CHAPTER 9 LINEAR MOMENTUM AND COLLISION

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Transcription:

CHAPTER 9 LINEAR MOMENTUM AND COLLISION Couse Outline : Linear momentum and its conservation Impulse and Momentum Collisions in one dimension Collisions in two dimension The center of mass (CM)

9.1 Linear Momentum and its conservation

Conservation of momentum Whenever there is two or more particles in an isolated system interact, the total momentum of the system remains constant. The total momentum of an isolated system at all times is equal to its initial momentum. The only requirement is that there is only internal forces in the system. The net external forces on a system is zero.

Example 1 A 60kg archer stands at rest on frictionless ice and fires a 0.5 kg arrow horizontally at 50 m/s. With what velocity does the archer move across the ice after firing the arrow? What if the arrow were fired in a direction that makes an angle Ө with the horizontal? Example 2 A car weighing 12 kn is driving due north at 30m/s. After driving around a sharp curve, the car is moving east at 13.6 m/s. What is the change in momentum of the car?

9.2 Impulse and Momentum

Example 3 A golf ball of mass 50 g is struck with a club. If the speed of the ball after the strike is 44 m/s above horizontal with certain angle, determine the magnitude of the impulse imparted to the ball during the collision. Example 4 A baseball of mass 0.145 kg is struck by a bat, with the details shown in figure. If the ball and bat are in contact for 1.5 m/s, find the impulse and the average force acted on the ball. Assume that the +x axis is horizontally to the right and the +y axis is vertically upward.

Collisions in one dimension The conservation of linear momentum is used to describe the situation of two particles collide. Collision means an event during two particles come close to each other in a short time and produce impulsive forces on each other. These impulsive forces are assumed that to be much greater than other external force presents. Due to the impulsive forces are internal, they does not change the total momentum of the system. Conclusion : The total momentum of an isolated system just before the collision = The total momentum of the system just after the collision.

Inelastic collisions The momentum is constant. The total kinetic energy is not the same for before and after the collision. Perfectly Inelastic If the particles collide and stick together and moves with a common velocity after collision. Eg. A bullet embedded in a block Inelastic When the colliding objects do not stick together, but some kinetic energy is lost, so the total kinetic energy after collision is less than before the collision.

Consider two particles of mass m 1 and m 2 move with initial velocity v 1i and v 2i along a straight line Two particles collide head one, stick together and the move with some common velocity v f after the collision. Momentum is conserved.

Elastic Collision The total momentum is conserved The total kinetic energy is conserved. Examples: Billiard-ball collisions. The collision of air molecule with the wall No mechanical energy is lost or gained.

Suppose that the mass and initial velocity of both particles are known, those equations in previous slides can used to solved for final speeds. Appropriate signs for V 1i and V 2i must be included in equations above. If the particle moves to the right (+v) If the particle moves to the left (-v)

Example 5 An 1800 kg car stopped at a traffic light is struck from the rear by a 900kg car. The two cars become entangled, moving along the same path as that of the originally moving car. If the smaller car were moving at 20 m/s before the collision, what is the velocity of the entangled cars after the collision? Suppose we reverse the masses of the cars. What if a stationary 900kg car is struck by a moving 1800 kg car? Is the final speed same as before?

Example 6 A proton of mass 1.66 x 10-27 kg moves with a velocity of 1.25 x10 6 m/s before it collides head-on with a helium atom at rest. The helium atom has a mass 6.64 x 10-27 kg and recoils with a speed of 5 x 10 5 m/s. If the collision is elastic find a) Final speed of the proton m 1 = 1.66 x 10-27 kg m 2 = 6.64 x 10-27 kg m 1 v 1i + m 2 v 2i = m 1 v 1f + m 2 v 2f v 1i = 1.25 x 10 6 m/s v 2i = 0, v 2f = 5 x 10 5 m/s (1.66 x 10-27 ) (1.25 x 10 6 ) = (1.66 x 10-27 ) v f + (6.64 x 10-27 ) (5 x 10 5 ) v f = -7.5 x 10 5 m/s a) The fraction of the proton s energy transferred to the helium

Example 7 A block of ma = 1.6 kg initially moving to the right with a speed of 4 m/s on a frictionless, horizontal track collides with a spring attached to a second block of mass m2=2.1 kg initially moving to the left with a speed of 2.5 m/s. The spring constant is 600 N/m. a) Find the velocities of the two blocks after the collision

b) During the collision, at the instant block 1 is moving to the right with a velocity of + 3 m/s, determine the velocity block 2. c) Determine the distance the spring is compressed at that instant.

Two Dimensional Collison

Based on the fig, after the collsion, Particle 1 moves with an angle Ө Particle 2 moves with an angle ф

If the collision is elastic, use the conservation of kinetic energy with V 2i = 0 to give: Only for Elastic Collision If the collision is inelastic, the kinetic energy is not conserved so this above equation cannot be applied.

Example 9 A proton collides elastically with another proton that is initially at rest. The incoming proton has an initial speed of 3.5x10 5 m/s. It makes a glancing collision with the second proton. After the collision, one proton moves off at an angle of 37⁰ to the original direction of motion and the second deflects at an angle of ф to the same axis. Find the final speeds of the two protons and the angle ф

9.5 The center of mass (CM) The center of mass (CM) is the average location of the system. The system moves as if the net external force were applied to a single particle located at CM. This behavior is independent of the other motion such as rotation or vibration of the system. CM of the system is located somewhere of the line joining the particles and is closer to the particle having larger mass.

Example 10 Find the center of mass for a system that consists of three particles at the corners of a right triangle as shown in figure.