7-6 Inelastic Collisions

Transcription

1 7-6 Inelastic Collisions With inelastic collisions, some of the initial kinetic energy is lost to thermal or potential energy. It may also be gained during explosions, as there is the addition of chemical or nuclear energy. A completely inelastic collision is one where the objects stick together afterwards, so there is only one final velocity.

2 Example 7-9 For the completely inelastic collision of two railroad cars that we considered in Example 7-3, calculate how much of the initial kinetic energy is transformed to thermal or other forms of energy. Before the collision, the total kinetic energy is 1 2 m v 2 A A = 1 2 (10,000. kg)(24.0 m/s)2 = 2.88x10 6 J After the collision, the total kinetic energy is 1 2 (m +m A B )v'2 = 1 2 (20,000. kg)(12.0 m/s)2 =1.44x10 6 J The energy transformed is KE i - KE f 2.88x10 6 J -1.44x10 6 J =1.44x10 6 J, half of the initial KE

3 Example 7-10 The ballistic pendulum is a device used to measure the speed of a projectile, such as a bullet. The projectile, of mass m, is fired into a large block (of wood or other material) of mass M, which is suspended like a pendulum. (Usually, M is somewhat greater than m.) As a result of the collision, the pendulum and projectile together swing up to a maximum height h. Determine the relationship between the initial horizontal speed of the projectile, v, and the maximum height h. (a) total p before impact = total p after impact mv = (m +M)v' (b) E just after impact = E at max height 1 2 (m +M)v'2 +0 = 0 +(m +M)gh v' = 2gh We insert this value for v' into our equation in part (a) to get v = m +M m v' = m +M m 2gh

4 7-7 Collisions in Two or Three Dimensions Conservation of energy and momentum can also be used to analyze collisions in two or three dimensions, but unless the situation is very simple, the math quickly becomes unwieldy. Here, a moving object collides with an object initially at rest. Knowing the masses and initial velocities is not enough; we need to know the angles as well in order to find the final velocities.

5 7-7 Collisions in Two or Three Dimensions Problem solving: 1. Choose the system. If it is complex, subsystems may be chosen where one or more conservation laws apply. 2. Is there an external force? If so, is the collision time short enough that you can ignore it? 3. Draw diagrams of the initial and final situations, with momentum vectors labeled. 4. Choose a coordinate system.

6 7-7 Collisions in Two or Three Dimensions 5. Apply momentum conservation; there will be one equation for each dimension. 6. If the collision is elastic, apply conservation of kinetic energy as well. 7. Solve. 8. Check units and magnitudes of result.

7 Example 7-11 Billiard ball A moving with speed v A =3.0 m/s in the +x direction strikes an equal-mass ball B initially at rest. The two balls are observed to move off at 45 degrees to the x axis, ball A above the x axis and ball B below. What are the speeds to the two balls after the collision? ' ' for x : mv A = mv A cos(45) +mv B cos(-45) for y : 0 = mvsin(45) +mvsin(-45) From the y equation, we get v B ' ' sin(45) = -v A sin(-45) = -v ' A sin45 -sin45 = v ' A Using this in the x equation, we have v A = v A ' v A ' = v B ' cos45 + v B ' = v A 2cos45 cos45 = 2v A ' = 3.0 m/s 2(0.707) cos45 = 2.1 m/s

8 7-8 Center of Mass In (a), the diver s motion is pure translation; in (b) it is translation plus rotation. There is one point that moves in the same path a particle would take if subjected to the same force as the diver. This point is called the center of mass (CM).

9 7-8 Center of Mass The general motion of an object can be considered as the sum of the translational motion of the CM, plus rotational, vibrational, or other forms of motion about the CM.

10 7-8 Center of Mass For two particles, the center of mass lies closer to the one with the most mass: where M is the total mass.

11 Example 7-12 Three people of roughly equal masses m on a lightweight (air-filled) banana boat sit along the x axis at positions x A =1.0 m, x B =5.0 m, and x C =6.0 m, measured from the left-hand end. Find the position of the CM. Ignore the mass of the boat. x CM = mx A +mx B +mx C = m(x A + x B + x C ) m +m +m 3m (1.0 m +5.0 m +6.0 m) x CM = = 12.0 m = 4.0 m 3 3 The CM is 4.0 m from the left - hand end of the boat.

12 7-8 Center of Mass The center of gravity is the point where the gravitational force can be considered to act. It is the same as the center of mass as long as the gravitational force does not vary among different parts of the object.

13 7-8 Center of Mass The center of gravity can be found experimentally by suspending an object from different points. The CM need not be within the actual object a doughnut s CM is in the center of the hole.

14 7-9 CM for the Human Body The x s in the small diagram mark the CM of the listed body segments.

15 7-9 CM for the Human Body The location of the center of mass of the leg (circled) will depend on the position of the leg.

16 7-9 CM for the Human Body High jumpers have developed a technique where their CM actually passes under the bar as they go over it. This allows them to clear higher bars.

17 Example 7-13 Determine the position of the CM of a whole leg (a) when stretched out, and (b) when bent at 90 degrees. Assume the person is 1.70 m tall. (21.5)(9.6) +(9.6)(33.9) +(3.4)(50.3) (a) x CM = = 20.4 units So the CM is 20.4 units from the hip joint, or = 31.7 units from the foot. The person is 1.70 m tall, this is (1.70 m)(31.7/100) = 0.54 m above the bottom of the foot. (21.5)(9.6) +(9.6)(23.6) +(3.4)(23.6) (b) x CM = =14.9 units (3.4)(1.8) +(9.6)(18.2) +(21.5)(28.5) y CM = = 23.0 units So the CM is (1.70 m)(14.9/100) = 0.25 m from the hip joint and (1.70 m)(23.0/100) = 0.39 m above the floor.

18 7-10 Center of Mass and Translational Motion The total momentum of a system of particles is equal to the product of the total mass and the velocity of the center of mass. The sum of all the forces acting on a system is equal to the total mass of the system multiplied by the acceleration of the center of mass: (7-11)

19 7-10 Center of Mass and Translational Motion This is particularly useful in the analysis of separations and explosions; the center of mass (which may not correspond to the position of any particle) continues to move according to the net force.

20 Momentum of an object: Newton s second law: Summary of Chapter 7 Total momentum of an isolated system of objects is conserved. During a collision, the colliding objects can be considered to be an isolated system even if external forces exist, as long as they are not too large. Momentum will therefore be conserved during collisions.

21 Summary of Chapter 7, cont. In an elastic collision, total kinetic energy is also conserved. In an inelastic collision, some kinetic energy is lost. In a completely inelastic collision, the two objects stick together after the collision. The center of mass of a system is the point at which external forces can be considered to act.

22 Homework - Ch.7 Questions # s 3, 4, 5, 7, 12, 15, 19 Problems # s 3, 5, 7, 11, 15, 17, 23, 25, 27, 31, 35, 41, 49, 57