Linear Momentum and Collisions Conservation of linear momentum

Transcription

1 Unit 4 Linear omentum and Collisions 4.. Conseration of linear momentum 4. Collisions 4.3 Impulse 4.4 Coefficient of restitution (e) 4.. Conseration of linear momentum m m u u m = u = u m Before Collision After Collision Consider that we are performing a collision experiment with two particles (not necessary identical particles) on a two-dimensional plane, say, smooth table. If the initial elocity ectors of the two particles were labeled as u and u respectiely, then after collision, their elocity were found to be and respectiely. The theory behind the collision During the collision, the forces act on each other are with the same magnitude but opposite in direction. This is the Newton s third law, it is about the action and reaction forces. They are always opposite in directions but they hae the same magnitudes (e.g. F = F ). Hence, we hae m t m m = t = m

2 Substituting, and rearrange the equation, we obtain m ( u) = m ( u ) or m( u) + m( u) = 0. That is, m u + m u = m + m The aboe expression is the conseration of linear momentum. Define the linear momentum of a particle as p, where p = m. We can rewrite the aboe equation as p = constant or in another form p i = 0. i Experimental facts After performing numerous trials with different initial elocities and final elocity being measured, it was found that () is always in opposite direction of, () = constant. We can repeat the experiment by changing different particles and we found that different particles hae different degree of resistance to change its magnitude of the elocity after the collision. We can check that the constant is gien by the ratio of m and m : m =, m where m and m are then called the inertia mass of the particles, which is a measure of the resistance to change the elocity magnitude during an interaction with another particle. From this experiment, we also discoer a conseration law if we define a physical quantity called momentum by: p = m. i i Example An ant lands on one end of a floating 4.75g stick. After sitting at rest for a moment, it runs toward the other end with a speed of 3.8cm/s relatie to the still water. The stick moes in the opposite direction at 0.cm/s. What is the mass of the ant? Still Water

3 The total momentum of the system before the ant runs on the stick is zero. By conseration of linear momentum, the total momentum of the system after the ant runs on the stick equals zero. Hence we can write p a + p s = 0, m a a + m s s = 0, Substituting a, s and m a, we obtain m a (3.80) ( 0.) = 0 m a = 0.5g Example A gun of mass fires a shell of mass m and recoils horizontally. If the shell traels with speed relatie to the barrel, find the speed with which the barrel begins to recoil if (a) the barrel is horizontal, (b) the barrel is inclined at an angle α to the horizontal. (a) V m Let the barrel be recoiling with speed V. The speed of the shell as it leaes the barrel is V. Before firing the shell, the gun is at rest and the total momentum is zero. By the conseration of momentum, m ( V) V = 0. Hence V = m / ( + m). (b) When the gun is inclined at angle α. The shell leaes the barrel with a elocity which is the resultant of two components, and V. By the conseration of momentum in the direction of recoil, m ( cos α V) V = 0. m Hence V = m cos α / ( + m). V α 3

4 Example Two men each of mass m ride on a moable platform which has mass and is initially at rest on a smooth track. Both of them take a leap from the platform simultaneously along the track with speed u relatie to the platform. (a) (b) (c) Find the speed of the platform just after they jump. If they jump one after one, find the final speed of the platform. Compare the final speed of platform in both cases. (a) Let V be the speed of platform just after both men jump and leae the platform. The conseration of linear momentum along the track implies 0 = m( u V ) V, which (b) mu giesv =. + m Let V be the speed of platform just after the first man leaes the platform. The conseration of linear momentum along the track gies 0 = mu ( V) ( + mv ), mu which impliesv =. + m Next, the second man leaps from the platform after the first man, we can write ( m + ) V = m( u V ) V, where V is the speed of platform just after the second man leaes the platform. After mu simplification, we hae V V =. Substituting V into the expression and + m mu mu thusv = +. + m + m (c) Note that V V > V. mu mu mu mu mu = + > + = = V + m + m + m + m + m, we obtain 4

5 4. Collisions () Elastic collision: A collision that the momentum and the kinetic energy are consered. mu + mu = m + m mu + mu = m + m, where u i and i are the speed of particles before and after the collision, i = and. () Inelastic collision: A collision that the momentum is consered but the kinetic energy is not consered. mu + mu = m + m mu + mu m + m (3) Perfectly inelastic collision: A collision that the colliding objects stick together after they hit each other. The momentum of the system is consered but the kinetic energy is not consered.. Example Two particles, whose masses are 5 kg and 7 kg are moing on the same line with speed 30m/s and 0 m/s, respectiely, when they collide. Assuming that the particles couple together after impact, find their common elocity after impact if they were (a) moing in the same direction, (b) moing in opposite directions. (a) By the conseration of momentum, (5 + 7)V = 5(30) + 7(0), we obtain V = 45/6 m/s 30m/s 5 kg 0m/s 7 kg (b) By the conseration of momentum, 30m/s 0m/s (5 + 7)V = 5(30) 7(0), we obtain V = 5/6 m/s. 5 kg 7 kg 5

6 Example A and B are two particles, of mass 4 kg and 8 kg respectiely, lying in contact on a smooth horizontal table, and connected by a string 3 m long. B is 7 m from the smooth edge of the table and is connected by a taut string passing oer the edge to a particle C of mass 4 kg hanging freely. If the system is released from rest, find the speed with which A begins to moe. A 4kg B 8kg T T C 4kg When B is in motion and less than 3 m from A, its acceleration is gien be the equations 4g T = 4 a, T = 8 a, where T is the tension in the string, and hence a = g = 3.7 m/ s 3 Hence, when B has moed 3 m, its elocity is gien by 9.8 = 3, 3 = 4.43 m/ s, and this will also be the elocity of the mass C hanging ertically. The impulse in the string joining B to C when the string AB becomes taut will gie a certain horizontal momentum to B and take away the same amount of ertical momentum from C. Hence, we may use the conseration of momentum as if all three particles were moing in the same straight line. If m/s be their common elocity after A has been brought into motion, we hae 6

7 ( ) = (8 + 4) 4.43, = 3.3 m/ s. This is therefore the speed with which A begins to moe. 4.3 Impulse Impulse is defined as the change in momentum, e.g. I = p = m mu = F ae t (Area under cure) The unit of impulse is Ns or kg m/s. F ae Area t Area = F t f = t F dt i ae Example A particle of mass lying on the ground is connected, by means of a light inextensible string passing oer a smooth pulley to a mass m. After the mass m has fallen through a height h, the string tightens and the mass begins to rise. Find the impulse applied to when the string tightens and the initial speed. The elocity of mass m just before the string tightens is gien by = 0 + gh i.e. = gh. By the conseration of momentum, we hae m + (0) = mv + V, m h m where V is the elocity of the system just after the string tightens. 7

8 Substituting the expression of, we obtain V = m gh. m+ If I is the impulse, we can write I = V (0) = m gh m+. Remarks: The impulse on m = As the elocity V is smaller than. m gh m gh mv m = m gh = m+ m+. Of course, it is negatie! 4.4 Coefficient of restitution (e) Before u u m I I m ( u > u ) After The aboe figure depicts two bodies. Since the total momentum is consered, we hae m u + m u = m + m. This one equation is not sufficient to calculate and and we hae recourse to Newton s experimental law. If the elocities both before and after impact are taken relatie to the same body, then, for two bodies impinging directly, their relatie elocity after impact is equal to a constant (e) times their relatie elocity before impact and in the opposite direction. e is known as the coefficient of restitution. = e (u u ) In the case of oblique impact, the result holds for the components the elocities in the direction of the common normal at impact. The alue of e has to be found by experiment and aries from 0 for completely inelastic bodies to practically for nearly perfectly elastic bodies. Note that the quantities u, u, and mentioned aboe are in the same direction. 8

9 Example A smooth sphere strikes an identical sphere initially at rest. If the elocity of the moing sphere before the impact is m/s at 45 o to the line of center AB, and e = 0.6, find the elocities of the spheres after the impact. The stationary sphere receies an impulse in the direction of AB. So this sphere moes in the direction of AB with elocity, after impact. Sphere A moes in the direction θ to the horizontal with final elocity u. Consider the momentum of the first sphere perpendicular to AB, m(sin 45) = mu sinθ (m: mass of the sphere) u sinθ = () u θ A B A B m/s Before impact After impact From the conseration of momentum along AB, and from Newton s law of restitution, m( cos 45) + 0 = mu cosθ + m 0.6(cos 45 0) = ucosθ. We obtain + ucosθ = cos 45 () and ucosθ =. cos 45 (3) From () and (3) we hae =.6 cos 45 and u cosθ = 0.4 cos 45 (4) i.e. =.3 and ucosθ = 0.83 Substitute () into (4), we hae tanθ = 5, hence we find o θ 79 and u =.44 m/ s Thus the elocity of the second sphere is.3 m/s along AB, and that of the first sphere is.44 m/s at 79 o to AB. 9

10 Example Three identical spheres are arranged as shown in figure. If sphere C is projected with elocity u while A and B are at rest. Gien that the coefficient of restitution is e for each sphere, find the subsequent elocities of each sphere. Show also that the condition for sphere C to pass through and beyond the two spheres A and B is e < /9. C A C A w u B B w Before impact After impact Let u be the elocity of sphere C before impact, the elocity of sphere C after impact and w the elocities of A and B after impact. By conseration of momentum: mu = m + mw cos 30 + w cos 30 = u () By Newton s law of restitution: Soling () and (), we get w cos30 e = w cos30 = eu cos30 0 u cos 30 u 3u = ( 3 e) and w= ( + e) 5 5 u Thus, the elocity of C after impact is ( 3 e) and the elocities of A and B after impact are 5 3 u the same as ( + e ). 5 If the sphere C passes through and beyond the two spheres A and B, then > wcos30 u 3u 3 ( 3 e) > ( + e) e> 3+ 3e e<. 9 () 0