Linear Impulse − Momentum KINE 3301 Biomechanics of Human Movement Linear Impulse − Momentum Chapter 8

Definitions Momentum: mass x velocity (units kg∙m/s) Conservation of Linear Momentum – The total linear momentum of a system of objects is constant if the net force acting on a system is zero. Elastic Collision: The objects collide and rebound. Inelastic Collision: The objects collide and stick together. Impulse (units N∙s) Constant force: Average force x time. Non-Constant force: Area under the force – time curve. Impulse – Momentum: The impulse is equal to the change in momentum.

Equations p = m v 𝑚 𝐴 𝑢 𝐴 + 𝑚 𝐵 𝑢 𝐵 = 𝑚 𝐴 𝑣 𝐴 + 𝑚 𝐵 𝑣 𝐵 Linear Momentum 𝑚 𝐴 𝑢 𝐴 + 𝑚 𝐵 𝑢 𝐵 = 𝑚 𝐴 𝑣 𝐴 + 𝑚 𝐵 𝑣 𝐵 Elastic Collisions 𝑣 𝐴 = 𝑚 𝐵 𝑢 𝐵 𝑚 𝐴 𝑣 𝐵 = 𝑚 𝐴 𝑢 𝐴 𝑚 𝐵 Inelastic Collisions 𝑚 𝐴 𝑢 𝐴 + 𝑚 𝐵 𝑢 𝐵 = (𝑚 𝐴 + 𝑚 𝐵 )𝑣 𝐽= 𝑡0 𝑡1 𝐹 𝑑𝑡 Impulse 𝐽= 𝐹 ∆𝑡 Impulse−Momentum 𝑡0 𝑡1 𝐹 𝑑𝑡=𝑚 𝑉 𝑓 −𝑚 𝑉 𝑖 𝐹 ∆𝑡 =𝑚 𝑉 𝑓 −𝑚 𝑉 𝑖

Linear Momentum p = m v p = (2 kg) (3 m/s) p = +6 kg∙m/s m = 2 kg The linear momentum (p) of an object is the product of it’s mass (m) and velocity (v). The units for linear momentum are kg∙m/s. m = 2 kg p = m v p = (2 kg) (3 m/s) p = +6 kg∙m/s v = 3 m/s p The vector for linear momentum points in the same direction as the velocity.

Conservation of Linear Momentum The total linear momentum of a system of objects is constant if the net force acting on a system is zero. The total linear momentum is defined by: 𝑚 𝐴 𝑢 𝐴 + 𝑚 𝐵 𝑢 𝐵 = 𝑚 𝐴 𝑣 𝐴 + 𝑚 𝐵 𝑣 𝐵 𝑢 is the initial velocity (before collision) 𝑣 is the final velocity (after collision) 𝑚 is the mass of the object

Collision Classifications Collisions are classified according to whether the kinetic energy changes during the collision. The two classifications are elastic and inelastic. In an elastic collision the total kinetic energy of the system is the same before and after the collision. In an a perfectly inelastic collision the total kinetic energy is still conserved but the two objects stick together and move with the same velocity.

Conservation of Linear Momentum 𝑚 𝐴 𝑢 𝐴 + 𝑚 𝐵 𝑢 𝐵 = 𝑚 𝐴 𝑣 𝐴 + 𝑚 𝐵 𝑣 𝐵 The equation above is usually rearranged for elastic and inelastic collisions as follows: Elastic Collisions 𝑣 𝐴 = 𝑚 𝐵 𝑢 𝐵 𝑚 𝐴 𝑣 𝐵 = 𝑚 𝐴 𝑢 𝐴 𝑚 𝐵 Inelastic Collisions 𝑚 𝐴 𝑢 𝐴 + 𝑚 𝐵 𝑢 𝐵 = (𝑚 𝐴 + 𝑚 𝐵 )𝑣

𝑣 𝐵 = 𝑚 𝐴 𝑢 𝐴 𝑚 𝐵 𝑣 𝐴 = 𝑚 𝐵 𝑢 𝐵 𝑚 𝐴 𝑣 𝐴 = (.3𝑘𝑔)(−2𝑚/𝑠) .8𝑘𝑔 Two billiard balls collide in a perfectly elastic collision. Ball A has a mass of 0.8 kg and an initial velocity (uA) of 3 m/s, ball B has a mass of 0.3 kg and an initial velocity (uB) of −2 m/s, determine the velocity of each ball after the collision. 𝑣 𝐵 = 𝑚 𝐴 𝑢 𝐴 𝑚 𝐵 𝑣 𝐴 = 𝑚 𝐵 𝑢 𝐵 𝑚 𝐴 𝑣 𝐴 = (.3𝑘𝑔)(−2𝑚/𝑠) .8𝑘𝑔 𝑣 𝐵 = (.8𝑘𝑔)(3𝑚/𝑠) .3𝑘𝑔 𝑣 𝐴 = −0.75𝑚/𝑠 𝑣 𝐵 =8.0 𝑚/𝑠

𝑚 𝐴 𝑢 𝐴 + 𝑚 𝐵 𝑢 𝐵 = (𝑚 𝐴 + 𝑚 𝐵 )𝑣 𝑣 = 2 m/s Two clay objects collide in an inelastic collision, object A has a mass of 0.8 kg and an initial velocity (uA) of 4 m/s, object B has a mass of 0.4 kg and an initial velocity (uB) of −2 m/s, determine the final velocity of A and B. 𝑚 𝐴 𝑢 𝐴 + 𝑚 𝐵 𝑢 𝐵 = (𝑚 𝐴 + 𝑚 𝐵 )𝑣 .8 𝑘𝑔 4 𝑚 𝑠 + .4 𝑘𝑔 −2 𝑚 𝑠 =(.8 𝑘𝑔+.4 𝑘𝑔)𝑣 𝑣 = 2 m/s

Computing Impulse 𝐽= 𝐹 ∆𝑡 𝐽= 𝑡0 𝑡1 𝐹 𝑑𝑡 Constant Force Impulse = Average Force x time Non−Constant Force 𝐽= 𝑡0 𝑡1 𝐹 𝑑𝑡 Impulse = area under force-time curve

Impulse Impulse (J) is defined as product of an average force ( 𝐹 ) and time (∆𝑡), or the area underneath the force time graph. The units for impulse are N∙s. 𝐽= 𝐹 ∆𝑡 𝐽= 𝑡0 𝑡1 𝐹 𝑑𝑡

Computing Impulse using Average Force Compute the Impulse (J) for the force shown below with an average force ( 𝐹 ) 95.6 N and time (∆𝑡) of 0.217 s. 𝐽= 𝐹 ∆𝑡 𝐽= 95.6𝑁 .217𝑠 𝐽=20.76 𝑁∙𝑠

Impulse-Momentum 𝑎= 𝑉 𝑓 − 𝑉 𝑖 Δ𝑡 𝐹=𝑚𝑎 𝐹=𝑚 𝑉 𝑓 − 𝑉 𝑖 Δ𝑡 The impulse – momentum relationship is derived from Newton’s law of acceleration. 𝑎= 𝑉 𝑓 − 𝑉 𝑖 Δ𝑡 𝐹=𝑚𝑎 𝐹=𝑚 𝑉 𝑓 − 𝑉 𝑖 Δ𝑡 𝐹 ∆𝑡 =𝑚 𝑉 𝑓 −𝑚 𝑉 𝑖 Impulse = change in momentum

A soccer player imparts the force shown below on a soccer ball with a mass of 0.43 kg and an initial velocity (Vi) of 0.0 m/s. After the force was applied the ball had a final velocity (Vf) of 23.02 m/s. The average force F of 90.8 N was applied for 0.109 s. Compute the impulse using both average force and the change in momentum.

A softball player imparts the force shown below on a softball with a mass of 0.198 kg and an initial velocity (Vi) of 0.0 m/s. After the force was applied the ball had a final velocity (Vf) of 50.51 m/s. The average force F of 31.74 N was applied for 0.315 s. Compute the impulse using both average force and the change in momentum.