PHYSICS 111 HOMEWORK SOLUTION #8

1 PHYSICS 111 HOMEWORKSOLUTION #8 March 24, particle of mass m moves with momentum of magnitude p. a) Show that the kinetic energy of the particle is:K=p22m(Do this on paper. Your instructor may ask you to turn in thiswork.) b) Express the magnitude of the particle s momentum in termsof its kinetic energy and )By definition, the momentum of a partcile moving with velocity~vis :~p=m~ magnitude isp= energy is :K=mv22=m(pm)22=p22mb)K=mv22, thereforev= 2 Kmp=mv=m 2Km= m22Km= object has a kinetic energy of 239 J and a momentum of kgm/s. Find the speed and the mass of the object. Let s use the expressions of problem 1Kp=mv22mv=v2v=2Kp=2 The mass cam be then obtained from momentum as :m=pv= one instant, a sled is moving over a horizontal surface ofsnow at m/s. After s has elapsed, the sled stops. Use amomentum approach to find the magnitude of the average frictionforce acting on the sled while it was moving. The sled started with a speed of m/s and will come to stop This change in speed or momentum is due to can rewrite Newton s second law by taking momentum change intoconsideration as follows: ~Fi=m~a=md~vdt=d(m~v)dt=d~pdtmoving right+y+x~N~fm~gIf we project this on the horizontal surface , the only force that remainsis friction f.

2 In average: f=pf pi t=mvf vi t= 17 0 girl is standing on a 159-kg plank. Both originally at reston a frozen lake that constitutes a frictionless, flat surface. The girlbegins to walk along the plank at a constant velocity of ~im/srelative to the plank. a) What is the velocity of the plank relative to the ice surface? b) What is the girl s velocity relative to the ice surface?a)The system{girl+plank}is in a frictionless environment for which momentumshould be conserved during adopt the following notations : the girl has mass m and velocity~vrelative to surface the plank has mass M and velocity~Vrelative to surface the girl s velocity relative to the plank is~vgrl/plk= ~iAt rest vector momentum is~0, during the motion this momentum ism~v+M~V. We should have :m~v+M~v=~0 or consequentlym~v= M~ already indicates that the girl and the plank are moving in opposite direc-tions. On the other hand, the girl s velocity relative to surface is an additionof her velocity relative to the plank and the velocity of the plank relative tothe surface :~v=~vgrl/plk+~Vm~v= M~Vm(~vgrl/plk+~V) = M~V (m+M)~V=m~vgrl/plk~V= mm+M~vgrl/plk= + ~i= ~iThe plank is moving with speed m/s in the opposite )The girl s velocity relative to the ice surface is:~v=~vgrl/plk+~V= ~i ~i= ~iv= blocks of masses m and 3m are placed on a frictionless, horizontalsurface.

3 A light spring is attached to the more massive block, and theblocks are pushed together with the spring between them as shownin the figure below. A cord initially holding the blocks together isburned; after that happens, the block of mass 3m moves to the rightwith a speed of~V3m= ~im/s a)What is the velocity of the block of mass m? (Assume rightis positive and left is negative.) b) Find the system s original elastic potential energy, taking m= kg. c) Is the original energy in the spring or in the cord? d) Explain your answer to part (c). e) Is the momentum of the system conserved in the bursting-apart process? f) Explain how that is possible considering there are large forcesacting. g)Explain how that is possible considering there is no motionbeforehand and plenty of motion afterward?7a)Momentum is conserved and we have :m~vm+ 3m~V3m=~0~vm= 3mm~V3m= 3~V3m= 3 ~i= ~i|~vm|= )Total energy is conserved, gravitation potential energy doesnt change but elas-tic potential energy changes after the spring is Ueli= (Kf Ki)0 Uelf= (12mv2m+123mV23m)Ueli=m2(v2m+ 3V23m)= ( + 3 )= )the original energy is in the )A force had to be exerted over a distance to compress the spring, transferringenergy into it by work.

4 The cord exerts force, but over no )the momentum of the system is conserved in the bursting-apart process andthat s what we used in the first )The forces on the two blocks are internal forces, which cannot change themomentum of the system the system is )Even though there is motion afterward, the final momenta are of equal magni-tude in opposite directions so the final momentum of the system is still a rubber ball is dropped from a height of m, itbounces off a concrete floor and rebounds to a height of m. a) Determine the magnitude and direction of the impulse deliv-ered to the ball by the floor. b) Estimate the time the ball is in contact with the floor to seconds. Calculate the average force the floor exerts on )Hitting the floor then bouncing up will cause a momentum change , the impulsedelivered to the ball by the floor is just this momentum Momentum=m~vafter m~vbeforeFalling from a m will give the speed at the moment the ball hits the floor:v2before 0 = 2a Hv2before= 2g Hvbefore= 2g H= 2 ~vbefore= ~jSimilarly, Bouncing up to will give the speed at the moment the ballbounces up:vafter= 2 ~vafter= ~j9 Finally,Impulse= Momentum=m~vafter m~vbefore= ( ~j+ ~j)= ~jThe impulse delivered amounts to with direction )In problem 2 we rewrote Newton s 2nd Law as: ~Fi=d~ average force the floor exerts on the floor is then~F= P t= ~j= ~jwith maginutude N and direction tennis player receives a shot with the ball ( 0 kg) travelinghorizontally at m/s and returns the shot with the ball travelinghorizontally at m/s in the opposite direction.

5 (Assume the initialdirection of the ball is in the x direction.) a) What is the impulse delivered to the ball by the tennis rac-quet? b)What work does the racquet do on the ball?a)We will use the same procedure as in problem impulse delivered to the ball is Momentum=m~vafter m~vbefore= ( ~i ( )~i)= ~ )The work done by the raquet on the ball can be calculated from the kineticenrgy change as:W= K=12m(v2f v2i)=12 ( )= 1 kg car traveling initially with a speed of m/sin an easterly direction crashes into the back of a 8 kgtruck moving in the same direction at veloc-ity of the car right after the collision is m/s to the east. a)What is the velocity of the truck right after the collision?(Give your answer to five significant figures.) b) What is the change in mechanical energy of the cartrucksystem in the collision? c) Account for this change in mechanical )Conservation of momentum of the{car+truck}system can be expressed as:m~Vci+M~Vti=m~Vcf+M~Vtf~Vtf=mM(~Vci ~Vcf) +~Vti=12408100(28 18)~i+ 20~i= ~iThe truck will keep moving east with speed of m/sb)The change in mechanical energy is computed through the kinetic energychange as no potential energy change takes place: E=12m(V2cf V2ci) +12M(V2tf V2ti)=12 1240 (182 252) +12 8100( 202)= 8301Jc)Most of the energy was transformed to internal energy with some being carriedaway by bullet is fired into a stationary block of wood having mass m= kg.

6 The bullet imbeds into the block. The speed of the bullet-plus-wood combination immediately after the collision is was the original speed of the bullet? (Express your answer withfour significant figures.)We use momentum conservation and we take into account that the bullet andthe block have the same speed after collisionm~vi+M~Vi=m~vf+M~ ~vi+ 0 = (m+M)~Vfvi=m+MmVf= + neutron in a nuclear reactor makes an elastic, head-on collisionwith the nucleus of a carbon atom initially at rest. a) What fraction of the neutron s kinetic energy is transferred tothe carbon nucleus? (The mass of the carbon nucleus is times the mass of the neutron.) b) The initial kinetic energy of the neutron is 13J. Findits final kinetic energy and the kinetic energy of the carbonnucleus after the s adopt the following notations : for the neutron, mass m,viandvfare the initial and final velocityrespectively. for the atom , mass M ,ViandVfare the initial and final fraction of the neutron s kinetic energy that s transferred to the carbonnucleus is just :12MV2F12mv2iConservation of momentum on the head-on collision gives:mvi=mvf+MVfvi vf=MmVf= 12Vf13 Conservation of kinetic energy gives :12mv2i=12mv2f+12MV2fv2i v2f=MmV2f= 12V2fandv2i v2fvi vf=vi+vf=12V2f12Vf=VfWe get the set of equations :vi vf= 12 Vfandvi+vf=Vfto finally obtainvi=132 Vfand the fraction transfered to the atom:12MV2F12mv2i= 12 (213)2= )The kinetic energy of the carbon nucleus is:12MV2F= 10 13= 10 14 JThe final kinetic energy of the neutron is :Kifinal= 10 13 10 14= 10 automobiles of equal mass approach an intersection.

7 One vehicleis traveling with velocity m/s toward the east and the other istraveling north with speed v2 . Neither driver sees the other. Thevehicles collide in the intersection and stick together, leaving parallelskid marks at an angle of north of east. Determine the initialspeedv2iof the northward-moving speed limit for both roads is 35 mi/h and the driver of thenorthward-moving vehicle claims he was within the speed limit whenthe collision occurred. Is he telling the truth?15~v1i~v2i~vfMomentum conservation :m~v1i+m~v2i= 2m~vf~v1i+~v2i= 2~vfWe can project the last equation on the West-East Direction to get:v1i= 2vfcos on the South-North direction to get :v2i= 2vfsin tan 13 tan driver is definitely lying about his object of mass kg, moving with an initial velocity of ~i,collides with and sticks to an object of mass kg with an initialvelocity of ~jm/s. Find the final velocity of the composite conservation of the two-object system will give us the final velocityafter collision:(m1+m2)~Vf=m1~v1i+m2~v2i~Vf=1m 1+m2(m1~v1i+m2~v2i)= + ( ~i ~j)= ~i ~ billiard ball moving at m/s strikes a stationary ball of thesame mass.

8 After the collision, the first ball moves at m/s at anangle of with respect to the original line of motion. Assumingan elastic collision (and ignoring friction and rotational motion), findthe struck ball s velocity after the rest~v1i~v1f~v2f Momentum is again conserved :m~v1i=m~v1f+m~v2f~v1i=~v1f+~v2fProjecti ons on the x and y-axis:v1i=v1fcos +v2fxv2fx=v1i v1fcos = cos 26= =v1fsin +v2fyv2fy= v1fsin = sin 26= of the struck ball is:v2f= + 200472= making an angle of: = arctan : +| |= 90 . This can be derived easily by looking at conservationof monentum and kinetic energy:~v1i=~v1f+~v2fv21i=v21f+v22f+ 2~v1f ~v2fand12mv21i=12mv21f+12mv22fv21i=v21f+ v22fThis implies~v1f ~v2f= 0 or +| |= 90 . unstable atomic nucleus of mass 10 26kg initially at restdisintegrates into three particles. one of the particles of mass 10 27kg, moves in the y direction with a speed of 106 particle of mass 10 27kg, moves in the x directionwith a speed of 106m/s.

9 Find the velocity of the third particle Find the total kinetic energy increase in the rest?~v1~v2~v3 Momentum is conserved :~0 =m1~v1+m2~v2+m3~v3~v3= 1m3(m1~v1+m2~v2)= 10 26( 6 106~j+ 4 106~i) 10 26= 106~i 106~jits maginitude isv3= + 106m/sb)The kinetic energy increase of the system is then computed as: Ki=12m1v21+12m2v22+12m3v2321= 62+ 42+ 1012 10 26= 10 13J22