Recall

Power

Example: P5.46

The acceleration is a=v/t=(1.75m/s)/(3.00s) = 0.583m/s². The net force on the elevator must be F_net=ma = (650kg)(0.583) = 379N. The net force is the force from the motor minus the force of gravity, so F_motor = F_net + mg = 379N + (650kg)(9.8) = 6750N. In 3s the elevator moves a distance x=1/2 at² = (.583)(3.00s)²/2 = 2.62m. The work done by the motor is then W_motor= F_motorx = (6750N)(2.62m) = 17,700J. The average power of the motor is P=W/t = 17700J/3.00s = 5900W

At t=3.00s, the instantaneous power is P=Fv = (6750N)(1.75m/s) = 11,800 W.

Chapter 6: Momentum and Collisions

Momentum

The component equations for momentum are: p_x = mv_x and p_y = mv_y.

Newton actually expressed his second law not as F=ma, but as F = Dp/Dt. Written this way, the second law basically says that the time rate of change of momentum of an object is equal to the net force acting on it.

Rewriting the above expression we get the impulse-momentum theorem: FDt = Dp = mv_f - mv_i.

The quantity FDt is called the impulse. The above expression says that the impulse acting on an object equals the change in momentum of the object. It is often not possible to know exactly what force is acting on an object. For instance, when a ball bounces off the floor, the force changes rapidly with time, from nearly zero when the ball first touches the floor to a large value as the velocity of the ball comes to zero, and reverses, then back to a small value as the ball begins to move up from the floor. But by using the impulse, we don't need to know exactly how the force varies of this short time, we can use the average force exerted on the ball during the time it is in contact with the floor.

Example: like P6.13

(a) Impulse = FDt = p_f-p_i = mv_f - mv_i = (1.3kg)(-10.0 - 9.8m/s) = 26kg m/s.

(b) F_avg = FDt/Dt = (26kg m/s)/(2.6x10^-3s) = 10⁴N.

Conservation of Momentum

Conservation of momentum is as fundamental in physics as conservation of energy, but better. Momentum is conserved in all isolated systems. A system is isolated if there is no outside influence (forces) acting on it. Examples of isolated systems are:

• a collision between two atoms of a gas,
• a rocket in space, plus the exhaust from its thrusters,
• a girl standing on a plank lying on a frictionless surface,
• the collision of "glider cars" on an air track,
• a satellite flying past a planet,
• or the collision of 2 vehicles on a roadway.

Consider a collision between two objects, 1 and 2. (Drawing of 2 objects before and after collision, indicating masses and velocities.) The impulse on object 1 is F₁Dt = m₁v_1f - m₁v_1i. Likewise for object 2, F₂Dt = m₂v_2f - m₂v_2i.

Recall Newton's third law: F_{2 on 1} = -F_{1 on 2}. In terms of the forces given above this means that F₁ = -F₂. (Although the impulse equation involves an average force, since F₁ equals -F₂ at every instant in time, then the same result holds for the averages.) Therefore, F₁Dt = -F₂Dt, or

m₁v_1f - m₁v_1i = -(m₂v_2f - m₂v_2i)

Putting the initial quantities on the left and the final quantities on the right yields:

m₁v_1i + m₂v_2i = m₁v_1f + m₂v_2f

This result is called conservation of momentum. It says that the initial momentum equals the final momentum. The key to this result is Newton's third law, and the fact that there are no other forces acting on the system. In this derivation, we used a system with two objects, but the result holds for systems with any number of objects.

Example: P6.15

A 730N man stands in the middle of a frozen pond of radius 5.0m. He is unable to get to the side because of a lack of friction between his shoes and the ice. To overcome this difficulty (since he just learned about conservation of momentum) he throws his 1.2kg physics textbook horizontally toward the north shore, at a speed of 5.0m/s. How long does it take him to reach the south shore?

Apply conservation of momentum. When he throws the textbook, he will get a push from the reaction force of the textbook on him.

m_bv_b = -m_mv_m

v_m = -m_bv_b/m_m = -(1.2kg)(5.0m/s)/((730N)/(9.8m/s2)) = -0.081m/s. (minus sign means the directions are opposite, I'll ignore it from here.) The time to move to the edge is t=(5.0m)/(0.081m/s) = 62s.

Collisions

Interactions between two objects often fall into the category of collisions -- even many things that don't seem to be related to collisions at all, for example, air pressure or glaucoma testing (see text). While momentum is strictly conserved in collisions (emphasizing again that there are no external forces), energy need not be conserved. Quick example: When a ball is dropped and bounces off the floor, it doesn't return to the same height it was dropped from. Energy is lost in the collision. Momentum is conserved, but you must consider the system of the ball and the Earth!

We call a collision where momentum is conserved but energy is not an inelastic collision. When the two objects stick together, like two train cars that hit and couple together, we say that the collision is perfectly inelastic.

When both momentum and energy are conserved, we say that the collision is elastic. Examples of truly elastic collisions are collisions between subatomic particles, or "collisions" between stellar objects. In the latter, the objects should not make physical contact, but we can treat their gravitational interaction as a collision.

Example: Inelastic Collision P6.25

Since the bullet embeds (sticks) in the ball, this is a perfectly inelastic collision. Initially we have 2 objects, and afterward we have one object with a mass that is the sum of the original masses.

(Drawing) m₁v₁+m₂v₂ = (m₁+m₂)v_f.

Thus v_f = (m₁v₁+m₂v₂) / (m₁+m₂) = (0.030kg)(200m/s)/(0.030kg+0.15kg) = 33.3m/s.

So the ball will rise to a height y = v²/2g = (33.3m/s)²/2(9.8m/s2) = 57m.

Example: Elastic Collision P6.30

Since this is an elastic collision, momentum and energy are conserved, so we have 2 equations:

m₁v_1i + m₂v_2i = m₁v_1f + m₂v_2f, and

1/2 m₁v_1i² + 1/2 m₂v_2i² = 1/2 m₁v_1f² + 1/2 m₂v_2f²

From the statement of the problem we know m₁ = 0.0100kg, m₂ = 0.0150kg, v_1i = 0.200m/s, and v_2i = -0.300m/s. We are left to find v_1f, v_2f. We have 2 equations and 2 unknowns. m₁v_1i + m₂v_2i = 0.00650 kg m/s. 1/2 m₁v_1i² + 1/2 m₂v_2i² = 0.000875J. Begin by eliminating v_2f = (m₁v_1i + m₂v_2i - m₁v_1f) / m₂ = 0.433 - 0.667 v_1f from the energy equation.

0.000875J = 1/2 m₁v_1f² + 1/2 m₂(0.433 - 0.667 v_1f)² = 0.00500 v_1f² + 0.00141 - 0.0100 v_1f + 0.00334 v_1f²

This results in a quadratic equation:

0.00834 v_1f² - 0.0100 v_1f + 0.000535 = 0