ACCESS– Summer 2014 Elastic and Inelastic Collisions - Numerical Integration

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ACCESS Summer 2014 Elastic and Inelastic Collisions - Numerical Integration. Gernot Laicher University of Utah - Department of Physics & Astronomy. - PowerPoint PPT Presentation

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Math 5900 Summer 2010 Lecture 1: Simple Harmonic Oscillations

ACCESS Summer 2014

Elastic and Inelastic Collisions - Numerical Integration

Gernot LaicherUniversity of Utah - Department of Physics & Astronomy

Newtons second law

..or in a more general form

Momentum

m=constant ( original form of Newtons second law.Re-arrange Newtons second law

A net Force Fnet applied for an infinitesimally small time interval dt results in a change of momentum dp.Net force applied over a time interval (tf-ti)

( Impulse or Impulse Integral I

Newtons second law :

Impulse = Change of Momentum.Example:

A golf ball (m=46g) is struck by a club which applies a constant force of 200N for a time period of 1/100 of a second. What is the impulse on the golf ball? If the ball is at rest initially, how fast will it be after it has been hit?

Change in momentum equals impulse

Conservation of Momentum:

For a closed system (no external forces applied to the system) the total linear momentum of the system is conserved. (The sum of all the individual momenta of the pieces that make up the system remains constant.)To approach a system with no applied external force, we can consider two gliders on an air track and just look at the linear momentum in the horizontal direction.

Demo: One glider is initially at rest; both gliders have equal mass.

Before the collision:

Total momentum:

V1i

V2i = 0

After the collision:

Total momentum:

V1f = 0

V2f = V1i

Mechanical energy of the system before and after the collision:

( Mechanical energy was conserved ( Totally elastic collision

Before the collision:

After the collision:

V2i = 0

V1i

Vf = V1i

Mechanical energy of the system:

Half the mechanical energy lost in collision (converted to heat) ( Inelastic collisionNumerical Integration

In general, applied forces are not constant during collision:

Impulse = Area under the force-time curve (the integral).

In lab activity: Force sensors ( Computer

Measurement not continuous - taken in discrete time intervals.

Estimate time integral with numerical integration.

Assume force remains constant over each time interval. Then calculate the sum of the rectangular areas (width ti and height Fi).

(Here, ti = t because we sample in equal time intervals.)

Alternative method: Trapezoidal Rule. SEQ CHAPTER \h \r 1Partition area under the function into n intervals (widths x). Label start of each partition as x0, x1, x2, x3, ... xn. ( f(x)

Width of intervals Width of intervals:

f(x)

Width of intervals:

f(x)

Width of intervals:

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x

x

xo x1 x2 x3 x4 x5 x6 x7 x8 x9

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x

x

xo x1 x2 x3 x4 x5 x6 x7 x8 x9

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x

x

xo x1 x2 x3 x4 x5 x6 x7 x8 x9

More partitions ( smaller time intervals Connect with straight lines f(xi) and f(xi+1) Area approximated by sum of n trapezoid areas.

w=width of trapezoid (= x)

c = length of the leg1= f(xi)

d = length of leg 2 = f(xi+1)

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Area Ai of ith trapezoid

Total area Atotal

Example Using Trapezoidal Rule:Integrate f(x) = x4 for 1 x 3 with n = 8. The width x of each partition is:

x0 = 1.00 f(x0) = (1.00)4 = 1.0000

x1 = 1.25 f(x1) = (1.25)4 = 2.4414(Rounding to 4 decimal places)

x2 = 1.50 f(x2) = (1.50)4 = 5.0625

x3 = 1.75 f(x3) = (1.75)4 = 9.3789

x4 = 2.00 f(x4) = (2.00)4 = 16.0000

x5 = 2.25 f(x5) = (2.25)4 = 25.6289

x6 = 2.50 f(x5) = (2.50)4 = 39.0625

x7 = 2.75 f(x7) = (2.75)4 = 57.1914

x8 = 3.00 f(x8) = (3.00)4 = 81.0000

Compared to exact result:

(Numerical integration within 1% of exact result)