Chapter 8: Linear Momentum and Collisions PDF

Chapter 8 : Linear Momentum and Collisions Linear Momentum and Force Linear momentum is defined as the product of a system’s mass multiplied by its velocity. In symbols, linear momentum is expressed as 𝑝 = 𝑚𝑣 Momentum is directly proportional to the...

Chapter 8 : Linear Momentum and Collisions Linear Momentum and Force Linear momentum is defined as the product of a system’s mass multiplied by its velocity. In symbols, linear momentum is expressed as 𝑝 = 𝑚𝑣 Momentum is directly proportional to the object’s mass and also its velocity. Thus, the greater an object’s mass or the greater its velocity, the greater its momentum. Momentum is a vector having the same direction as the velocity. The SI unit for momentum is kg. m/s 1. Calculate the momentum of a 110-kg football player running at 8.00 m/s. 2. Calculate the momentum of a 2000-kg elephant charging a hunter at a speed of 7.50m/s. 3.What is the mass of a large ship that has a momentum of 1.60 x 109 kg. m/s, when the ship is moving at a speed of 48.0 km/hr? Momentum and Newton’s Second Law Momentum was deemed so important that it was called the “quantity of motion.” Newton actually stated his second law of motion in terms of momentum: The net external force equals the change in momentum of a system divided by the time over which it changes. Using symbols, this law is ∆𝑝 𝐹𝑛𝑒𝑡 = ∆𝑡 𝐹𝑛𝑒𝑡 is the net force, ∆𝑝 is change in momentum, ∆𝑡 𝑖𝑠 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑡𝑖𝑚𝑒.  Note that the change in momentum ∆𝑝 is given by, ∆𝑝 = ∆𝑚𝑣 4. A runaway train car that has a mass of 15,000 kg travels at a speed of 5.4m/s down a track. Compute the time required for a force of 1500 N to bring the car to rest. Impulse The effect of a force on an object depends on how long it acts, as well as how great the force is. In a very large force acting for a short time had a great effect on the momentum. A small force could cause the same change in momentum, but it would have to act for a much longer time. For example, if the ball were thrown upward, the gravitational force would eventually reverse the momentum of the ball. Quantitatively, the effect we are talking about is the change in momentum.  ∆𝑝 By rearranging the equation 𝐹𝑛𝑒𝑡 = to be ∆𝑡 ∆𝑝 = 𝐹𝑛𝑒𝑡 ∆𝑡 Impulse is the same as the change in momentum. Forces are usually not constant. Forces vary considerably even during the brief time intervals considered. It is, however, possible to find an average effective force 5. What is the average force exerted on a 0.0300-kg bullet to accelerate it to a speed of 600 m/s in a time of 2.00 ms (milliseconds)? 6. A car moving at 10 m/s crashes into a tree and stops in 0.26 s. Calculate the force the seat belt exerts on a passenger in the car to bring him to a halt. The mass of the passenger is 70 kg. 7. A professional boxer hits his opponent with a 1000-N horizontal blow that lasts for 0.150 s. Calculate the impulse imparted by this blow. Conservation of Momentum Both cars are coasting in the same direction when the lead car (labeled m2) is bumped by the trailing car (labeled m1) The only unbalanced force on each car is the force of the collision. Car 1 slows down as a result of the collision, losing some momentum, while car 2 speeds up and gains some momentum. We shall now show that the total momentum of the two-car system remains constant.  Conservation of Momentum Principle 𝑝𝑡𝑜𝑡 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 ′ 𝑝𝑡𝑜𝑡 = 𝑝𝑡𝑜𝑡 (𝑖𝑠𝑜𝑙𝑎𝑡𝑒𝑑 𝑠𝑦𝑠𝑡𝑒𝑚) An isolated system is defined to be one for which the net external force is zero 𝐹𝑛𝑒𝑡 = 0 Subatomic Collisions and Momentum The conservation of momentum principle not only applies to the macroscopic objects, it is also essential to our explorations of atomic and subatomic particles. Momentum is found to be a property of all subatomic particles including massless particles such as photons that compose light. Momentum relates to wave properties and plays a fundamental role in what measurements are taken and how we take these measurements. The conservation of momentum principle is used to analyze the masses and other properties of previously undetected particles, such as the nucleus of an atom and the existence of quarks that make up particles of nuclei. Elastic Collisions in One Dimension An elastic collision is one that also conserves internal kinetic energy. Internal kinetic energy is the sum of the kinetic energies of the objects in the system. Figure illustrates an elastic collision in which internal kinetic energy and momentum are conserved.  Truly elastic collisions can only be achieved with subatomic particles, such as electrons striking nuclei. Macroscopic collisions can be very nearly, but not quite, elastic—some kinetic energy is always converted into other forms of energy such as heat transfer due to friction and sound. One macroscopic collision that is nearly elastic is that of two steel blocks on ice. Another nearly elastic collision is that between two carts with spring bumpers on an air track. Icy surfaces and air tracks are nearly frictionless, more readily allowing nearly elastic collisions on them.  The equation for conservation of momentum for two objects in a one- dimensional collision is 𝑝1 + 𝑝2 = 𝑝1′ + 𝑝2′ 𝐹𝑛𝑒𝑡 = 0 or 𝑚1 𝑣1 + 𝑚2 𝑣2 = 𝑚1 𝑣1′ + 𝑚2 𝑣2′ (𝐹𝑛𝑒𝑡 = 0) where the primes (') indicate values after the collision. By definition, an elastic collision conserves internal kinetic energy, and so the sum of kinetic energies before the collision equals the sum after the collision. 1 1 1 1 Thus: 𝑚1 𝑣12 + 𝑚2 𝑣22 = 𝑚1 𝑣1′ 2 + 𝑚2 𝑣2′ 2 (elastic two object collision) 2 2 2 2 𝑣1 + 𝑣1′ = 𝑣2 + 𝑣2′ (simplified KE equation) 8. A 4kg ball moving east at speed of 5 m/s strikes 2kg ball at rest. Calculate the velocities of two balls assuming perfectly elastic collision. Inelastic Collisions in One Dimension An inelastic collision is one in which the internal kinetic energy changes (it is not conserved). This lack of conservation means that the forces between colliding objects may remove or add internal kinetic energy. Work done by internal forces may change the forms of energy within a system. For inelastic collisions, such as when colliding objects stick together, this internal work may transform some internal kinetic energy into heat transfer. Or it may convert stored energy into internal kinetic energy, such as when exploding bolts separate a satellite from its launch vehicle.  The two objects come to rest after sticking together, conserving momentum. But the internal kinetic energy is zero after the collision. A collision in which the objects stick together is sometimes called a perfectly inelastic collision because it reduces internal kinetic energy more than does any other type of inelastic collision. In fact, such a collision reduces internal kinetic energy to the minimum it can have while still conserving momentum.  An inelastic one-dimensional two-object collision. Momentum is conserved, but internal kinetic energy is not conserved. (a) Two objects of equal mass initially head directly toward one another at the same speed. (b) The objects stick together (a perfectly inelastic collision), and so their final velocity is zero. The internal kinetic energy of the system changes in any inelastic collision and is reduced to zero in this example. 9. Suppose a clay model of a koala bear has a mass of 0.200 kg and slides on ice at a speed of 0.750 m/s. It runs into another clay model, which is initially motionless and has a mass of 0.350 kg. Both being soft clay, they naturally stick together. What is their final velocity? 10. A train car are coupled together by being bumped into one another. Suppose if two loaded train cars are moving towards one another, the first car has mass of 150,000 kg and a velocity of 0.300 m/s and second car has mass of 110,000 kg and a velocity of -0.120 m/s (minus indicates the direction of motion). What is their final velocity? Collisions of Point Masses in Two Dimensions One-dimensional collisions; during such collisions, the incoming and outgoing velocities are all along the same line. But what about collisions, such as those between billiard balls, in which objects scatter to the side? These are two-dimensional collisions. One complication arising in two-dimensional collisions is that the objects might rotate before or after their collision. For example, if two ice skaters hook arms as they pass by one another, they will spin in circles. To avoid rotation, we consider only the scattering of point masses— that is, structureless particles that cannot rotate or spin.  A two-dimensional collision with the coordinate system chosen so that is initially at rest and is parallel to the -axis. This coordinate system is sometimes called the laboratory coordinate system. Because many scattering experiments have a target that is stationary in the laboratory, while particles are scattered from it to determine the particles that make-up the target and how they are bound together. The particles may not be observed directly, but their initial and final velocities are.  Conservation of Momentum along the x-axis, 𝑚1 𝑣1 = 𝑚1 𝑣1′ cos 𝜃1 + 𝑚2 𝑣2′ cos 𝜃2 Conservation of Momentum along the y–axis, 0 = 𝑚1 𝑣1′ sin 𝜃1 + 𝑚2 𝑣2′ sin 𝜃2 11. When serving a tennis ball, a player hits the ball when its velocity is zero (at the highest point of a vertical toss). The racquet exerts a force of 540 N on the ball for 5.00 ms, giving it a final velocity of 45.0 m/s. Using these data, find the mass of the ball.