---- WITH SOLUTIONS ----

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1 PHYS 101 MIDTERM October 24 th, WITH SOLUTIONS ---- The exam comprises two parts: 8 short-answer questions, and 4 problems. A formula sheet is attached to the back of the exam. Calculators are allowed. Answer all the short-answer questions with a few words or a phrase, but be concise, please! For the problems, your grade will be calculated with the best two problems. Show your work. The short answer problems are worth two points each, and the problems are worth 10 points each. Put all answers in the red and white answer booklets provided; you may keep this exam. Good luck! Short answer questions (answer all): you should not need to do any calculations for these questions. Answer in a few words, a short phrase, or a simple sketch. 1) [2 pts] Draw a velocity-vs-time graph for a ball that you throw vertically upwards and then catch some time later. You catch the ball at the same height that you released it from. Answer: As below: 2) [2 pts] In the diagram below, you re looking downwards at a ball being whirled around the head of your instructor in uniform circular motion. Which trajectory best represents what would happen if the string connecting the ball broke at point P? Explain briefly why.

2 Answer: Path (B) from the law of inertia (Newton s first law) and that fact that no forces act in the horizontal plane after the string breaks. The path is thus tangential to the circle. 3) [2 pts] In the diagram below, the block slides down a frictionless track from a height h. Express the final speed at point A in terms of h, r, and d. Consider that h>>r so that the block does not leave the track at the top of the loop. Answer: v = (2gh) from conservation of energy; neither d nor r matter. 4) [2 pts] A student falls out of a tree onto a trampoline. The trampoline sags 0.65 m before it launches the student back into the air. At the very bottom, where the sag is the greatest, what is the student s velocity? Is there an acceleration at that moment, and if so, in what direction is the student s acceleration? Answer: the student s velocity at the bottom is zero, and the acceleration is non-zero and directed upward. 5) [2 pts] An object is placed on a simple spring scale (which usually indicates the weight of the object). Even if the scale is calibrated correctly and working properly, there are circumstances where it will not indicate the correct weight (mg) of object. What are those circumstances?

3 Answer: It will not indicate the correct weight of an object in any non-inertial frame, such as an accelerating elevator. 6) [2 pts] Two students, having just written their Physics 101 final exam, are standing on a balcony throwing snowballs to the street below and pondering their future. One throws a snowball upwards at 30 o, while the second student throws a heavier snowball downwards at 30 o but at the same speed. Which has the largest speed on impact? Answer: they will have the same speed, by conservation of energy. 7) [2 pts] The same two students, fresh from their snowball antics, are in front of a concrete wall. Student A is on a frictionless skateboard, and is throwing small balls at the wall to propel himself backwards. Student B asserts that the balls don t need to hit the wall, they just need to be thrown. Student A disagrees and states that the balls have to hit the wall to push against something. Who s correct, and why? Answer: Student B is correct. From conservation of momentum, as the student throws the balls (imparting momentum to them), he will move in the opposite direction. They don t need to strike anything. 8) [2 pts] Out on a Sunday stroll, you find yourself on a bike path with a difficult choice. You can t avoid an accident, but you can choose either to get hit head-on by an adult on a bike, or by a child (also on a bike) moving at twice the speed. The mass of the adult+bike is twice the mass of the child+bike. Which collision do you prefer, and why? Answer: you d prefer to hit the adult. Although the momentum of the two is the same, there is less energy in the adult s motion. Further, her momentum transfer p to you will be less, because the larger mass in the final state means your share of the initial momentum will be smaller. Problems (you will be graded on the best TWO out of four): 1) [10 pts] William Tell is preparing to shoot the apple from the top of his youngest child (the others, sadly, have fallen victim to his poor aim). The apple is at the same height as his crossbow and arrow, namely 1.75 m above the level ground. The distance is 75 m, and William knows that the launch speed of his crossbow is 84 m/s. a) At what angle upwards from the horizontal should he aim the arrow to hit the apple? b) If he misses the apple laterally (that is, his arrow is at the correct height but slightly to the side), what is the distance d that the arrow goes beyond his son before it hits the ground?

4 Solution: use the range equation R = v 2 sin(2 )/g to get sin(2 ) = 0.104, and so 2 =5.98 o or 2 =174.0 o. Thus =2.99 o or =87.0 o. To calculate d, note that for the second part of the trajectory v oy = - v sin = m/s for the 3 o trajectory (the negative sign means that the initial y velocity at this moment is downwards). So we have, for the time to reach the ground: 0 m = 1.75 m 4.4m/s t g t 2 giving t = 0.25 s The horizontal distance covered by the arrow during this time, d, is d = 84 cos t = 21.0 m For the steeper angle, the numbers are t = sec and d = m. Good for bonus points; not good for William Tell s child! 2) [10 pts] A pinball machine has a ball launcher with a spring of force constant k=120 N/m, on a surface angled upwards from the horizontal at 10 o. The spring is initially compressed by 5.0 cm and the ball s mass is 100 g. Ignore friction (as well as the mass of the plunger). Find the speed of the ball as it leaves the spring when the plunger is released. Solution: at maximum compression, the energy in the spring is ½ kx 2 with x=5.0 cm, so E= 0.15 J. If we take that position to be zero, then when the ball is released 5.0 cm up the ramp, there is no more energy in the spring and it has been transferred to a combination of kinetic energy ½ mv 2 of the ball and the increased height of the ball mgh:

5 giving 0.15 J = ½ mv 2 + mg (0.05 m sin 10 o ) v = 1.68 m/s 3) [10 pts] An amusement park ride has a rotating cylinder platform 8.0 m in diameter, with 10.0 kg seats suspended by 2.5 m chains. When the platform rotates, the chains make an angle of 28 o from the vertical. a) What is the speed of each seat? b) Draw the free-body diagram of a 40.0 kg child in the seat when the system is rotating, and find the tension in the chain. Solution: the free body diagram has the tension in the chain T (at the angle ) and mg (vertically downward) only. The vertical components must balance, so T cos = mg and the unbalanced horizontal component T sin must be equal to mv 2 /r. Thus: mv 2 /r = mg tan The radius of the circular motion is: r = 4.0 m + L sin = 5.17 m so v = 27.0 m/s (independent of mass!) Finally, using m = 50 kg (10 kg of seat and 40 kg of child), we get: T = mg/cos = 555 N 4) [10 pts] James Bond, on his eternal flight from the bad guys, skis down a frictionless hill 10.0 m high. Just at the cliff edge (cliff height 6.0 m), Bond grabs the Bond Girl (BG), who is waiting motionless there. At what horizontal distance d do the two land? Bond s mass is 75 kg, the BG s mass is 60 kg, and the collision is perfectly inelastic.

6 Solution: We use conservation of energy in Bond s initial descent, then conservation of momentum (the collision is perfectly inelastic since the two stick together), then finally projectile motion. In the initial descent, we have mgh = ½ mv 2 so Bond s speed v as he hits BG is v B = (2gh) = 14.0 m/s The collision conserves momentum so m B v B = (m B + m BG ) v B+BG v B+BG = 7.78 m/s For the distance d, we must calculate the fall time since B+BG take off horizontally, this is just: 6.0 m = ½ g t 2 t = 1.11 sec and so the distance d is d = 7.78 m/s t = 8.61 m