Exam III review:RIII-1. A projectile is fired straight up and then it comes back down to its original height. There is air resistance as the projectile is moving. During the entire flight of theprojectile, the total work done by the force of gravity is..A: zero. B: positive C: negativeAnswer: Zero. The work done by gravity is negative on the way up, positive on the way down. The total work done is zero.During the flight of the projectile, the total work done by the force of air resistance is...A: zero. B: positive C: negativeAnswer: Negative. The force of friction always opposes the motion, so the work done byfriction is negative on the way up and on the way down.During the flight of the projectile, the total work done by the net force is...A: zero. B: positive C: negativeAnswer: Negative. Work done by net force = Work done by gravity and by friction, so Wnet = Wgrav+Wfric. Wgrav=0 and Wfric<0 (from above questions) so Wnet<0.RIII-2. A pendulum is launched in two different ways. During both launches, the bob is given an initial speed of 3.0 m/s and the same initial angle from the vertical. On launch 1,the speed is upwards, on launch 2, the speed is downwards.Which launch will cause the pendulum to swing the largest angle from the equilibrium position on the left side? Assume no friction.A: Launch 1 B: Launch 2C: Both launches give the same maximum displacement.Answer: Both launches give the same maximum displacement. Apply conservation of energy, Etot = KEi + PEi = KEmax = PEmax. For both launch 1 and 2, the initial KE and the initial PE are the same. So the Etot is the same for both launches.1 2RIII-3 A block of mass m with initial speed v slides up a frictionless ramp of height h inclined at an angle  as shown. Assume no friction.True A or False B: Whether the block makes it to the top of the ramp depends on the mass of block and on the angle .Answer: false. The block will make it to the top if (1/2)mv2 > or = mgh. The m’s cancel.Whether the block makes it to the top depends only on v and h.RIII-4 A block with an initial speed vo slides up a rough incline. A student wishes to know the speed v when the block is at a height h. The friction between the block and the incline has coefficient of kinetic friction K.True(A) or False (B). omv mv mgh= +2 21 12 2Answer: False. There is friction and heat generated in this situation. So KEi + PEi = KEf + PEf + Q (Q = heat generated = |Wfric | )RIII-5 The tip of the nose of a pogo stick rider moves along the path shown. The maximum compression of the pogo stick spring is shown.vmhhvmvov = ?KAt what point(A, B, C or D) is the elastic potential of the spring energy a maximum?Answer: D, when the spring is fully compressedAt what point is the gravitational potential energy a maximum?Answer: A, at the maximum height.RIII-6 A mass m is placed into a spring-loaded gun, with the spring compressed by an amount x. The gun is fired so the mass is pushed forward and slides down a frictionless surface of height h1, and then up to a final height h2 as shown. What is the final kinetic energy of the mass m when it is at its final height h2? (NOTE! This question is asking for the KE, not the speed v). A) 211 22kx mg(h h ) B) 22 1kx 2mg(h h ) AxBCDh2 h1 m v xC) 22 1kx 2g(h h )m D) 22 1kx 2g(h h )m E) None of these is the correct expression for the KE. Answer:(A) i i f f2 21 11 22 22 21 11 22 2KE PE KE PE0 mgh kx mv mghmv kx mgh mgh+ = ++ + = += + -RIII-7 A pendulum consists of a mass m at the end of a string of length L. When the string is vertical, the mass has a speed vo, as shown. What is the maximum height h, above the lowest point, to which the mass swings?A)22gvoB) vgo2C) mvgo22D)vgo22E) None of theseAnswer: vgo22 Apply conservation of energy: i i f f21o22oKE PE KE PEmv 0 0 mghv 2gh (the m's cancel)+ = ++ = +=RIII-8Ball 1 of mass m moving right with speed v bounces off ball 2 with mass M (M > m), andthen moves left with speed 2v. vohmWhat is the magnitude of the impulse I = p of ball 1?A: mv B: 2mv C: 3mv D: (1/2)mv E: None of these By how much did the momentum of ball 2 decrease?A: mv B: 2mv C: 3mv D: (1/2)mv E: zero Answers: Lets choose right as the +x direction. p = pf - pi = -2mv - mv = -3mv.p 3mv . Or graphically,   p p pf i  The momentum of ball 2 decreased by 3mv By conservation of momentum, the change in momentum of ball 2 must be equal in magnitude (and opposite in sign) to the change in momentum of ball 1. So the momentum of ball 2 decreases by 3mvRIII-9 A bullet of mass m traveling horizontally with initial speed v strikes a wooden block of mass M resting on a frictionless table. The bullet buries itself in the block, and the block+bullet have a final speed vf. Fill in the blank: The total kinetic energy of the bullet+block after the collision is ___________ the total KE before the collision.A) greater than B) less than C) equal toTrue(A) or False(B): In the problem above, the collision was elastic.Answer: KE was lost in the collision. There was lots of friction when the bullet hit the block and so lots of heat was generated. The collision was NOT elastic.RIII-10 Car 1 and Car 2 are moving along X-axis road and collide head-on at Origin junction. Before the collision, Car 1 was moving right (toward Positive City) and Car 2 pipfpm v m 2v M M M m v Before M+m vf Afterwas moving left (towards Negativeville). The colliding cars stick together and the mangled pile of metal slides to the right after the collision. Consider the following statements:I. Before the collision, the magnitude of Car 1's momentum was greater than Car 2's.II. Before the collision, the kinetic energy of Car 1 was greater than Car 2's.III. The mass of car Car1 was greater than Car 2's.Which statements must be true always?A) I only B) I and II only C) I, II, and III D) None of the three statements above must always be true.Answer: Only I is true always. Momentum is conserved in collisions always. Since the wreckage slide right, the total momentum must be to the right. So right-going car 1 must have the greater momentum. The masses and KE's could be various relative values.RIII-11 A rider in a "barrel of fun" finds herself stuck with her back to the wall. The free-body diagram is shown.Is the following statement definitely true for this situation? sN=mgA: Yes, it is certainly true. B: No, it might not be true.Answer: No, it