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Momentum and Impulse

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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

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

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**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

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**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

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**Check Your Understanding**

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

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**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

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

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

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

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**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.

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**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.

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**Check Your Understanding**

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)

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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

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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 = ?

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**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

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**Conservation of Momentum!**

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**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...

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

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**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 =

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

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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

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**Solution: p before = p after Givens: m (cannon) = 300 kg**

m (cannonball) = 10 kg v (cannonball) = 200 m/s v (cannon) = ?

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**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

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**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.

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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?

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**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) = ?

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**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

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**More Complicated Momentum Conservation**

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**Collisions do not always take place in a nice neat line:**

Often, collisions take place in 2 or 3 dimensions:

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**Momentum is always conserved.**

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

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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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Momentum & Impulse Think of P as in Pmomentum. Momentum & Impulse Momentum = m x v, it’s a vector, P = m x v Remember F = ∆ P/ ∆ time = m ∆v/∆t = ma Impulse.

Momentum & Impulse Think of P as in Pmomentum. Momentum & Impulse Momentum = m x v, it’s a vector, P = m x v Remember F = ∆ P/ ∆ time = m ∆v/∆t = ma Impulse.

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