1、Chapter 6 MomentumCan we solve conveniently all classical mechanical problems with Newtons three laws?No,the problems such as collisions.This busy image was recorded at CERN(欧洲粒子物理研究所),in Geneva,Switzerland Four galaxies colliding taken by using Hubble Space Telescope.Non-touched collisions6-1 How t
2、o analyze a collision?In a collision,two objects exert forces on each other for an identifiable(可确认的)time interval,so we can separate the motion into three parts.Before,during,and after the collision.During the collision,the objects exert forces on each other,these forces are equal in magnitude and
3、opposite in direction.1)We usually can assume that these forces are much larger than any forces exerted on the two objects by other bodys in the environment.The forces vary with time in a complex way.2)The time interval during the collision is quite short compared with the time during which we are w
4、atching.These forces are called“impulsive forces(冲力)”.Characteristics of a collision6-2 Linear Momentum To analyze collisions,we define a new dynamic variable,the“linear momentum”as:(6-1)The direction of is the same as the direction of .The momentum (like the velocity)depends on the reference frame
5、of the observer,and we must always specify this frame.Pm vP vP FmadPFdt The equivalence of and depends on the mass being a constant.Any conditions for existence of above Eq.?Can be related to?P dvdmvdPFmamdtdtdt dPFdt(6-2)F6-3 Impulse(冲量)and Momentum(动量)In this section,we consider the relationship b
6、etween the force that acts on a body during a collision and the change in the momentum of that body.Fig 6-6 shows how the magnitude of the force might change with time during a collision.tFF(t)Fig 6-6avFitft0 From Eq(6-2),we can write the change in momentum as To find the total change in momentum du
7、ring the entire collision,we integrate over the time of collision,starting at time (the momentum is )and ending at time (the momentum is ):(6-3)dPFdt itftffiiPttPdPFdt fP iP The left side of Eq(6-3)is the change in momentum,The right side defines a new quantity called the impulse.For any arbitrary f
8、orce ,the impulse is defined as (6-4)A impulse has the same units and dimensions as momentum.From Eq(6-4)and(6-3),we obtain the“”:(6-5)fiPPP fittJF dt FJ fiJPPP Notes:1.Eq(6-5)is just as general as Newtons second law 2.Average impulsive force avfiJFtPP avF3.The external force may be negligible,compa
9、red to the impulsive force.Sample problem 6-1 A baseball(棒球)of mass 0.14 kg is moving horizontally at a speed of 42m/s when it is struck by the bat it leaves the bat in a direction at an angle above its incident path and with a speed of 50m/s (a)find the impulse of the force exerted on the ball.(b)a
10、ssuming the collision lasts for 1.5ms what is the average force.(c)find the change in the momentum of the bat.35Sample problem 6-2 A cart of mass m1=0.24kg moves on a linear track without friction with an initial velocity of 0.17 m/s.It collides with another cart of mass m2=0.68 kg that is initially
11、 at rest.The first cart carries a force probe that registers the magnitude of the force exerted by one cart on the other during the collision.The output of the force probe is shown in Fig.6-9.Find the velocity of each cart after the collision.Fig 6-948122410F(N)T(ms)68 6-4 Conservation of Momentumdt
12、tFPPif)(0,0)(ifPPtFif0dtPd,namely or is a constant.P1)When the net external force acting on a system is zero,the total momentum of the system is conserved.(6-12)2)The internal forces(内力)of a system do not change the momentum of the system.gbm2m000bgvv00p0gbbmgm vvgbm2m2gbvv 3)Equation(6-12)is called
13、 the law of conservation of linear momentum.It is a general result,valid for any type of interaction between the bodies.4)Because we derived the law using Newtons law,the law is valid in any inertial frame of reference.How about the law when using different inertial frames?Pmay be different,but are
14、conserved.6-5 Two Body CollisionsJ 1）Two-body collision If the two bodies are isolated from environment,the total momentum of the two-body system is conserved before and after collision:(6-15)Another way of writing Eq(6-15)is (6-16)or (6-17)ffiivmvmvmvm22112211)()(222111ififvvmvvm21PP a)Equation(6-1
15、7)can be written as .This equality follows directly from Newtons third law.b)In some collision,the bodies stick together and moves with a common final velocity ,Eq(6-15)becomes21JJfiivmmvmvm)(212211fvc)Often we have a“head-on”collision(正面碰撞),in this case Eq(6-15)can be written (6-19)fxfxixixvmvmvmvm
16、22112211Sample problem 6-8 A puck(冰球)is sliding without friction on the ice at a speed of 2.48m/s.It collides with a second puck of mass 1.5 time that of the first and moving initially with a velocity of 1.86m/s in a direction away from the direction of the first puck(Fig 6-15).After the collision,the moves at a velocity of 1.59m/s in a direction at a angle of from its initial direction.Find the speed and direction of the second puck.40501m Solution:From Fig 6-15 and conservation of momentum,we
