Momentum And Impulse Physics Games : Pipeng Free Online Software

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Pipeng Free Online Software : Momentum And Impulse Physics Games

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Pipeng : Physics Translation Motion : Momentum And Impulse Physics Homework BETA Game Module

Momentum And Impulse Physics Games

Description : Physics translation motion games : momentum and impulse

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Module List

BETA : Translation : Momentum 01 : Change Of Linear Momentum And Impulse : Single Mass : Beta Physics Homework Game

Description : Change of momentum and impulse: single mass, linear.

Discussion : The change of momentum dp is equal to the impulse J.

dp = m * (vf - vi)

rearranging

m = dp / (vf - vi)

vf = dp / m + vi

vi = vf - dp / m

Impulse

J = F * t = dp

rearranging

F = J / t = dp / t

t = J / F = dp / F

where

F = force amplitude

m = mass

dp = change in momentum

vi = initial velocity amplitude

vf = final velocity amplitude

t = elapsed time

J = impulse

BETA : Translation : Momentum 02 : Change Of Vector Momentum And Impulse : Single Mass : Beta Physics Homework Game

Description : Final momentum vector from the sum of the initial momentum vector pi and an impulse vector J.

Discussion : In order to add two vectors pi and J they must first be converted into cartesian values which are summed to find the resultant momentum cartesian values px and py. Values px and py can then be converted back into the final vector. See Figure Momentum And Impulse Vector Sum

px = pi * cosd(θi) + J * cosd(θj)

py = pi * sind(θi) + J * sind(θj)

pf = √(px² + py²)

θf = atan2d(py, px)

To convert from pf and θf back to J and θj

px = pf * cosd(θf)

py = pf * sind(θf)

define custom variables

Jx = px - pi * cosd(θi)

Jy = py - pi * sind(θi)

Vector J

J = √(Jx² + Jy²)

θj = atan2d(Jy, Jx)

For the special case where θf = 0 and θi and θj are known, find J and pi

px = pf

py = 0

define custom variable

aox = -sind(θj) / sind(θi)

Vectors J and pi

J = pf / (ioj * cosd(θi) + cosd(θj)

pi = ioj * J

where

pi pf and J = momentum / impulse amplitudes

θi θf θj = momentum / impulse angles

px and py = cartesian values of final momentum vetor

Jx Jy = cartesian values of vector J

ioj = ratio of pi / J

Note : To find pi and J from pf where θf ≠ 0. Rotate all of the vectors by -θf so that θf = 0;. The angles are in degrees. You are recommended to use the trigonometry degree functions sind(), cosd(), asind(), acosd() and atan2d().

BETA : Translation : Momentum 03 : Change Of Linear Momentum And Impulse : Force Versus Time Graph Or Chart : Beta Physics Homework Game

Description : Change of linear momentum and impulse : force versus time graph or chart.

Discussion : The change of momentum is equal to the impulse, the area under the force time graph line (or the integral). See Figure Impulse : Force verus Time Chart

The game has 3 sets of questions with 4 questions per set.

J, pi, pf, vf

m, J, F, vi

pi, pf, J, t3

The basic equations are :

J = F (t2 + t3 - t1) / 2

pi = m vi

pf = pi + J

vf = pf / m

rearranging

m = pf / vf = pi / vi

J = pf - pi

F = 2 J / (t2 + t3 - t1)

vi = pi / m

pf = m vf

t3 = 2 J / F + t1 - t2

where

vi = initial velocity

vf = final velocity

pi = initial momentum

pf = final momentum

m = mass

J = impulse

t1 t2 t3 = time 1 2 3

BETA : Translation : Momentum 04 : Conservation Of Linear Momentum : Two Stationary Masses Separate : Beta Physics Homework Game

Description : Conservation of linear momentum : two stationary masses separate.

Discussion : If two stationary masses are separated by a force time impulse, the impulse acts equally on both masses. The force is assumed constant for the duration of the impulse. The sign of the velocity is ignored (ie both velocities are positive). See Figure Impulse : Two Masses Separate

The game has 3 sets of questions with 3 questions per set.

J, va, vb

va, F, mb

J, t, ma

The basic equations are :

J = F t

va = J / ma

vb = J / mb

rearranging

ma = J / va

mb = J / vb

F = J / t

t = J / F

J = va ma = vb mb

where

ma = mass a

mb = mass b

va = velocity of mass a

vb = velocity of mass b

J = impulse

F = force

t = time

BETA : Translation : Momentum 11 : Conservation Of Linear Momentum : Two Masses Joined After Collision : Beta Physics Homework Game

Description : Conservation of linear momentum for two masses joined after the collision.

Discussion : Momentum is conserved before and after the collision, ie it is constant. Both masses have the same velocity after the collision. See Figure Conservation of Linear Momentum - Masses Joined After Collision

p = ma * va + mb * vb = vf * (ma + mb)

rearranging

vf = p / (ma + mb)

ma = p / vf - mb

mb = p / vf - ma

vb = (p - ma * va) / mb

va = (p - mb * vb) / ma

where

ma mb = mass of a and b

va and vb = initial velocity of masses a nd b

p = total momentum

vf = final velocity

BETA : Translation : Momentum 12 : Conservation Of Linear Momentum : Two Masses Separate After Collision : Beta Physics Homework Game

Description : Conservation of linear momentum for two masses separate after the collision.

Discussion : Momentum is conserved before and after the collision, ie it is constant. Therefore the change in momentum of mass a is equal and opposite to the change in momentum of mass b. See Figure Conservation of Linear Momentum - Masses Separate After Collision

dpa = paf - pai

dpb = - dpa

rearranging

paf = pai + dpa

pbf = pbi + dpb

pai = paf - dpa

pbi = pbf - dpb

where

pai pbi = initial momentum of a and b

paf pbf = final momentum of mass a nd b

dpa dpb = change in momentum of mass a and b

BETA : Translation : Momentum 21 : Conservation Of Vector Momentum : Moving Mass Collides With Stationary Mass : Beta Physics Homework Game

Description : Final momentum vectors of a stationary mass and a moving mass after collision, with the masses separated.

Discussion : Momentum is conserved during the collision. The total momentum of masses a and b after the collision is equal to the initial momentum of the moving mass. See Figure Moving Mass Collides With A Stationary Mass

Vectors pa and pb - define custom variable

aob = -sind(θb) / sind(θa)

Vectors

pb = pi / (aob * cosd(θa) + cosd(θb)

pa = aob * pb

Vectors pb and pi

pb = - pa * sind(θa) / sind(θb)

pi = pa * cosd(θa) + pb * cosd(θb)

where

pi = initial momentum amplitude

pa pb = final momentum of mass a and b

θa θb = angle of momentum vectors pa and pb relative to pi

Angles θa and θb will have different sign depending on whether they are clockwise or anticlockwise. The angles are in degrees. You are recommended to use the trigonometry degree functions sind() and cosd().

BETA : Translation : Momentum 22 : Conservation Of Vector Momentum : Three Stationary Masses Separate : Beta Physics Homework Game

Description : Final momentum vectors of three stationary masses after separation.

Discussion : Momentum is conserved during the collision. The total momentum of a b and c after the collision is equal to zero. See Figure Three Stationary Masses Separate

Vectors pa and pb - define custom variable

aob = -sind(θb) / sind(θa)

Vectors

pb = pc / (aob * cosd(θa) + cosd(θb)

pa = aob * pb

Vectors pb and pc

pb = - pa * sind(θa) / sind(θb)

pc = pa * cosd(θa) + pb * cosd(θb)

where

pa pb pc = final momentum of mass a b and c

θa θb = angle of momentum vectors pa and pb relative to pc

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