Chapter 7 Linear Momentum Linear momentum is the

Chapter 7: Linear Momentum • Linear momentum is: – the product of mass and velocity – Represented by the variable p – Equal to mv, where m is the mass of the object and v is its speed. (p = mv) – Conserved – A vector

More on the momentum vector • Because velocity is a vector, momentum is a vector. • The direction of the momentum depends on the direction of the velocity. • The magnitude of the vector is p = mv.

The units of momentum • Because p = mv, we would expect the units of momentum to be units of mass x units of velocity. • Indeed, the units of momentum in SI units is kg*m/s, this unit has no special name. • However, this tells you that you need kilograms and meters per second.

Example • A 100, 000 kg truck is traveling east at a speed of 20 m/s. Find the magnitude and direction of the momentum vector.

Solution • Remember the magnitude of momentum is p = mv, so p = (100, 000 kg)(20 m/s) = 2, 000 kgm/s. • The direction of the momentum vector is the same direction of the velocity vector, so east in this example.

Changing momentum • The only way to change the momentum of an object is to either change its mass or change its velocity. • Remember that a change in velocity is called acceleration, which requires a force. • So, changing momentum requires a force.

Momentum and Newton’s 2 nd • Let’s start with ΣF = ma – a = Δv/Δt • So ΣF = m Δv/Δt – Δv = v – v 0 • So ΣF = m(v – v 0)/ Δt = – p = mv • So ΣF = Δp/Δt

Example of momentum change • Water leaves a hose at a rate of 1. 5 kg/s at a speed of 20 m/s and is aimed at the side of a car. The water stops when it hits the car. What is the force exerted on the water by the car?

Solution • Every second 1. 5 kg of water moves 20 m/s. • This means the water has a momentum of p = mv = (1. 5 kg)(20 m/s) = 30 kgm/s, which goes to 0 when it hits the car (because v = 0). • F = Δp/Δt = (pfinal – pinitial)/Δt = (0 – 30 kgm/s )/1 s = -30 N

Question • What happens if a car hits a semi head on?

Conservation of momentum • Earlier I told you that momentum is conserved. • What that means is “the total momentum before a collision equals the total momentum after” • Momentum before = momentum after • m 1 v 1 + m 2 v 2 = m 1 v 1’ + m 2 v 2’ • The ‘ is read as “prime” and means after.

The law of conservation of momentum • The law states this “The total momentum of an isolated system of bodies remains constant. ” • System = a set of objects that interact with each other • Isolated system = the only forces present are those between objects in the system.

Example • A 10, 000 kg railroad car traveling at a speed of 24. 0 m/s strikes an identical car at rest. If the cars lock together as a result of the collision, what is their speed afterward?

Solution • We start with pinitial = pfinal • pinitial = m 1 v 1 + m 2 v 2 = (10, 000 kg)(24 m/s) + (10, 000 kg)(0 m/s) = 240, 000 kgm/s • pfinal = (m 1 + m 2)v’ (it’s m 1 + m 2 because the cars linked up and became one object in the eyes of physics) • (m 1 + m 2)v’ = 240, 000 • v‘ = 240, 000 / 20, 000 = 12 m/s

Example • Calculate the recoil velocity of a 5. 0 kg rifle that shoots a 0. 050 kg bullet at a speed of 120 m/s.

Solution • Momentum is conserved, so start with pi = pf • What is pinitial? • Ask yourself the following: – What is the starting speed of the bullet before firing? – What is the starting speed of the rifle before firing? • pinitial = m. Bv. B + m. Rv. R = 0

Solution Cont. • Now we need to set up pfinal and set it equal to 0. • pfinal = m. Bv. B’ + m. Rv. R’ = (0. 050 kg)(120 m/s) + (5. 0 kg)(v’R) = 0 • Solve for v’R • v ’R =

Interpretation of data • Does our answer make sense? • The rifle is moving 100 times slower than the bullet. How does that work?

Impulse • During a collision the force on an object usually jumps from 0 to very high in a very short amount of time and then abruptly returns to 0. • Let’s start with ΣF = Δp/Δt • And solve it for Δp, • Δp = FΔt = impulse

When do we care about impulse? • Impulse is very helpful when we are working with large forces that occur in a very short amount of time. • Examples: – A bat hitting a ball – Two particles colliding – Brief body contact

Example • Calculate the impulse experienced when a 70 kg person lands on firm ground after jumping from a height of 3. 0 m. Then estimate the average force exerted on the person’s feet by the ground, if the landing is stiff-legged and again with bent legs. With stiff legs the body moves 1 cm during impact. With bent legs the body moves 50 cm.

Solution • We don’t know F so we can’t solve for impulse directly. But we know that impulse = Δp, and Δp = mv 2 – mv 1. • This means we need to find v 1 (we know that v 2 will be 0) • We can find v using conservation of energy: ΔKE = - ΔPE 1/2 mv 2 – 0 = -mgΔy

Solution • Algebra gives us • Δp = 0 – (70 kg)(7. 7 m/s) = -540 Ns

Straight Legged • In coming to a rest the body goes from 7. 7 m/s to 0 in a distance of 0. 01 m. • The average speed during this period is (7. 7 m/s + 0)/2 = 3. 8 m/s = v • Δt = d/v = 0. 01 m / 3. 8 m/s = 2. 6 E-3 s • Impulse = FΔt = -540 Ns so, • F = -540 Ns / 2. 6 E-3 s = 2. 1 E 5 N

Bent Legs • This is done just like the straight leg except d = 0. 50 m so Δt = 0. 50 m / 3. 8 m/s = 0. 13 s so, • F = 540 Ns/0. 13 s = 4. 2 E 3 N

Elastic Collisions • An elastic collision is a collision in which kinetic energy is conserved. • This means both kinetic energy and momentum are conserved. • This is handy, because it gives us 2 equations we can solve simultaneously to find the two unknowns (the speed of each object after the collision)

The Math • The two equations we need to solve are: • v 1 – v 2 = v’ 2 – v’ 1 (derived from conservation of kinetic energy) and • m 1 v 1 + m 2 v 2 = m 1 v’ 1 + m 2 v’ 2 (the conservation of momentum equation • The strategy is to solve the first equation for either v’ 2 or v’ 1 plug that into the second equation.

Example • A billiard ball of mass m moving with speed v, collides head-on with a second ball of equal mass at rest (v 2 = 0). What are the speeds of the two balls after the collision, assuming it is elastic?

Solution • Conservation of momentum gives us: • mv = mv’ 1 + mv’ 2, which we can divide by m to get: • v = v’ 1 + v’ 2 (call this *) • Now we use the first equation, v 1 – v 2 = v ’ 2 – v ’ 1 , • v = v’ 2 – v’ 1 (call this #) • * - # gives us 0 = 2 v’ 1, so v’ 1 = 0

Solution • We can now substitute v’ 1 = 0 into v = v’ 2 – v’ 1 and solve for v’ 2 • v ‘ 2 = v + v ’ 1 = v + 0 = v • To summarize, • Before collision: v 1 = v and v 2 = 0 • After collision: v’ 1 = 0 and v’ 2 = v

Inelastic Collisions • An inelastic collision is a collision in which kinetic energy is not conserved. • If it is not conserved, then either KEf < KEi or KEf > KEi • In the former case, the energy of the objects is wasted as heat energy, sound energy, potential energy, or crushing the objects. • In the later case, chemical or nuclear potential energy is released. (Think explosives)

Completely Inelastic Collisions • When two objects completely stick together as a result of the collision, the collision is said to be completely inelastic. • When this happens, the conservation of momentum becomes m 1 v 1 + m 2 v 2 = (m 1 + m 2)vf

Ballistic Pendulum • 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 block of mass M, which is suspended like a pendulum (M > m). As a result of the collision, the pendulum-projectile combination swings up to a maximum height h.

Solving the Ballistic Pendulum • Let us determine the relationship between the initial speed of the projectile, v, and the height h. • mv = (m + M)v’ (i) • KE 1 + PE 1 = KE 2 + PE 2 or • ½(m + M)v’ 2 + 0 = 0 + (m + M)gh (ii) so • v‘ =

Solution Continued • Combining (i) and (ii) gives us