Momentum and impulse Book page 73-79

Transcription

1 Momentum and impulse Book page 73-79

2 Definition The rate of change of linear momentum is directly proportional to the resultant force acting upon it and takes place in the direction of the resultant force

3 p=mv and the change in momentum v=m v From Newton s 2nd law F_net=ma and a=(v u)/ t we get the expression F=(m(v u))/ t=m v/ t= p/ t The longer the time of collision, the smaller the force exerted F t = m v = p This quantity is called Impulse J Momentum is a vector quantity

4 Example A ball of mass 0.25kg is moving to the right at a speed of 7.4ms 1. It strikes a wall at 90 0 and rebounds from the wall with a speed of 5.8ms 1. Calculate the change in momentum. V = - 7.4ms ms 1 Solution p = m v 2 v 1 p = 0.25 ( 5.8 (- 7.4)) = 3.3kgms 1 [left]

5 Graphical interpretation Real graph Idealized graph The area under the graph = F t= Impulse If we know the impulse we can find the change in speed Impulse J = m v v = J m

6 Example Water is poured from a height of 0.50 m on to a top pan balance at the rate of 30 litres per minute. Estimate the reading on the scale of the balance. Solution h = 0.50m rate = 30l /min = 30kg / min = 0.6 kg / sec Assumption: water bounces off the top of the balance horizontally mgh = 1 2 mv2 OR v = 2gh = = 3.2ms 1 p = m v = = 1.6N Reading on scale = m = W g = = 0.160kg v 2 = u 2 + 2as v = 2as = 2x = 3.2ms 1

7 Example KE = 1 2 mv2 = m m v2 2m KE = p2 2m The graph right shows how the momentum of an object of mass 40kg varies with time A) calculate the force acting on the object B) The change in KE over the 10s of motion Solution Gradient = p t = F = = 20N KE = p2 2m = = 500J Can you think of examples where theory of impulse has been used to good effect?

8 Law of conservation of momentum In a closed system, linear momentum is always conserved Closed system: no external forces are acting If the net force of a system is zero, then there is no change of momentum of the system Momentum before = momentum after The momentum gained by one object is equal to the momentum lost by another object

9 3 rd law: F AB = F BA 2 nd law: m A a A = m B a B m(v A u A ) t = m(v B u B ) t Since time is the same for both m A v A m A u A = m B u B m B v B m B u B + m A u A = m A v A + m B v B Momentum before = momentum after This approach cannot always be used

10 Example Sand is poured vertically at a constant rate of 400 kg s -1 on to a horizontal conveyor belt that is moving with constant speed of 2.0 m s -1. Find the minimum power required to keep the conveyor belt moving with constant speed. Solution P = Fv = 800 x 2 = 1600W Rate = 400 kg s -1 v belt = 2.0 m s -1 p = mv = = 800kgms 1 Force on belt = 800N = force of friction between sand and belt This force accelerates the sand to the speed of the surveyor belt

11 Momentum in collisions Completely inelastic Because they stick together, they must have equal velocity

12 Case 5: Head on collisions of two identical masses where one is at rest perfect elastic collision m 1 u = m 1 v 1 + m 2 v 2 KE is conserved: 1 2 m 1u = 1 2 m 1v m 2v 2 2 m 1 u 2 1 = m 1 v m 2 v 2 2 m 1 u 1 = m 1 v 1 + m 2 v 2 Rearrange m 1 u 2 1 m 1 v = m 2 v 2 and m 1 u 1 m 1 v 1 = m 2 v 2 m 1 u 2 1 v 2 1 = m 2 v 2 2 and m 1 (u 1 v 1 ) = m 2 v 2 Take ratios m 1 u 2 1 v 2 1 m 1 (u 1 v 1 ) = (u 1 v 1 )(u 1 +v 1 ) (u 1 v 1 ) = m 2v 2 2 m 2 v 2 v 2 = u 1 + v 1 and v 1 = v 2 u 1 Substitute back into m 1 u 1 m 1 v 1 = m 2 v 2

13 continued m 1 u 1 m 1 v 1 = m 2 v 2 v 2 = u 1 + v 1 and v 1 = v 2 u 1 m 1 u 1 m 1 v 1 = m 2 (u 1 + v 1 ) and m 1 u 1 m 1 (v 2 u 1 )= m 2 v 2 m 1 u 1 m 2 u 1 = m 1 v 1 + m 2 v 1 and 2 m 1 u 1 = m 1 v 2 + m 2 v 2 u 1 (m 1 m 2 ) = v 1 (m 1 + m 2 ) and 2 m 1 u 1 = v 2 (m 1 + m 2 ) v 1 = m 1 m 2 m 1 + m 2 u 1 v 2 = 2 m 1u 1 m 1 + m 2

14 Inelastic collisions energy lost Energy is lost 1 mass is stationary m 1 u 1 = (m 1 +m 2 )v 1 v 1 = m 1u 1 m 1 +m 2 Energy loss: E i = 1 2 m 1u 1 2 E f = 1 2 (m 1+m 2 )v 1 2 = 1 2 (m 1+m 2 ) Take ratio of energies m u 1 1 m 2 m 1 +m 2 = 1 u m 1 +m 1 2 KE initial 1 = 2 m 2 1u 1 KE final 1 m1 2 2m1+m2 u 1 = (m 1+m 2 ) m 1 If we know the mass, we can find the ratio of energies

15 Two initially stationary masses gain energy If the masses are equal, then the speeds are the same with one velocity the negative of the other If the masses are not equal, then m 1 m 2 = v 2 v 1 Momentum before Momentum after 0 m 1 v 1 + m 2 v 2 m 1 v 1 + m 2 v 2 = 0 m 1 v 1 = m 2 v 2

16 Example A railway truck B of mass 2000Kg is at rest on a horizontal track. Another track A of the same mass is moving with a speed of 5.0ms 1 collides with the stationary truck and they link up and move together. Find the speed with which the two trucks move off and the loss of KE on collision = 4000v v = 2.5ms 1 Momentum before KE before = 1 2 mv2 = = 25000J KE after = 1 2 (m 1 + m 2 )v 2 = = 12500J KE = = 12500J Momentum after 2000 x 0 (m 1 + m 2 ) v 2000 x 5 ( )v Energy lost - Dissipated to surroundings - Sound - Most heat up coupling b/w trucks

17 Example A billiard ball of mass 100g strikes the cushion of a billiard table with 10ms 1 at an angle of 45 0 to the cushion. It rebounds at the same speed and angle to the cushion. What is the change of momentum of the billiard ball? p = mv mu p is a vector need vector addition p = mv + ( mu) Reverse mu R = mv 2 + mu 2 = m 2 (v 2 + u 2 ) = = = 1.4kgms 1 At 90 0 away from the cushion R -mu mv mu mv 45 45