object.
When breaking blocks or boards, the swift strike takes place over a short period of time. This increases the force, thus breaking the object.

If the halfback experienced a force of 800 N for 0.9 seconds to the north, determine the impulse J = F ( t ) = m D v 800N ( 0.9s ) = 720 N*s the impulse was 720 N*s or a momentum change of 720 kg*m/s Given: F = 800 N t = 0.9 s Find : Impulse (J)

Impulse Question #2 A 0.10 Kg model rocket’s engine is designed to deliver an impulse of 6.0 N*s. If the rocket engine burns for 0.75 s, what is the average force does the engine produce? J = F ( t ) = m D v 6.0 N*s = F (0.75s) 6.0 N*s/ 0.75s = F 8.0 N = F Given: F = 800 N t = 0.9 s Find : Average Force

Impulse Question # 3 A Bullet traveling at 500 m/s is brought to rest by an impulse of 50 N*s. What is the mass of the bullet? J = F ( t ) = m D v 50 N*s = m ( 500 m/s – 0 m/s ) 50 kg-m/s 2 *s / 500 m/s = m .1 kg = m Given: v = 500 m/s J = 50 N*s Find : m = ?

the impulse equals the momentum change
Summary the impulse experienced by an object is the (force) x (time) the momentum change of an object is the (mass) x (velocity change) the impulse equals the momentum change

Conservation of Momentum!

Conservation of Momentum:
In all collisions or interactions, momentum of a system is always conserved. You cannot gain or lose any momentum - what you started with (total) is what you will end with! You may have previously learned about conservation of mass or energy from chemistry class...

Since momentum is a vector quantity, direction must be taken into account to see that momentum truly is conserved.

Conservation of Momentum Problems:
When solving problems involving the conservation of momentum, the most important thing to consider is: Total momentum before collision Total momentum after collision =

Sample Problem: A 300 kg cannon fires a 10 kg projectile at 200 m/s. How fast does the cannon recoil backwards? BOOM

Solution: The momentum of the projectile must equal the momentum of the cannon. They must be equal since they must cancel each other out. p before = p after BOOM

Solution: p before = p after Givens: m (cannon) = 300 kg
m (cannonball) = 10 kg v (cannonball) = 200 m/s v (cannon) = ?

vcannon = -6.67 m/s p before = p after 0 = mcannonvcannon + mprojvproj
0= (300 kg) (vcannon) + (10kg) (200m/s) -(2000kgm/s) vcannon = 300 kg vcannon = m/s

Q: Why does the cannon move so much slower compared to the projectile?
A: It is much more massive, more inertia. Q: What does the negative sign indicate? A:The cannon moves in the opposite direction compared to the projectile.

Another Problem: A 5kg fish swims toward and swallows a 1kg fish at rest (it wasn’t paying attention). The big fish initially swims at 1m/s. How fast will it be swimming after having lunch?

Solution: p before = p after Givens: m (big fish) = 5 kg
m (small fish) = 1 kg v (big fish before) = 1 m/s v (little fish before) = 0 v (total after) = ?

m1v1 + m2v2 = m1v1’ + m2v2’ p before = p after
(5kg) (1m/s) + (1 kg) (0m/s) = (6kg) (v) 6 kg represents the combined mass of the fish 5kgm/s = 6kg (v) v = 0.83 m/s

More Complicated Momentum Conservation

Collisions do not always take place in a nice neat line:
Often, collisions take place in 2 or 3 dimensions:

Momentum is always conserved.
Although the mathematics needed to show this may be complicated, the general idea can easily be conveyed.

Another Example: One ball collides into another. By using momentum vector components, you can predict the result: After impact: Before impact: Total P before Y components cancel out X components add up to previous P

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