PHYS 131 MIDTERM October 31 st, 2008

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1 PHYS 131 MIDTERM October 31 st, 2008 The exam comprises two parts: 8 short-answer questions, and 4 problems. Calculators are allowed, as well as a formula sheet (one-side of an 8½ x 11 sheet) of your own making. 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 three 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] a) a system of particles is known to have zero kinetic energy. What can you say about the momentum of the system? b) A system of particles is known to have zero momentum. What can you say about the kinetic energy of the system? Solution: a) If the KE is zero, then all speeds are zero, so the momentum is zero. b) if the momentum is zero, this could be (but is not necessarily) the result of zero speeds. So in case b) there is no general statement that can be made. 2) [2 pts] You drop a ball from a high balcony and it falls freely (and there s no air resistance). Does the ball s kinetic energy increase by equal amounts in equal times, or by equal amounts in equal distances? Explain. Solution: during the fall, the PE is converted to KE. PE is equal to mgh where h is the height, so the amount of energy going into KE is increasing by equal amounts for equal heights. This is not the same as equal amounts in equal times, because the fall distance d is given by d= ½ gt 2. 3) [2 pts] Two drivers are on a highway at the same speed, side by side. At the same instant, they each see an obstacle on the highway and start to brake (ignore reaction times). Driver 1 hits the brakes, locks the wheels, and skids to a stop. Driver 2 hits the brakes and applies the brakes to the verge of locking, but so that the wheels never lock up. Which driver stops in the shorter distance?

2 Solution: Driver 1 is using kinetic friction to stop (his wheels are skidding). Driver 2 is using static friction, and µ s > µ k, so driver 2 will stop in a shorter distance. 4) [2 pts] A skier is being pulled up a slope (angle θ) at constant speed by a cable angled at φ from the slope, as shown in the diagram. Draw the free-body diagram on the skier (don t ignore friction). Write down the sum of forces in the x and y directions (two equations), using the coordinate system indicated. Solution: The FBD has mg downwards, F N in the +ve y direction, F f in the ve x direction, and T (tension in the cable) along the cable direction. In x, we have: F x,net = T cos(φ) F f mg sin(θ) In y, we get: F y,net = F N + T sin(φ) mg cos(θ) 5) [2 pts] Two balls of clay of equal mass are suspended from the ceiling on massless strings of equal length. One of them is pulled to an angle of 45 o and released, and it swings down to hit the other. The two of them stick together. To determine the angle to which the balls rise on the other side, would you invoke a) conservation of mechanical energy, b) conservation of momentum, c) both, d) either but not both, or e) none of the above? DON T solve the problem just explain which approach [a) through e)] is the correct one. Solution: c) both would be used: conservation of momentum in the initial collision, where the two balls stick together. Then conservation of mechanical energy in the swing up to some new position. 6) [2 pts] Tarzan swings through the jungle on a vine. At the bottom of his swing, is the tension in the vine greater to, equal to, or less than his weight mg? Solution: The tension T is greater than mg. We must have T mg = mv 2 /r so we get T = mg + mv 2 /r > mg.

3 7) [2 pts] A projectile is launched over horizontal ground at an angle between (but excluding) 0 o and 90 o. a) Is there any point on the trajectory where velocity and acceleration are parallel? If so, where? b) is there any point on the trajectory where velocity and acceleration are perpendicular to each other? If so, where? a) No. v and a are never parallel. b) Yes. At the uppermost point of the trajectory (where v is horizontal), v and a are perpendicular to each other. 8) [2 pts] The figure below shows the position-vs-time graph for a moving object. Positive coordinates are to the right. At which point (or points) a) is the object moving the fastest? b) is the object moving to the left? c) is the object at rest? d) is the object turning around? Solution: a) fastest means highest speed, so largest (magnitude of) slope. I d say A (although D might be possible too). b) moving to the left means negative velocity, so C and D work. c) The object is at rest at B and E. d) the object is turning (velocity changing sign) at B.

4 Problems (you will be graded on the best THREE out of four): 1) [10 pts] A cowboy wishes to display his shooting skills to his friends. He asks a gentleman, standing 8.0 m from him, to toss a coin, and tells his friends that he can shoot the coin in the air. Having done his homework, he knows that the muzzle velocity of his Colt revolver is 73 m/s, and that the man will toss the coin upwards with a speed of 3.2 m/s. The cowboy shoots at the same moment as the man tosses the coin, and the height above the ground of his revolver is the same as the man s hand, 1.0 m. a) At what angle from the horizontal does the cowboy aim? Solution: taking the shooting height to be zero, v b to be the bullet s speed, θ to be the aiming angle above the vertical, and v c to be the coin s initial speed, we have: y b = v b sin(θ) t ½ gt 2 y c = v c t ½ gt 2 The two must be equal, so we get v b sin(θ) = v c, so θ = 2.51 o. b) At what height above the ground does the bullet strike the coin? Solution: The horizontal velocity of the bullet is v b cos(θ) = 72.9 m/s, so the time of impact is t=8.0m/72.9 m/s = sec. At that time, the height of the coin (and the height of the bullet too) is: y b = v b sin(θ) t ½ gt 2 = 0.29 m (above the shooting point), y above ground = 1.29 m. 2) [10 pts] A child is sledding on a semi-circular hill as shown, starting from rest at the top of the hill. Assume a frictionless surface. At some point down the hill the sled leaves the surface. a) Draw a free-body diagram for the sled at an arbitrary angle θ. Solution: the FBD has only two forces: F N (the normal force) radially outwards, and mg vertically downwards. b) When the sled leaves the surface, the normal force is zero. Calculate where this happens (ie, at what value of the angle θ). (Hint: at that instant consider the motion to be uniform circular motion).

5 Solution: The net radial force at an angle θ is: F net = F N mg sin(θ) = - mv 2 /r When the sled leaves the surface, F N is zero, so we get: mg sin(θ) = mv 2 /r sin(θ) = v 2 /(rg) (*) We can calculate v 2 from energy conservation: at angle θ the sled has lost a height h = r r sin(θ) and so ½ mv 2 = mgh gives v 2 = 2gh = 2gr (1-sin(θ)) Putting this into (*) and solving, gives ultimately: sin(θ) = 2/3 3) [10 pts] In the Supernova model demonstrated in class, a small rubber ball is dropped on top of a larger, more massive, ball, from a height h above the floor. Assume the more massive ball (of mass M) reaches the floor first and bounces elastically, and then the two balls collide (essentially at floor height) elastically. Find the height h to which the lighter ball (mass m) rises after the collision, in terms of h. Find the rebound height h in the case where M=100g, m=50g, and h=1.0 m. Hint: in a 1-D elastic collision between m 1 moving at v 1 and m 2 moving at v 2, the final velocities v 1, v 2 are given by: v 1 = ((m 1 -m 2 )/(m 1 +m 2 ) v 1 + 2m 2 /(m 1 +m 2 ) v 2 v 2 = 2m 1 /(m 1 +m 2 ) v 1 + ((m 2 -m 1 )/(m 1 +m 2 )) v 2 Solution: At the floor level, we have Mgh = ½ mv 2, so the speed of each ball is v = (2gh) After the collision of the more massive ball, it is moving upwards at this speed, and the small ball is still moving downwards at this speed. Taking positive to be upwards, we have a collision between M moving at v and m moving at v. Using the equations above with M as ball 1 and m as ball 2, we need to find v 2. v 2 = 2M/(M+m) v (m M)/(m+M) v v 2 = (3M m)/(m+m) v = (3M m)/(m+m) (2gh) Giving: v 2 = 7.38 m/s and a rebound height of: H = v 2 2 / 2g = 2.78 m

6 4) [10 pts] A water park has a stunt where a water-skier, pulled by a towboat, skies up a frictionless ramp and over a tank filled with piranha. The ramp is 2.0 m high, and the tank is 5.0 m wide. What minimum speed must the skier have at the base of the ramp (where she drops her tow rope) to clear the tank and thus survive the stunt? Solution: She takes off from the jump horizontally, so her drop time from a height of 2.0 m is: t = (2h/g) = sec Thus, to clear the piranha tank, she must have a minimum horizontal speed as she leaves the jump of: v min = 5.0/0.639 m/s = 7.83 m/s From conservation of energy, her initial speed at the bottom of the ramp must have been at least v where: ½ mv 2 mg h = ½ mv min 2 And so: v = 10.0 m/s