Nov. 27, 2017 Momentum & Kinetic Energy in Collisions In our initial discussion of collisions, we looked at one object at a time, however we'll now look at the system of objects, with the assumption that the system is closed. This allows to determine the results of the collision without knowing any details about the collision. We'll also be interested in the total kinetic energy of the system of the two colliding bodies. If the total kinetic energy is unchanged by the collision, then the kinetic energy is conserved, and we have an elastic collision. If some of the kinetic energy is transferred to another form (thermal, acoustic, etc), then the kinetic energy is not conserved and we have an inelastic collision. We can approximate some situations as elastic even though their might be some energy loss (usually very small). However, the inelastic collision always involves a loss in kinetic energy of the system. The greatest loss occurs when the two objects stick together, in what we call a completely inelastic collision.
Inelastic Collisions in One Dimension Since we have a closed system, the law of the conservation of momentum applies giving us: In a completely inelastic collision, the objects stick together, and therefore have the same final velocity (V), given us: or more simply, if object 2 is stationary: and the final velocity is given by:
Inelastic Collisions in One Dimension (cont'd) Velocity of the Center of Mass In a closed system, the velocity of the center of mass CANNOT be changed by a collision. Why? (Because it is closed, there is no added mass or external force to cause it to change) We can relate the to the total linear momentum by writing We know the total linear momentum is conserved so that the total can also be written by: Now, solving to find the gives us:
Elastic Collisions in One Dimension In our elastic collisions we have the conservation of kinetic energy giving us: Stationary Target If our object one collides with a stationary object 2, the we have the conservation of linear momentum and kinetic energy giving us: (linear momentum) (kinetic energy) If we know the masses of the objects and the initial velocity of object one, then with two equations, we can find the other two unknowns (v 1f and v 2f ). SPECIAL CASES: Equal masses > If m 1 = m 2, then and Massive target > If m 2 m 1, then and Massive projectile > If m 1 m 2, then and
Moving Target Elastic Collisions in One Dimension (cont'd) Now let us look at a collision in which the projectile and target are both moving. Our momentum and kinetic energy equations are: (linear momentum) (kinetic energy) Again, we can solve these two equations for our unknowns final velocities (v 1f and v 2f ) and get the equations:
Collisions in Two Dimensions When two objects collide, the impulse between them determines the direction in which they travel. If their collision is not head on, they will not end up traveling along their initial axis. However, if the system is closed and isolated, the linear momentum must still be conserved: and if the collision is also elastic, then the total kinetic energy is also conserved: The first (linear momentum) equation is best if we can rewrite it for the x and y components of the momentum. So, according to our figure above: (horizontal component) (vertical component)
Systems with Varying Mass: A Rocket All the systems we have worked with have had a constant total mass. However, in a rocket, the total mass it does not. We'll handle the varying mass by applying Newton's second law to the system that includes the rocket and its ejected combustion products. The mass of this system does not change. Finding the acceleration Since our system consists of the rocket and its exhaust products released during the time interval dt, the system is closed and isolated, and the total linear momentum must be conserved, so that: We'll let the mass and velocity of our rocket be M and v, respectively, at some time t. After a time interval of dt, the rocket has velocity v + dv, and mass M + dm, where dm is the change in mass (negative quantity). The exhaust products released by the rocket during the interval dt have mass dm and a velocity U relative to the inertial reference frame. dm M + dm U v + dv The linear momentum of the system becomes: We can simplify this using the relative speed v rel between the rocket and its exhaust products. Substituting this equation for U into our first equation (and simplifying algebraically) give us: and dividing both sides by dt We can replace dm/dt with R, where R is the mass rate of fuel consumption (positive), and dv/dt with a to get The units of the right side are newtons (N). We will call the Rv rel term the thrust of the rocket engine and replace it with T.
Systems with Varying Mass: A Rocket (cont'd) Finding the velocity Let's now address how the velocity of the rocket changes as it consumes fuel. Solving for dv gives us Integrating leads us to Evaluating the integral now becomes
Example Ans: 7.1 m/s A 2.3 kg object traveling at 6.1 m/s collides head on with a 3.5 kg object traveling in the opposite direction at 4.8 m/s. If the collision is perfectly elastic, what is the final speed of the 2.3 kg object?
Example A plate falls vertically to the floor and breaks up into three pieces, which slide along the floor. Immediately after the impact, a 320 g piece moves along the x axis with a speed of 2.00 m/s and a 355 g piece moves along the y axis with a speed of 1.50 m/s. The third piece has a mass of 100 g. In what direction does the third piece move? You can neglect any horizontal forces during the crash. Ans: from the x axis
Example Ans: 0.32 J A 1.2 kg spring activated toy bomb slides on a smooth surface along the x axis with a speed fo 0.50 m/s. At the origin, O, the bomb explodes into two fragments. Fragment 1 has a mass of 0.40 kg and a speed of 0.90 m/s along the negative y axis. In the figure, what is the energy released by the explosion?
Example Ans: 7.1 m/s A 2.3 kg object traveling at 6.1 m/s collides head on with a 3.5 kg object traveling in the opposite direction at 4.8 m/s. If the collision is perfectly elastic, what is the final speed of the 2.3 kg object?
Example A plate falls vertically to the floor and breaks up into three pieces, which slide along the floor. Immediately after the impact, a 320 g piece moves along the x axis with a speed of 2.00 m/s and a 355 g piece moves along the y axis with a speed of 1.50 m/s. The third piece has a mass of 100 g. In what direction does the third piece move? You can neglect any horizontal forces during the crash. Ans: from the x axis
Example Ans: 0.32 J A 1.2 kg spring activated toy bomb slides on a smooth surface along the x axis with a speed fo 0.50 m/s. At the origin, O, the bomb explodes into two fragments. Fragment 1 has a mass of 0.40 kg and a speed of 0.90 m/s along the negative y axis. In the figure, what is the energy released by the explosion?