2012_1061_Lecture_Ch_09

1 1 1 1 2 2 2 2 ma va mb vb ma v a mb v b 2 2 2 2

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Unformatted text preview: − → d→ − p F= dt •  The momentum changes by − = − dt →→ dp F •  If we integrate over the duration of the collision ￿ i f tf − → − → F dt = J ti ￿ tf − → − → J= F dt →→ → → d− = − f − − i = ∆− = p p p p Impulse ￿ ti Conservation of Energy and Momentum •  In the case of elastic collisions we have conservation of kinetic energy and linear momentum. 1 1 1 1 2 2 ￿2 ￿2 mA vA + mB vB = mA v A + mB v B 2 2 2 2 Elas.c Collision KA + KB = K ￿ A + K ￿ B + thermal and other forms of energy Elastic Collisions in One Dimension •  Conservation of momentum reads: mA vA + mB vB = mA v ￿ A + mA v ￿ B •  For an elastic collision we have conservation of kinetic energy 1 1 1 1 2 2 ￿2 ￿2 mA vA + mB vB = mA v A + mB v B 2 2 2 2 •  If we know the masses and velocities before collision we can solve for the velocities after collision. 2 2 2 2 mA (vA − v ￿ A ) = mB (v ￿ B − v ￿ B ) mA (vA − v ￿ A ) = mB (v ￿ B − vB ) •  Taking the ratio we have vA + v ￿ A = v ￿ B + vB vA − vB = −(v ￿ A − v ￿ B ) [head − on (1 − D) elastic collision] Examples in one dimension Examples of Elastic collisions in 2 Dim Inelastic Collisions •  The kinetic energy in Inelastic collisions is not conserved •  If two objects stick together the collision is inelastic Center of Mass Center of Mass (2) xCM mA xA + mB xB mA xA + mB xB = = mA + mB M xCM xCM m(xA + xB ) (xA + xB ) =m = 2m 2 m1 x1 + m2 x2 + . . . + mn xn = = m1 + m2 + . . . + mn n ￿ mi xi i=1 M Center of Mass xCM = ￿ mi xi , M ￿ CM = r yCM = ￿ ￿ ￿ mi ri M mi y i , M zCM = ￿ mi zi M xCM 1 = M yCM 1 = M zCM 1 = M ￿ CM = r 1 M ￿ ￿ ￿ ￿ x dm y dm z dm ￿ dm r...
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This note was uploaded on 02/03/2014 for the course PHYSICS 1061 taught by Professor Tsankov during the Fall '09 term at Temple.

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