ENGINEERING COUNCIL DYNAMICS OF MECHANICAL SYSTEMS D5 TUTORIAL LINEAR IMPULSE AND MOMENTUM On copletion of this ttorial yo shold be able to do the following. State Newton s laws of otion. Define linear oent and iplse. State the law of conseration of oent. Define the coefficient of restittion. Sole probles inoling the collision of bodies. Sole probles inoling pile driers. D.J.DUNN freestdy.co.k
. NEWTON'S LAWS OF MOTION. A body at rest or with nifor otion will reain at rest or contine with nifor otion ntil it is acted on by an external force.. An external force will case the body to accelerate or decelerate. 3. Eery force has an eqal and opposite reaction.. EXPLANATION Iagine a person on an ice rink with absoltely no friction between the skates and the ice. If he was oing, he wold be nable to neither slow down nor speed p. The person cold only change his otion if an external force was applied to hi. This is an exaple of the first law. In reality the external force is obtained by finding friction with the ice by digging the skates into the ice and pshing or braking. This force prodces changes in the otion of the skater. Using friction to enable hi to either accelerate or decelerate is an application of the second law. Next, iagine the person stationary on the ice. In his hands he has a heay ball. If he threw the ball away, he wold oe on the ice. In order to throw the ball away he st exert a force on the ball. In retrn, the ball exerts an eqal and opposite force on the person so he oes away in the opposite direction to the ball. This is an exaple of the third law. The sae principles apply to a space ehicle. There is no friction in space and the only way to change the otion of a space ehicle is to eject atter fro a rocket so that the reaction force acts on the ehicle and changes its otion. The law which has the greatest significance for s is the nd. law so let's look at this in detail. D.J.DUNN freestdy.co.k
. NEWTON'S nd LAW OF MOTION We sally think of the second law as stating Force = ass x acceleration. In fact it shold be stated in a ore fndaental for as follows. The IMPULSE gien to a body is eqal to the change in MOMENTUM. This reqires s to ake a few definitions as follows. IMPULSE IMPULSE is defined as the prodct of force and the tie for which it is applied. Iplse = Force x Tie = Ft WORKED EXAMPLE No. A ehicle has a force of 400 N applied to it for 0 seconds. Calclate the iplse? IMPULSE = Ft = 400 x 0 = 8000 N s MOMENTUM MOMENTUM is defined as the prodct of ass and elocity. Moent = Mass x Velocity = kg /s WORKED EXAMPLE No. A ehicle of ass 5 000 kg changes elocity fro /s to 8 /s. Calclate the change in oent. Initial oent = = 5 000 x = 0 000 kg /s Final oent = = 5 000 x 8 = 40 000 kg /s Change in oent = 40 000-0 000 = 30 000 kg /s D.J.DUNN freestdy.co.k 3
REWRITING THE LAW Fro the stateent Iplse = change in oent, the second law can be written as Ft = This eqation ay be rearranged into other fors as follows. F = /t If the ass is constant and since acceleration = /t this becoes F = /t = a This is the ost failiar for bt if the ass is not constant then we ay se F = /t = elocity x ass flow rate This for is sed in flid flow to sole forces on pipe bends and trbine blades. The for we are going to se is F = /t = rate of change of oent This for of the law is sed to deterine the changes in otion to solid bodies. Let's se this to stdy what happens to bodies when they collide. 3. COLLISIONS When bodies collide they st exert eqal and opposite forces on each other for the sae period of tie so the iplse gien to each is eqal and opposite. Since the iplse is eqal to the rate of change of oent, it follows that each body will receie eqal and opposite changes in their oent. It frther follows that the total oent before the collision is eqal to the total oent after the collision. This reslts in the law of conseration of oent. 3. THE LAW OF CONSERVATION OF MOMENTUM. The total oent before a collision is eqal to the total oent after the collision. Consider two bodies of ass and oing at elocities and in the sae direction. After collision the elocities change to and respectiely. Figre The initial oent = + D.J.DUNN freestdy.co.k 4
D.J.DUNN freestdy.co.k 5 Figre The Final oent = + By the law of conseration we hae...() - - by ltiply rearrange each ter by by diide 3. ENERGY CONSIDERATIONS The law of conseration of oent is tre regardless of any energy changes that ay occr. Howeer, in order to sole the elocities, we st consider the energy changes and the easiest case is when no energy is lost at all. The only energy for to be considered is kinetic energy....(3)... Factorising gies each ter by ltiply Sbstitteeqation () into ()...() eery terby ltiply lost then no K.E. is If AFTER the collision Total K.E. before the collision Total K.E. ( - ) is the relatie elocity between the two bodies before they collide. ( ) is the relatie elocity between the after the collision.
It follows that if no energy is lost then the relatie elocity before and after the collision is eqal and opposite. This eans that the elocity with which they approach each other is eqal to the elocity at which they separate. If energy is lost in the collision then it follows that ( - ) > - ( - ) In order to sole nerical probles when energy is lost we se the COEFFICIENT OF RESTITUTION which is defined as e...(4) These eqations st be reebered. WORKED EXAMPLE No.3 A ass of 00 kg oes along a straight line at /s. It collides with a ass of 50 kg oing the opposite way along the sae straight line at 0.6 /s. The two asses join together on colliding to for one ass. Deterine the elocity of the joint ass. Figre 3 The noral sign conention st be sed naely that otion fro left to right is positie and fro right to left is negatie. = 00 kg = /s = 50 kg = -0.6 /s Initial oent = (00 x) + {50 x (-0.6)} = 0 kg /s Final oent = 0 kg /s (consered) After collision the ass is 50 kg and the elocity is. Final oent = 50 = 0 = 0/50 = 0.04 /s The cobined ass ends p oing to the left at 0.04 /s. Notice that becase the asses joined together, eqation 3 was not needed. D.J.DUNN freestdy.co.k 6
WORKED EXAMPLE No.4 A ass of 5 kg oes fro left to right with a elocity of /s and collides with a ass of 3 kg oing along the sae line in the opposite direction at 4 /s. Assing no energy is lost, deterine the elocities of each ass after they bonce. Figre 4 = 5 kg = /s = 3 kg = - 4 /s Initial oent = 5 x + 3 x (-4) = - kg /s Final oent = 5 + 3 = -...(a) Relatie elocity before collision = ( - ) = - (-4) = 6 /s (coing together). Since no energy is lost the coefficient of restittion is.0. Relatie elocity after collision = ( - ) = -(6) = -6 /s (parting)...(b) The elocities ay be soled by cobining the siltaneos eqation a and b. One way is as follows. Eqation (b) x 3 + eqation (a) yields 3-3 = -8 5 + 3 = - 8 + 0 = -0 hence = -.5 /s and = + 3.5 /s Figre 5 D.J.DUNN freestdy.co.k 7
WORKED EXAMPLE No.5 Repeat exaple bt this tie there is energy lost sch the coefficient of restittion is 0.6. This tie we hae = 5 kg = /s = 3 kg = - 4 /s Initial oent = 5 x + 3 x (-4) = - kg /s Final oent = 5 + 3 = -...(a) Relatie elocity before collision = ( - ) = - (-4) = 6 /s (coing together). Relatie elocity after collision = ( - ) = -(0.6)(6)= - 3.6 /s (parting)...(b) The elocities ay be soled by cobining the siltaneos eqation a and b. One way is as follows. 5 + 3 = - hence = -0.4-0.6 + = -3.6 = -3.6 + = -0.4-0.6 hence = -.6 /s = +.0 /s D.J.DUNN freestdy.co.k 8
SELF ASSESSMENT EXERCISE No.. A ass of 0 kg traels at 7 /s and collides with a ass of kg traelling at 0 /s in the opposite direction along the sae line. The coefficient of restittion is 0.7. Deterine the elocities after collision. (Ans. -0. /s and 8.687 /s). A ass of 0 kg traels at 8 /s and collides with a ass of 0 kg traelling along the sae line in the sae direction at 4 /s. The coefficient of restittion is 0.8. Deterine the elocities after collision. (Ans. 3. /s and 6.4 /s) 3. A railway wagon of ass 3 000 kg traelling at 0.4 /s collides with a stationary wagon of ass 3 500 kg and becoes copled. Deterine the coon elocity after collision. (Ans. 0.94 /s) 4. A ass of 4 kg oes at 5 /s along a straight line and collides with a ass of kg oing at 5 /s in the opposite direction. The coefficient of restittion is 0.7. Calclate the final elocities. (Ans. -0.67 /s and 6.33 /s) D.J.DUNN freestdy.co.k 9
4. PILE DRIVERS A particlar application of the preceding section is to pile driers. These are deices which drop large asses onto a pile in order to drie the pile into the grond. The pile has no initial oent and the otion gien to it is qickly decelerated by the frictional resistance as it oes into the grond. The elocity of the drier is obtained by graitational acceleration and conersion of potential energy into kinetic energy. WORKED EXAMPLE No.6 A pile drier has a ass of 00 kg and falls 3 onto the pile which has a ass of 00 kg. The coefficient of restittion is 0.7. Calclate the elocity of the pile and the drier iediately after ipact. Figre 6 Initial potential energy of drier = gz Kinetic energy at ipact = / Eqating gz = / = (gz) ½ = ( x 9.8 x 3 ) / = 7.67 /s Since this is down, by noral conention it is negatie. = -7.67 /s Initial elocity of the pile = 0 Initial oent = -00 x 7.67 = -767. kg /s Final oent = + 00 + 00 = -767. kg /s...() Initial relatie elocity = - = -7.67 /s coing together. Final relatie elocity = - = -0.7(-7.67) = 5.37 /s parting. - = 5.37 /s...() Soling eqations () and () we hae =.03 /s (pwards) = - 4.35 /s (downwards) D.J.DUNN freestdy.co.k 0
WORKED EXAMPLE No.7 Using the sae data as for proble No.6, deterine the height to which the drier rebonds. The pile is drien into the grond 0.08. Deterine the aerage grond resistance assing nifor deceleration. The K.E. of the drier is reconerted into P.E. so the height to which it rebonds is z = /g = (.03)/( x 9.8) = 0.053 Fro the laws of otion coered in earlier ttorials, for nifor deceleration x = at / = / a 0.08 = (-4.35) /a hence a = 8.6 /s Since the pile is decelerating, then strictly a = - 8.6 /s Fro Newton's nd Law of otion we hae F = a = 00 (-8.6) = - 6 650 N The negatie sign indicates F is a resistance and not a help to otion. D.J.DUNN freestdy.co.k
SELF ASSESSMENT EXERCISE No.. a) A ass oes along a straight line with elocity before colliding with a ass. Following the ipact, the two asses contine to oe along the sae line with elocities and respectiely. Show that = ( - e)/( + ) = ( + e)/( + ) The coefficient of restittion is defined as e = ( - )/ b) A pile of ass 500 kg is drien into the grond by a ra of ass 000 kg. On the final stroke, the ra descends 5 freely nder graity and the pile is drien 0.05 into the grond. The coefficient of restittion is 0.5. Deterine i. the elocity of the ra jst before ipact. (Ans. 9.9 /s) ii. the elocities of the ra and pile iediately after ipact. (Ans. 4.95 and 9.9 /s both downwards) iii. the resisting force of the grond assing nifor deceleration. (Ans. 98 kn) i. the energy lost in the ipact. (Ans..3 kj). A projectile of ass 30 graes traels horizontally at elocity. It ebeds itself in a stationary sand bag of ass 0 kg which is sspended on a rope. The centre of ass of the bag is. below the point of sspension. After being strck, the bag swings freely throgh an arc of 30 o. Deterine the elocity and kinetic energy lost in the ipact. (Answers 593.8 /s and 5 73 J) Hint... the easiest way to do this is by se of potential and kinetic energy as well as oent theory). Figre 7 D.J.DUNN freestdy.co.k
3. a) A ass oes along a straight line with elocity before colliding with a ass oing in the sae direction at elocity (hence > ). Following the ipact, the two asses contine to oe along the sae line with elocities and respectiely. Show that ( )< -( ) b) A pile of ass 600 kg is drien into the grond by a ra of ass 300 kg. On the final stroke, the ra descends 3 freely nder graity and the pile is drien into the grond. The coefficient of restittion is 0.8 and the resisting force of the grond is a constant ale of 40 kn. Deterine i. The depth of penetration of the pile into the grond. (Ans. 0.059 ) ii. The height to which the ra rebonds. (Ans. 0. ) iii. The energy lost in the ipact. (Ans. 4 J) Note that the coefficient of restittion is defined as ( - )= -e( - ) D.J.DUNN freestdy.co.k 3