G12 Momentum P1

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Information about G12 Momentum P1

Published on May 13, 2010

Author: kwarne

Source: slideshare.net


A set of slides created to teach G12 Momentum P1 to learners at Bishops Diocesan College in Cape Town.

Newton’s Third Law



MOMENTUM AND ITS CONSERVATION MOMENTUM:(Symbol : p) Definition: Momentum of an object is the product of the mass and velocity of a moving body. p = m. v units: kg, m.s'1 N.B. Since velocity is a vector quantity, momentum is also a vector quantity.

THE CONSERVA TION OF LINEAR MOMENTUM Definition: It states that the total linear momentum of an isolated system remains constant in both magnitude and direction. Alternatively: In an isolated system the total momentum before collision equals the total momentumafter the collision. Total pbefore = Total pafter . Pa + pb = Pa2 + Pb2 (Ma.va + mb.vb ) before= (mava + mbvb)after (0.1)(+10) + (0.1)(-10) = (0.1).(0)+(0.1).(0) +1 + (-1) = 0 0 = 0 Total momentum conserved!! +10km.hr-1 -10km.hr-1 0 km.hr-1 A B

LAW OF MOMENTUM Definition: The applied resultant Force is equal to the rate of change of momentum, and that this change is in the direction of the applied Force. Fres = mv - mu t This is derived from Newton's 2nd Law,

IMPULSE Impulse is the product of the resultant force being applied for a certain time interval. Fres = ma = m v/t = mv-mu/t Fres t = mv - mu Impulse = change in momentum

Collisions – Elastic & Inelastic Objects may collide in one of two ways: 1. Combine on impact 2. Separate on impact. In EITHER CASE collisions can be: a. Elastic (Ek is conserved and no energy lost) b. lnelastic (Ek is not conserved - some is converted to heat, sound etc) N.B. Momentum is conserved in any collision between two objects, but kinetic energy is conserved only in a perfectly elastic collision.

CHANGE IN MOMENTUM change in momentum = final momentum - initial momentum p = mvf - mvi unit: k,m,s.1 N.B. Momentum is conserved in any collision between two objects

PRACTICAL CONSIDERATIONS Allow a trolley, on a "friction compensated" slope, to run down the slope before releasing a heavy mass piece vertically anta the moving trolley. mass of trolley = 0.8 kg; mass added 1,6 kg frequency of timer ~ 50 Hz ave. V before = s/t = 0.012111/0.02 = 06 m.s-1 ave. V after = s/t = 0.004/0.02 = O 2 m s-1 Momentum before collision p =mv = 0,8 x 0,6 = 0,48 kg.m.s' Momentum after mass added p = mv = 2,4 x 0.2 ~ 0,48 kg.m.s-1 Le. momentum before collision = momentum after coliision and thus momentum is conserved.

"EXPLOSION' BETWEEN TROLLEYS apparatus as seen in sketch. Assume frictionless surface and arrange blocks and loaded trolleys so that they hit the blocks simultaneously ie. t = 1 Distance x1 represents V1 and Distance X2 represents V2 Since distance s  V as time is constant) Momentum of M1 = momentum M2 Le. M1xX1 = M2xX2

v1 Vertical component Horizontal component Resolving vectors into components graphically

Momentum in two dimentions 0.1kg 4.27m.s-1 5m.s-1 41o 22.8o Calculate the impulse experienced by the ball if it is in contact with the ground for 0.05s.

Momentum in two dimensions 0.1kg 4.3m.s-1 5m.s-1 41o 22.8o Resolve into components.

Momentum in two dimensions 0.1kg 4.8m.s-1 5m.s-1 41o 22.8o Solve separately for components.

Momentum in two dimensions 0.1kg 4.27m.s-1 5m.s-1 41o 22.8o Solve separately for x & y components.

Momentum in 2d… ? m.s-1 2kg  B 3m.s-1 A A 2kg 60o I PB after collision Solution II vB after collision 1.5m.s-1 @

Momentum in 2d… ? m.s-1 2kg  B 3m.s-1 A A 2kg 60o 1.5m.s-1 @ Calculate: I PB after collision II vB after collision

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