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1 University of California, Berkeley Physics 7A, Section 1, Spring 2009

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1 University of California, Berkeley Physics 7A, Section 1, Spring 2009 1 University of California, Berkeley Physics 7A, Section 1, Spring 2009 (Yury Kolomensky) SOLUTIONS TO THE SECOND MIDTERM Maximum score: 100 points 1. (10 points) Blitz (a) hoop, disk, sphere This is a set of simple questions to warm you up. The (b) disk, hoop, sphere problem consists of five questions, 2 points each. (c) sphere, hoop, disk 1. Circle correct answer. Imagine a basketball bouncing on a court. After a dozen or so (d) sphere, disk, hoop bounces, the ball stops. From this, you can con- (e) hoop, sphere, disk clude that Answer: (d). The object with highest moment of (a) Energy is not conserved. inertia has the largest kinetic energy at the bot- (b) Mechanical energy is conserved, but only tom, and would reach highest elevation. The ob- for a single bounce. ject with the smallest moment of inertia (sphere) (c) Total energy in conserved, but mechanical will therefore reach lowest altitude, and the ob- energy is transformed into other types of ject with the largest moment of inertia (hoop) energy (e.g. heat). will be highest. (d) Potential energy is transformed into kinetic 4. A particle attached to the end of a massless rod energy of length R is rotating counter-clockwise in the (e) None of the above. horizontal plane with a constant angular velocity ω as shown in the picture below. Answer: (c) 2. Circle correct answer. In an elastic collision ω (a) momentum is not conserved but kinetic en- ergy is conserved; (b) momentum is conserved but kinetic energy R is not conserved; (c) both momentum and kinetic energy are conserved; (d) neither kinetic energy nor momentum is Which direction does the vector ~ω point ? conserved; (e) none of the above. (a) Tangential to the circle, in the counter- clockwise direction Answer: (c) (b) Along the radius, towards the center of the 3. Circle correct answer. A hoop, a uniform disk, circle and a uniform sphere, all with the same mass (c) Along the radius, away from the center of and outer radius, start with the same speed and the circle roll without sliding up identical inclines. Rank the objects according to how high they go, least (d) Normal to the page, towards you to greatest. (e) Normal to the page, away from you 2 Answer: (d), according to the right-hand rule 50% efficient, i.e. half of the energy from burn- ing diesel fuel goes to locomotive motion, and 5. Clean-and-jerk is an Olympic weightlifting half goes to waste. Burning one gallon of diesel event. In the first phase (clean), a competitor fuel produces about 150 MJ of energy. How stands up with a heavy barbell from the squat many gallons of diesel is spent on the 8 km trip position, lifting the bar by about 1 m in less than up Grapevine Grade ? 1 second. The Olympic record for this event is 263.5 kg (580.3 pounds), set by Hossein Reza Zadeh of Iran in 2004 Olympics in Athens. Es- 2. Solution timate (to 1 digit) the power delivered to the (a) There are four forces acting on the truck and the barbell by Zadeh during this lift in horsepower trailer: weight mg, directed downward, normal force (1 hp = 750 W). from the surface, air drag (pointing along the surface in the direction opposite to the velocity of the truck), Answer: 3 hp. 4 hp would also be an allowed and the propulsion force Fpr. The propulsion force is answer (e.g., if you used g = 10m/s2, or ac- the force of static friction between the truck’s tires and counted for the elevation of the weightlifter’s ground, and it prevents the wheels from slipping. Since body). the velocity of the truck is constant, the forces are bal- anced. The balance of forces along the slope leads to 2. (15 points) Grapevine Grade the equation: Grapevine grade is a famous place on interstate I-5 be- tween Bakersfield and Los Angeles, where a 5 mile (8 Fdrag + mg sin θ = Fpr = 7.3 kN km) stretch of the road has an average incline of 6% ◦ (3.4 ). Imagine a heavy duty pickup truck towing a The power provided by the force Fpr (which ultimately trailer up Grapevine Grade at the constant speed of 65 comes from the truck’s drivetrain) is mph (29 m/s). At this speed, the air drag force acting on P = F v = 2.1 · 105 J = 280hp the truck is about Fdrag = 1.5 kN. The total mass of the pr truck and the trailer is m = 10 tons (104 kg). (b) The useful work done by the propulsion force is (a) (10 points) Estimate the power output of the truck’s engine (assume that all power goes into W = FprL = 58 MJ locomotive motion of the truck). Express you an- swer in Horsepower (1 hp = 750 W) One gallon of diesel fuel produces 0.5∗150 MJ = 75 MJ of useful work. So the entire trip would use up 60/75 = (b) (5 points) The modern diesel engines are about 0.8 gal of diesel. 3. (20 points) Dangerous Demo You may remember this demonstration from earlier in the semester. A 200 g wooden block is placed on top of a muzzle of a 0.22” rifle, mounted vertically next to a meter stick (see picture below). The rifle fires a 3 g bullet into the block. The bullet stops inside the block, and we observe that the block (with the bullet inside) lifts up to the elevation h = 240 cm above the muzzle (this is not what happened in class as we used a low-velocity lighter bullet). This is a convenient way to measure the muzzle velocity (velocity with which it exits the rifle) of the bullet. 3 (a) (10 points) What velocity does the bullet have just before it hits the block ? That is the muzzle velocity. (b) (10 points) Obviously, the rifles are not normally used this way. Usually, you would hold the rifle in your hands, press the butt end against your shoulder, and fire. When the bul- let is fired, you feel the rifle recoil, i.e. the butt of the rifle strike you in the shoulder. Why does the rifle recoil ? What average force does the rifle exert on the shoulder while the bullet accelerates ? You will need to know the length of the rifle barrel (68.5 cm). 3. Solution Energy conservation requires (a) This part should be familiar, since it has been worked out in class when we discussed inelastic col- Uinitial + Kinitial = Ufinal + Kfinal lisions. Plugging in the values, we can find V : First, momentum conservation relates the velocity v of the bullet just before the collision to the velocity V V = 2gh of the block-bullet combination just after the collision: p (which is the standard answer from kinematics as well), pbefore = mv = pafter = (m + M)V and plug it into Eq. (1): where m = 3 g is the mass of the bullet, and M = 200 g m + M v = 2gh = 464 m/s . is the mass of the block. The collision is fully inelastic: m p the bullet is embedded in the block. So if we know the (b) The recoil of the rifle is another demonstration of velocity V of the block after the collision, we can find momentum conservation. Initially, the rifle and the bul- out the velocity of the bullet: let are at rest, and their total momentum is zero. Af- m + M v = V (1) ter the bullet is fired and before it and the rifle have m a chance to interact with the surrounding bodies (e.g. Since we know the elevation of the block, we can re- your shoulder), the momentum of the system has to re- late it to the velocity V using the conservation of energy. main the same. So if the bullet is traveling fast away At the muzzle, the potential energy of the block-bullet from you, the rifle has to recoil towards you, to keep the combo is zero, and the kinetic energy is total momentum zero. Ignoring the powder gas (which m + M 2 escapes through the holes on the side of the muzzle and Kinitial = V 2 carries no forward momentum – I did not tell you about The block rises to the elevation h = 2.4 m, where its it in the problem, and it’s safe to ignore this), the mo- kinetic energy is zero, and its potential energy is mentum of the bullet is equal in magnitude and opposite Ufinal = (m + M)gh in direction to the momentum the rifle picks up. 4 You can also think about it in terms of the 3rd New- up by the bullet, if we ignore the powder gases), and ton’s law. Since the bullet is accelerated, there must be a ∆t is the time it takes for the bullet to fully accelerate force applied to it. The source of this force, ultimately, (i.e. leave the barrel). The change in momentum is (by is the rifle, or more precisely, the gun powder gas that magnitude) pushes against the rifle and accelerates the bullet. By 3rd Newton’s law, the force that accelerates the bullet ∆p = mv has to have a pair – the force acting on the rifle in the opposite direction. To find the time interval ∆t, we can use kinematics To compute the force acting on the shoulder, we will equations for the bullet: put a 3rd Newton’s law to good use. The force that ac- celerates the bullet is equal and opposite to the force ∆t2 l = hai that is applied to the rifle and, through the buttstock, 2 transfered to the shoulder.
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