Answers for Physics 208A. Test 4 27 May/98

You may use one sheet of helpful information for this test. This information should be made by the student and could contain useful information like conversion factors and formulae as well as example problems. You will be permitted one sheet for each of the tests. No books or notebooks are permitted. This test concentrates on chapters 6 and 9.

You may leave your answers in unrationalized form. eg Kinetic E = 1/2 * 4 *30^2 Joules. There is no need to use a calculator to multiply out the numbers. The quantities must be in a consistent set of units. The preferred units are the International units….. m, sec, joules and watts. Answers with mixed units will be marked-off.

1.An athlete of 50 Kg and one of 100 Kg compete in jumping. Both athletes are trained to maximize their jumping ability.

a)What is the ratio of jumping energy that each athlete can output. Please show the argument for your answer.

The muscle force depends on the area and the work it can do depends on the length. Since work = force*length, the work a muscle can do is dependent on the volume of the muscle which is proportional to the mass. The energy available for jumping is proportional to the mass. Energy/energy’ = m/m’= 50/100 = 1/2

b) Use the energy conservation argument to estimate the relative maximum height that each athlete can raise their CM.

To reach a height h, requires a total energy of PE = mgh. . E/E’ = mgh/m’g h’ = m/m’ . The m and m’ cancel from each side so that the potential jumping height is independent of the mass of the athlete. The larger power of the bigger athlete is neutralized by the need for larger power to reach the same height.

2. A top cyclist can output mechanical power at the rate of 300 watts for many hours.

How much time is necessary to burn off the calories of a Big Mac lunch that contains 700 Kcalories. One Kilocalorie = 4200 Joules. What fraction of the Big Mac comes out as mechanical energy and what fraction comes out as heat?

The big Mac lunch contains 700 Kcalories * 4200Joule/Kcalorie = 2.9 E6 Joules.

To find the time to burn this, we note that the power output is 300 watts but that this is only about 20% of the energy used. The other 80% is emitted as heat. The athlete is burning food energy at a rate of Power = 300/0.2 = 1500 watts = 1500 Joules/sec.

The time interval to burn the big mac is t = energy/power = 2.9E6/1500 seconds. = 2000 seconds= 33 minutes.

3. Assuming that you can compress your air Jordans by about 2cm when you jump on them from a height of 0.5m. These air Jordans have soles that are completely elastic. You have a mass of 70 Kg.

a) How much energy is stored in the Air Jordans at their maximum compression?

The energy stored by the air jordans is just the kinetic energy produced by the jump from 0.5m. This energy PE = mgh = 70*10*0.5=350J.

b) How high will you bounce up if you keep your legs still during the bounce?

If the shoes are elastic, and your body is held stiffly and doesn’t absorb any energy by deformation, you should bounce straight back up to 0.5m. Energy is conserved.

c) If you can normally jump 0.4m, how high can you reach if you jump and use the elasticity of the shoes together.

If you time your jump perfectly, you will add the jump kinetic energy to the energy released by the shoes. Taken together, this will give you a jump of PE = mgh = mgh’+mgh’’ = mg(h’+h’’) = mg(0.5+o.4). h = 0.9 m.

4. Estimate how many pounds of muscle are required to launch a 70kg athlete on a 0.6 meter vertical jump? Make the assumption that the upper and lower body is involved and that the jumping motion is accomplished in 0.2 seconds while the preparation for the jump has lowered the CM by 0.3 m.

Taken the preparation distance and height of jump, the total height change is 0.9m. The muscles supply this energy in 0.2 seconds. This means that the power must be Power = mgh/time = 70*10*0.9/0.2 = 3150 watts. The muscle power density is 400 watts/Kg. The amount of muscle required is m = Power/power-density =3150/400 Kg = 7.8 Kg.or 17 pounds of muscle.

5. If the best 60 Kg athlete can press 150 kg of mass, estimate the maximum press for the best 90 kg athlete.

Since the force a muscle can produce is proportional to the area, the maximum weight that an athlete can lift is proportional to the mass^(2/3). F/F’ = (m/m’)^2/3 .

F = F’*(m/m’)^(2/3) = 150*(90/60)^(2/3) = 196 Kg.

6. Tennis

a) For an incoming tennis ball with speed 120 mph, what must the racket speed be in order to deliver a 120 mph

return?

The rule of thumb for the average tennis racket is that Vlaunch = 3/2 Vracket + 1/2Vball.

Vracket = 2/3{Vlaunch - 1/2 Vball } = 2/3{120 –1/2*120} = 2/3*60 = 40 mph. You only need a modest racket speed to launch a devastating return by feeding on the incoming velocity.

b) If you had a very high mass racket plus completely elastic racket – ball collision, what racket speed would you need to deliver a 120 mph serve?

For a massive complete elastic collision with a ball, the racket needs to be ½ of the speed of the launch. The racket speed needs to be 60 mph even for this idealized racket. For real rackets, the racket speed must be larger. {From the point of view of an insect on the racket face, the ball is arriving at 60mph. For the elastic collision with the massive racket, the ball departs at 60mph. From the courtside observer, the ball’s launch speed is 120 mph.}

7. If a tackler and ball carrier approach head on each with a speed of 10m/s, what is the velocity of the two when the tackler holds on the the carrier? The ball carrier has a mass of 80 Kg and the tackler has a mass of 100 Kg. Explain your answer.

In this totally inelastic collision, some of the initial kinetic energy is lost by the deformation of the players. However, momentum must be conserved. The intital momentum was p 1 + p2 = 80*10 – 100*10 = -200 kg-m/s. This must be equal to the net momentum after the collision. p1’ + p2’= (m1+m2)v’ = -200. V’ = -200/(180) = -1.1m/s. This is in the direction of the intital momentum of the tackler.

8. Muscle control hierarchy. A weight lifting session with high reps of small weights will not develop very large muscle masses. Explain this in terms of the muscle control hierarchy and the process of muscle building.

For small forces, the muscle control system will call upon only the type I [slow twitch] muscles. It does NOT involve the type II muscles that are the fast twitch type. If the type II muscles are not exercised, they will not be developed. Since large musculature involves development of the type II muscles, big muscles cannot be obtained by repetitions of small forces.

9. Force and Energy.

a) Characterize the percentage of maximum muscle force at 75% of rest length and at 50% of rest length.

The maximum force is obtained at 120% of the rest length of the muscle. It scales linearly so that at 50% of rest length there is no force produced. At 75% length, the maximum force is only about 2/3 of maximum force.

b) About how much more work is done if the athlete prestretches to 120% of rest length compared to starting from rest length.

The work is the area under the curve. Since the height grows with L, the work grows as L^2. Take care that we only count the length over 50% of the rest length. The ratio of work is E/E’ = (70/25)^2 = 8. The amount of work by stretching a muscle is very large compared with one of only a small stretch.