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CHAPTER CHAPTER 99 LINEAR LINEAR MOMENTUM MOMENTUM ,, IMPULSE IMPULSE AND AND COLLISIONS COLLISIONS 103 Phys 1 9.1 Linear Momentum The principle of energy conservation can be used to solve problems that are harder to solve just using Newton’s laws. It is used to describe motion of an object or a system of objects. A new concept of linear momentum can also be used to solve physical problems, especially the problems involving collisions of objects. Linear momentum of an object whose mass is m and is moving at a velocity of v is defined as ur r p = mv 103 Phys 2 Page 1 Consider two particles interact with each other By Newton’s 3rd law: r r F21 = − F12 r r F21 + F12 = 0 r r m 1a1 + m 2a 2 = 0 r r dv dv m1 1 + m 2 2 = 0 dt dt r r d (m 1v 1 ) d (m 2v 2 ) + =0 dt dt r r d (m 1v 1 + m 2v 2 ) =0 dt ur r Linear momentum p ≡ mv ( d p1 + p 2 dt ) =0 r r r r i.e., Total momentum p tot = ∑ p = p 1 + p 2 remains constant 103 Phys 3 What can you tell from this definition about momentum? Momentum is a vector quantity. The heavier the object the higher the momentum The higher the velocity the higher the momentum Its unit is kg.m/s What else can use see from the definition? Do you see force? 103 Phys 4 Page 2 ur r p = mv (kg·m/sec.) p x = mv x , p y = mv y , p z = mv z VECTOR Relationship to force: dp d (mv) = = m d v + v dm dt dt dt dt = ma Mass usually doesn’t change with time = ∑F Improved form of 2nd Law: r d pr F ∑ = dt Probably was Newton’s original form of 2nd Law. Works even when mass does change, e.g., rockets. 103 Phys 5 Momentum is conserved in the absence of a net external force. r r r p tot = p 1 + p 2 : Constant ⇒ Conservation of Total Momentum r r r r ∴ p 1 i + p 2 i = p 1f + p 2 f 103 Phys 6 Page 3 Momentum Conservation r r ∆P FEXT = ∆t r ∆P =0 ∆t r FEXT = 0 ¾ The concept of momentum conservation is one of the most fundamental principles in physics. ¾ This is a component (vector) equation. - We can apply it to any direction in which there is no external force applied. ¾ You will see that we often have momentum conservation even when energy is not conserved. 103 Phys 7 Example 9.1 An archer standing at rest on frictionless ice. He fires an arrow horizontally. m1 = 60 kg , m2 = 0.5 kg , v2 = 50 m/s What is the velocity of the archer after firing the arrow? r r m 1v 1f + m 2v 2f = 0 m r r v 1f = − 2 v 2f = − 0.42 m/s m1 If the arrow were shot at an angle θ with theh horizontal, what is the recoil velocity m 1v 1f + m 2v 2f cos θ = 0 v 1f = − m2 v cos θ m 1 2f READ Examples 9.2 103 Phys 8 Page 4 Example: Figure Figure below below shows shows aa 2.0 2.0 kg kg toy toy race race car car before before and and after after taking taking aa turn turn on on aa track. track. Its Itsrspeed speed isis 0.50 0.50 m/s m/s before before the the turn turn and and 0.40 0.40 m/s m/s after after the the turn. turn. What What isis the the change change ∆ ∆ P in in the the linear linear momentum momentum of of the the car car due due to to the the turn? turn? r v i = − (0.50 m / s ) jˆ and r v f = (0.40 m / s ) iˆ r r Pi = M v i = (2.0 kg )( −0.50 m / s ) jˆ = ( −1 kg ⋅ m / s ) jˆ r r Pf = M v f = (2.0 kg )(0.40 m / s ) iˆ = (0.80 kg ⋅ m / s ) iˆ r r r ∆ P = Pf − Pi r ∆ P = (0.80 kg ⋅ m / s ) iˆ − ( −1.0kg ⋅ m / s ) jˆ = (0.8 iˆ + 1.0 jˆ ) kg ⋅ m / s 103 Phys 9 Example 9.2 A type of particle, neutral kaon (K0) decays (breaks up) into a pair of particles called pions (π+ and π-) that are oppositely charged but equal mass. Assuming K0 is initially produced at rest, prove that the two pions must have momenta that are equal in magnitude and opposite in direction. This reaction can be written as K0 K0 → π + +π − Since this system consists of a K0 in the initial state which results in two pions in the final state, the momentum must be conserved. So we can write Since K0 is produced at rest its momentum is 0. pπ+ π+ π− pπ− p K 0 = pπ + + pπ − p K 0 = pπ + + pπ − = 0 pπ + = − pπ − Therefore, the two pions from this kaon decay have the momenta with same magnitude but in opposite direction. 103 Phys 10 Page 5 9.2 Impulse and Momentum The change of momentum in a given time interval vi vf F fk = 0 a = ∴ ∆t vf -vi v -vi , Q F = ma = m( f ) ∆t ∆t F ∆t = m v f − m v i = p f − p i = ∆ p Impulse = force times time=change in momentum G G JG G G G G JG ∆ p mv − mv 0 m v − v 0 ∆v = = =m = ma = ∑ F ∆t ∆t ∆t ∆t ( ) r r r I = F ∆t = ∆ p 103 Phys 11 Impulse and Momentum A classic photograph taken by Arthur Edgerton at MIT in 1935 showing the moment of impact between bat and softball. The huge force exerted by the bat on the ball causes severe distortion of the ball as it is hit. 103 Phys 12 Page 6 Impulse and Momentum • Newton’s 2nd Law: r ∆ r F= ( p) ∆t r r ∆p = F ∆t r r r r ∆p = pf − pi = F ∆t A classic photograph taken by Arthur Edgerton at MIT in 1935 showing the moment of impact between bat and softball. The huge force exerted by the bat on the ball causes severe distortion of the ball as it is hit. r r I = F ∆t Impulse ÆImpulse-momentum theorem 103 Phys 13 Impulse and Momentum r tf r I = ∫ Fdt ti r r In general, F = F ( t ) r ÆImpulse = area under F, t curve r 1 tf r F = ∫ Fdt ∆t ti r r I = F∆t Simple case: constant Force r r I = F∆t 103 Phys 14 Page 7 Example; (a) Calculate the impulse experienced when a 70 kg person lands on firm ground after jumping from a height of 3.0 m. Then estimate the average force exerted on the person’s feet by the ground, if the landing is (b) stiff-legged and (c) with bent legs. In the former case, assume the body moves 1.0cm during the impact, and in the second case, when the legs are bent, about 50 cm. We don’t know the force. How do we do this? Obtain velocity of the person before striking the ground. K E = − ∆ PE 1 2 mv = − mg ( y − yi ) = mgyi 2 Solving the above for velocity v, we obtain v = 2 gyi = 2 ⋅ 9.8 ⋅ 3 = 7.7 m / s Then as the person strikes the ground, the momentum becomes 0 quickly giving the impulse I = F ∆t = ∆p = p f − pi = 0 − mv = = −70 kg ⋅ 7.7 m / s = −540 N ⋅ s 103 Phys 15 Example; cont’d In coming to rest, the body decelerates from 7.7m/s to 0m/s in a distance d=1.0cm=0.01m. 0 + vi 7.7 = = 3.8 m / s 2 2 d 0.01m The time period the collision lasts is = 2.6 × 10 − 3 s ∆t = = 3.8 m / s v I = F ∆t = 540 N ⋅ s Since the magnitude of impulse is I 540 The average force on the feet during = = 2.1× 105 N F= this landing is ∆t 2.6 × 10 −3 2 2 How large is this average force? Weight = 70 kg ⋅ 9.8m / s = 6.9 × 10 N The average speed during this period is v= F = 2.1× 105 N = 304 × 6.9 × 10 2 N = 304 × Weight If landed in stiff legged, the feet must sustain 300 times the body weight. The person will likely break his leg. 103 Phys 16 Page 8 Example ; cont’d What if the knees are bent in coming to rest? The body decelerates from 7.7 m/s to 0 m/s in a distance d=50 cm=0.5 m. The average speed during this period is still the same v = 0 + vi 7.7 = = 3.8 m / s 2 2 d 0.5 m = = 1.3 × 10 − 1 s 3.8 m / s v I = F ∆t = 540 N ⋅ s Since the magnitude of impulse is I 540 The average force on the feet during = = 4.1× 103 N F= −1 this landing is ∆t 1.3 × 10 2 2 How large is this average force? Weight = 70 kg ⋅ 9.8m / s = 6.9 × 10 N The time period the collision lasts changes to ∆t = F = 4.1× 103 N = 5.9 × 6.9 × 10 2 N = 5.9 × Weight It’s only 6 times the weight that the feet have to sustain! So by bending the knee you increase the time of collision, reducing the average force exerted on the knee, and will avoid injury! 103 Phys 17 Example 9.3 A golf ball of mass 50 g is struck by a club. The force exerted on the ball by the club varies from 0, at the instant before contact, up to some maximum value at which the ball is deformed and then back to 0 when the ball leaves the club. Assuming the ball travels 200m, estimate the magnitude of the impulse caused by the collision. vB θB m The range R of a projectile is v B2 sin 2θ B R = = 200m g Let’s assume that launch angle θi=45o. Then the speed becomes: Note the deformation of the ball due to the large force from the club v B = 200 × g = 1960 = 44m / s 103 Phys 18 Page 9 Considering the time interval for the collision, ti and tf , initial speed and the final speed are v i = 0 (immediately before the collision) v f = 44 m / s(immediately after the collision) Therefore the magnitude of the impulse on the ball due to the force of the club is ur I = ∆ p = m v Bf − m v Bi = 0 + 0.05 × 44 = 2.2 kg ⋅ m / s What is the average force on the ball during the collision with the club? F = I = 200 N ∆t 103 Phys 19 Page 10