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**Chapter 7: Impulse and Momentum**

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Impulse and momentum There are many situations in which the force acting on an object is not constant. The force varies with time

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Impulse and momentum To describe how a time varying force affects the motion of an object, we need to use impulse and momentum.

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**Definition of Momentum**

The linear momentum of an object, p, is the product of the object’s mass, m, and velocity, v. p=mv Momentum is a vector whose direction is the same as the velocity UNITS: kg m/s

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Definition of Impulse The impulse, J, of a force is the product of the average force, F, and the time interval, ∆t, during which the force acts. J=F∆t Impulse is a vector whose direction is the same as the force UNITS: kg m/s

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**Impulse-Momentum Theorem**

Recall, ΣF=ma (Newton’s 2nd Law) We also know that a= 𝑣𝑓−𝑣0 ∆𝑡 Putting them together, we get Σ𝐹=𝑚 𝑣𝑓−𝑣0 ∆𝑡 Or, Σ𝐹∆𝑡=𝑚𝑣𝑓 −𝑚𝑣0 Impulse Change in momentum

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Example 1: A baseball (m=0.14kg) has an initial velocity of v0=-38m/s as it approaches a bat. The bat hits the ball, giving it a final velocity vf=+58m/s. A) determine the impulse applied to the ball by the bat B) assume ∆t=1.6x10-3sec, find the average force exerted by the bat.

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**Principle of Conservation of Linear Momentum**

When two objects collide, two types of forces act on the system Internal forces – forces that the objects within the system exert on each other External forces – forces exerted on the objects by agents external to the system

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**Principle of Conservation of Linear Momentum**

An isolated system is a system in which the sum of external forces is zero (no net external force). The bat hitting the ball is an example of an isolated system. If we considered air resistance, it would NOT be an isolated system. The total linear momentum of an isolated system is constant(conserved). This makes it very easy to solve problems when we just care about the “before” and “after” we don’t care about the “during” when objects interact.

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Warning! One of the most common errors in momentum problems is a failure to use the appropriate signs. The momentum component, p, has the same sign conventions as velocity; negative momentum is to the left or downward.

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Example 2 A 4-kg ball traveling westward at 25 m/s hits a 15-kg ball at rest. The 4-kg ball bounces east at 8.0 m/s. What is the speed and direction of the 15-kg ball?

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Example 3 Starting from rest, two skaters push off against each other on ice where friction is negligible. One is a 54-kg woman and one is a 88-kg man. The woman moves away with a speed of m/s. Find the recoil velocity of the man.

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Example 4 A 15-kg medicine ball is thrown at a velocity of 20 km/hr to a 60-kg person who is at rest on ice. The person catches the ball and slides with the ball across the ice. Determine the velocity of the person and the ball after the collision.

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Example 5: While playing basketball a student lost his balance after making a lay-up and collided with the padded wall behind the basket. His 74-kg body decelerated from 7.6 m/s to 0 m/s in 0.16 seconds. a. Determine the force acting upon the student. b. If the student had hit the concrete wall moving at the same speed, his momentum would have been reduced to zero in seconds. Determine what the force on his body would have been for such an abrupt collision.

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Types of collisions: Collisions are often classified according to whether the total kinetic energy changes during a collision: Elastic collision: one in which the total KE of the system after the collision is equal to the total KE of the system before the collision (i.e., KE is conserved) Inelastic collision: energy is lost during the collision. In a perfectly inelastic collision, the objects stick together after colliding.

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Explosions We know that total system momentum is conserved for collisions between objects in an isolated system. For collisions occurring in isolated systems, there are no exceptions to this law. This same principle of momentum conservation can be applied to explosions. In an explosion, an internal impulse acts in order to propel the parts of a system (often a single object) into a variety of directions.

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Explosions After the explosion, the individual parts of the system have momentum. If the vector sum of all individual parts of the system could be added together to determine the total momentum after the explosion, then it will be the same as the total momentum before the explosion. total system momentum is conserved. A firecracker placed in a stationary object will cause the object to explode. Before the explosion, the total system momentum = 0. Upon explosion, the object will break into a variety of fragments. The vector sum of the momenta of the individual fragments = 0.

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Warm-Up 2 (#6 p. 217) A baseball (m=149g) approaches a bat horizontally at a speed of 40.2m/s and is hit straight back at a speed of 45.6m/s. If the ball is in contact with the bat for a time of 1.10ms, what is the average force exerted on the ball by the bat? Choose the direction of the incoming ball as the positive direction. -11,600N

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Warm-Up 3 (#7 p. 217): A space probe is traveling in outer space with a momentum that has a magnitude of 7.5x107 kg m/s. A retrorocket is fired to slow down the probe. It applies a force to the probe that has a magnitude of 2.0x106N and a direction opposite to the probe’s motion. It fires for a period of 12 s. Determine the momentum of the probe after the retrorocket ceases to fire. 5.1x107 kg m/s

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Collisions in 2-D Collisions usually occur in two or three dimensions. The balls are moving in two dimensions (x- and y-directions). The momentum in the x-direction must be conserved. And, the momentum in the y-direction must be conserved. Find the final velocity (magnitude and direction) of Ball-1. A firecracker placed in a stationary object will cause the object to explode. Before the explosion, the total system momentum = 0. Upon explosion, the object will break into a variety of fragments. The vector sum of the momenta of the individual fragments = 0.

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**Collisions in 2-D Initial Final**

x: (.26)(.54) + (.15)(-.9 sin(50°)) = (.26)(.7 cos(35°)) + (.15)(vf1x) vf1x = .63 m/s y: (.26)(0) + (.15)(-.9 cos(50°)) = (.26)(-.7 sin(35°)) + (.15)(vf1y ) vf1y = .117 m/s Pythagorean gives: vf1 = .64 m/s And tanθ = .12/ θ= 10.78

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Assignment Worksheet P. 217 # 2, 4, 5, 8(a&b only) 16, 17, 20, 21, 23, 28, 30, 32 Challenge problems (optional): 29, 33-35

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Warm-Up 1 Suppose you are standing on the edge of a dock and jump straight down. If you land on sand your stopping time is much shorter than if you land on water. Which of the following statements is correct A) in bringing you to a halt, the sand exerts a greater impulse on you than does the water B) in bringing you to a halt, the sand and the water exert the same impulse, but the sand exerts a greater average force C) in bringing you to a halt, the sand and the water exert the same impulse on you, but the sand exerts a smaller average force.

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