1. 1 6. Center of mass, Collisions 6.1. The center of mass To describe the motion of a system of particles (also for their continuous distribution – a solid body), one introduces the concept of the center of mass. A baseball bat thrown into the air moves in a complicated manner but one of its points, the center of mass (CM), moves in a parabolic path. CM moves as if all the system’s mass were concentrated there and all external forces were applied to it (to be proved). We define the position of CM by a vector equation (6.1) For a solid body summation in Eq.(6.1) must be replaced by integration (6.2) where is a mass density. (figure from HRW)

2. 2 For a uniform body (ρ=const), position of its CM depends on the body’s shape only (6.3) Examples 1.Two particles of masses m 1 and m 2 separated by distance d. The position of CM we find placing both particles on an x axis. From Eq.(6.1) one obtains Putting x 1 =0, we have 2.The center of mass of a triangle. In this case there is no need to use Eq.(6.3), but dividing a triangle into segments it can be concluded that CM is placed at the intersection of triangle medians. The center of mass, cont.

3. 3 For a system of n particles one can write from a CM definition (6.1) (6.4) Differentiating both sides of Eq.(6.4) vs. time gives (6.5) By differentiating Eq.(6.5) in respect to time one obtains (6.6) Eq.(6.6) can be rewritten as (6.6a) The right side of Eq.(6.6a) includes all forces, both the internal forces between the particles and the external forces acting from tha outside. From the Newton’s third law the sum of all internal forces reduces to zero and the right side of Eq.(6.6a) is the vector sum of all external forces, the net force. In this case one obtains (6.7) This proves the statement given at the beginning of this chapter. 6.2. Motion of a system of particles Neton’s second law applied to the center of mass

4. 4 Eq.(6.5) can be rewritten as (6.8) Thus, the linear momentum of a system of particles (RHS of above equation) is equal to the product of the total mass of the system and the velocity of the center of mass: (6.9) Calculating the time derivative of Eq.(6.9) we have (6.10) From above relation and taking into account Eq.(6.7) one obtains (6.11) From (6.11) it follows, that if is zero, then (6.12) The linear momentum of the system is changed by the net external force (Newton’s second law for a system of particles). The momentum of a system is not changed if it is isolated (law of momentum conservation for a system of particles). 6.3. Momentum of a system of particles

5. 5 The object’s momentum can be changed by the collision with other object. The forces during collision act for a short time and are large. An example is the collision of a baseball with a baseball bat. From the Newton’s second law (6.13) The net change in the momentum is obtained after integrating Eq.(6.13) in time Δt or(6.14) The change in objet’s mometum is equal to the force impuls. 6.4. Collisions (figure from HRW) In the figure to the left we see the magnitude of force varying with time for the collision of a ball with a bat. The area under the curve is equal to the force impulse in the collision.

6. 6 In many cases we consider the system isolated in which the objects mutually collide. In that case the vector of linear momentum of the system cannot change. The total kinetic energy of colliding bodies can be conserved and in this case we have an elastic collision. When the kinetic energy of the system is not conserved, a collision is called inelastic. If the colliding bodies stick together, the collision is called completely inelastic. Two bodies collide moving along an x axis. From the conservation of a linear momentum (6.15) The total kinetic energy is also conserved (6.16) Collisions, cont. Elastic collision in one dimension

7. 7 Solving the simulaneous equations (6.15) and (6.16) for u 1 and u 2 one obtains When the particles have the same masses, they exchange the velocities after collision Collisions, cont. Completely inelastic collision in one dimension From the conservation of linear momentum we have (6.17) and hence the velocity of sticked together masses is It is easy to check that the kinetic energy of the body after the collision is less than the sum of kinetic energies of the bodies before the collision.