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7.1 Impulse and Momentum Change From Fav = DP/Dt = ma , one obtains Fav Dt (Impulse) = DP (Change in Momentum) which can be used to calculate the impulse or the average force during the collision.
7.2
Varying Impact Force: impact
force or collision force is very intense and last for only a
very short in duration. It rises from zero to a maximum value
and falls off to zero in milliseconds.
From the definition Fav = DP/Dt, the change of momentum in a collision can be calculated as DP = Fav Dt In a two-body collision system, the collision force is internal to the system. By considering the two bodies involved in a collision as a single system, the net force acting on the system is zero (why?), i.e., Fav = 0 leading to DP = Pi - Pf = 0, Pi = Pf = constant ---------------------------------------------- (1) The momentum of each body changes in a collision, but the total momentum of the system (two bodies) is constant as shown in (1).
7.5 Collisions Collisions involve internal forces, no external forces. The simplest case of collisions involves two-body colliding head on. During collisions, body A exerts a force on body B, and body B exerts an equal and opposite force on body A. No external forces are involved in the collision. (A) Elastic Collision: Kinetic Energy is Conserved Consider a two-body head-on collision. If the collision is elastic, no permanent deformation occurred to either bodies. No kinetic energy is transformed into potential energy associated with any deformation, i.e., the total kinetic energy after the collision is the same as the total kinetic energy before the collision. This is why the kinetic energy is conserved in an elastic collision. Two bodies with masses m1 and m2 undergo head-on elastic collision. The velocities of the two bodies before collision are V1i and V2i = 0 (m2 was at rest before collision) respectively. Determine their velocities, V1f and V2f, after the collision. Conservation of momentum: the total momentum before and after the collision is the same, i.e., m1V1i + m2V2i = m1V1f + m2V2f -------------------------------------------------- (1) Kinetic energy is conserved if the collision is elastic (no metal bending or configurational change). The total kinetic energy before and after the collision is the same: (1/2)m1 (V1i)**2 + (1/2)m2 (V2i)**2 = (1/2)m1 (V1f)**2 + (1/2)m2 (V2f)**2 ------ (2) Equations (1) and (2) can be solved for the two unknowns V1f and V2f: Since V2i = 0, (1) and (2) are reduced to m1V1i + 0 = m1V1f + m2V2f ------------------------------------------------------- (1') (1/2)m1 V1i**2 + 0 = (1/2)m1 V1f**2 + (1/2)m2 V2f**2 -------------------------- (2') (1') becomes m1 (V1i - V1f) = m2V2f --------------------------------------------------- (1") (2') becomes m1 [V1i**2 - V1f**2] = m2 V2f**2 which can be factored to m1 (V1i - V1f)(V1i + V1f) = m2V2f**2 ------------------------------------ (2") Divide (2") by (1") leads to (V1i + V1f) = V2f ------------------------------------------ (3) Substitute (3) into (1") leads to V1f = (m1 - m2)/(m1 + m2) V1i ---------------------- (4) Substituting (4) into (3) leads to V2f = 2m1 /(m1 + m2) V1i ------------------------- (5)
(B) Inelastic Collisions: momentum is conserved, but kinetic energy is not. Equation (1) is valid, but equation (2) is no longer valid for inelastic collisions. We now have one equation with two unknowns. We cannot solve the problem unless an additional condition or equation is given or identified. In the case of a complete inelastic head-on collisions, the two bodies stuck together so that they move at the same velocity Vf = V2f = V1f . Thus, (1) is reduced to Vf = (m1V1i + m2V2i)/(m1 + m2)
HW Chap 7: 11, 27, 39, 59 |