Phys 201 1st EditionExam # 3 Study Guide Lectures: 15 - 20Lecture 15 (October 27)A 70kg soccer player kicks a .5kg ball, giving it a speed of 14m/s. The contact time of the foot on the ball is 0.1s. a) What is the force on the ball? b) What is the force on the foot?Answer: a) 70N b) 70NExplaination: a) Kicking the ball changes the balls momentum. Therefore, we can use the momentum equation to solve for the force. FΔt=Δρ Δρ=ρf-ρi =mVf - mVi = (14)(.5)-0 =7FΔt=Δρ  FΔt=7kgm/s  F=7/Δt =7/.1= 70Nb) There is no need to calculate the force on the foot. Consider Newton’s third law (For every action there is an equal and opposite reaction). This means that the force on the foot will be thesame as the force on the foot, just in the opposite direction.Lecture 15 (October 27)A 1668.0 g mass is on a horizontal surface with μk = 0.37, and is in contact with a massless spring with a force constant of 619.0 N/m which is compressed. When the spring is released, it does 2.51 J of work on the mass while returning to its equilibrium position. a) Calculate the distance the spring was compressed. b) What is the velocity of the mass as it loses contact with the spring? Answer: a).09 m b) 1.534 m/s Explanation: a) PEspring=.5kx22.51=.5(619)x2X2=.0081X=.09b) The easiest was to calculate the velocity of the mass is from its kinetic energy, which is equal to the work done by the compressed minus the work done by friction. The work done by friction is equal to the force of friction times the distance that the spring was compressed.KE=Ws-Wf.5mV2 =Ws-fxF=(.37)(1.668)(9.81)F=6.04N.5(1.668)V2=2.51-(6.04(.09))V2=2.3758V=1.535Lecture 16 (October 29)A croquet mallet delivers an impulse of 8.77 N s to a 0.64- kg croquet ball initially at rest. What is the speed of the ball immediately after being struck?Answer: 13.7m/sExplanation: Impulse (p)= Ft  p=mV 8.77=(.64)V V=13.7m/sLecture 16 (October 29)A 0.145- kg baseball traveling 33.50 m/s is stopped in 0.193 s by a catcher's mitt. A) What is the average acceleration of the ball? b) What is the average force on the catcher's mitt?Answer: a) 173.58m/s b) 25.17NExplanation: a) a=V/t a=33.5/.193 a= 173.575m/s b) The force that the ball is thrown is the same as the force the ball exerts on themitt. Plug your value for acceleration into the equation; F=ma F=.145(173.575) F= 25.168N. Lecture 16 (October 29)A 7.85- g bullet from a 9-mm pistol has a velocity of 374.0 m/s. It strikes the 0.745- kg block of a ballistic pendulum and passes completely through the block. If the block rises through a distanceh = 13.96 cm, what was the velocity of the bullet as it emerged from the block?Answer: 216.93m/sExplanation:Use the conservation of momentum equation for this problem: m1V1i=m2V2 +m1V1f where m1 and V1 I are the mass and initial speed of the bullet, m2 and V2 are the mass and velocity of theblock, and V1f is the final speed of the bullet, which is what you're trying to find. Before you solve for V1f, you need to solve for V2. Use conservation of energy:KE=PE .5mV2=mgh.5(.745)V2 = (.745)(9.81)(.1396m)V=√(2.739) = 1.655m/sPlug this value into the conservation of momentum equation to solve for Vfm1V1i=m2V2 +m1V1f (.00785)(374)=(.745)(1.655) +(.00785)Vf Vf=216.93m/sLecture 17 (November 3)An Escort and a Camaro traveling at right angles collide and stick together. The Escort has a mass of 1200 kg and a speed of 20 km/h in the positive x direction before the collision. The Camaro has a mass of 1500 kg and was traveling in the positive y direction. After the collision, the two move off at an angle of 54 o to the x axis. What was the speed of the Camaro?Answer: 2.20×101 km/h Explanation: Use the mass and velocity of the cars before the collision to find the velocity of the cars after the collision. The angle describes the direction that they move off at after the collision. Use the x and y components of this final velocity to figure out the initial speed of the Camaro. M1 VfV1 VyVxM1=1200kg V2V1=5.56m/s M2M2=1500P1=Pxm1V1=(m1+m2)Vx [m1/(m1+m2)]V1=VxVx=[1200/(1500+1200)]5.56=2.471m/sTan(θ)=Opp/adjTan(54)=Vy/2.471Vy=3.4m/sPiy=PfyM2V2=(m1+m2)Vfy 1500V2=(1200+1500)(3.4)V2=6.12m/s = 22.032 Km/hLecture 17 (November 3)A 30-caliber rifle with a mass of 3.84 kg fires a 7.13- g bullet with a speed of 606 m/s with respect to the ground. What kinetic energy is released by the explosion of the gunpowder that fires the bullet?Answer: 1312JExplanation: m1V1=m2V23.84V1=(.00713)(606)V1=1.125m/sKE=.5m1V12 + .5m2V22KE=.5(3.84)(1.1252) + .5(.00713)(6062)KE=1311.626Lecture 17 (November 3)An 84 kg ice hockey player standing on a frictionless sheet of ice throws a 5.4 kg bowling ball horizontally with a speed of 3.3 m/s. With what speed does the hockey player recoil?Answer: .212m/sExplanation: Use conservation of momentum (m1V1=m2V2). The hockey player's recoil is V1. m1V1=m2V2 84(V1)=(5.4)(3.3) V1=17.82/84 V1=.212m/sLecture 18 (November 05)A distant planet has a mass of 0.6800 ME and a radius of 0.8000 RE. What is the ratio of the escape speed from this planet to the escape speed from the earth? Answer: .922Explanation: The equation for the escape velocity on Earth is Vesc= √(2MG/Re). To find the ratio of this new planet, just substitute these values in for M and R. All other variables in this equation are the same, and therefore irrelevant.√(M/R)√(.68/.8)=.922 Lecture 19 (November 10)Two children are playing on a balanced seesaw. One child with a mass of 39 kg sits 1.3 m from the center. Where on the other side must the second child sit if her mass is 32 kg? Answer: 1.58mExplanation: Torque is equal to force times radius or distance from the center. For the seesaw tobalance, the torques of the two children must be equal. T1=T2F1r1=F2r2(39)(9.81)(1.3)=(32)(9.81)rr=1.58m (from the center)Lecture 20 (November 12)A torque of 16Nm is applied to a heavy wheel whose moment of inertia is 48 kgm2. a) What is the angular acceleration of the wheel? b) If the wheel was initially at rest and the torque is applied for 5 s, what will be the rotational frequency ω of the wheel at the end of the 5s?Answer: a) 3.33×10-1 rad/s2 b) 1.67 rad/sExplanation: a) Torque=Iα16=48αα= .3333b)