linear impulse − momentum
DESCRIPTION
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 . - PowerPoint PPT PresentationTRANSCRIPT
Linear Impulse − Momentum
Chapter 8
KINE 3301Biomechanics of Human Movement
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.
Equationsp = m v
𝑚𝐴𝑢𝐴+𝑚𝐵𝑢𝐵=𝑚𝐴𝑣𝐴+𝑚𝐵 𝑣𝐵
𝑣𝐴=𝑚𝐵𝑢𝐵
𝑚𝐴𝑣𝐵=
𝑚𝐴𝑢𝐴
𝑚𝐵
Elastic Collisions
𝐽=𝐹 ∆ 𝑡 𝐽=∫𝑡 0
𝑡 1
𝐹 𝑑𝑡
𝑚𝐴𝑢𝐴+𝑚𝐵𝑢𝐵=(𝑚¿¿ 𝐴+𝑚𝐵)𝑣¿Inelastic Collisions
𝐹 (∆ 𝑡 )=𝑚𝑉 𝑓 −𝑚𝑉 𝑖∫𝑡 0
𝑡1
𝐹 𝑑𝑡=𝑚𝑉 𝑓 −𝑚𝑉 𝑖
Impulse
Impulse−Momentum
Linear Momentum
Linear Momentum• 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 kgv = 3 m/sp = m v
p = (2 kg) (3 m/s)p = +6 kg 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:
𝑚𝐴𝑢𝐴+𝑚𝐵𝑢𝐵=𝑚𝐴𝑣𝐴+𝑚𝐵 𝑣𝐵
is the initial velocity (before collision) is the final velocity (after collision) 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 Momentum𝑚𝐴𝑢𝐴+𝑚𝐵𝑢𝐵=𝑚𝐴𝑣𝐴+𝑚𝐵 𝑣𝐵
𝑣𝐴=𝑚𝐵𝑢𝐵
𝑚𝐴𝑣𝐵=
𝑚𝐴𝑢𝐴
𝑚𝐵
𝑚𝐴𝑢𝐴+𝑚𝐵𝑢𝐵=(𝑚¿¿ 𝐴+𝑚𝐵)𝑣¿
The equation above is usually rearranged for elastic and inelastic collisions as follows:
Elastic Collisions
Inelastic Collisions
𝑣𝐴=𝑚𝐵𝑢𝐵
𝑚𝐴
𝑣𝐵=𝑚𝐴𝑢𝐴
𝑚𝐵
𝑣𝐴=(.3𝑘𝑔)(−2𝑚 /𝑠)
.8𝑘𝑔𝑣𝐵=
(.8𝑘𝑔)(3𝑚/ 𝑠).3𝑘𝑔
𝑣𝐵=8.0𝑚/𝑠
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.
𝑚𝐴𝑢𝐴+𝑚𝐵𝑢𝐵=(𝑚¿¿ 𝐴+𝑚𝐵)𝑣¿
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.
( .8𝑘𝑔 )(4𝑚𝑠 )+( .4 𝑘𝑔 )(−2𝑚𝑠 )=( .8𝑘𝑔+.4𝑘𝑔)𝑣
= m/s
Constant Force
𝐽=𝐹 ∆ 𝑡
𝐽=∫𝑡 0
𝑡 1
𝐹 𝑑𝑡
Non−Constant Force
Computing Impulse
Impulse = Average Force x time
Impulse = area under force-time curve
Impulse (J) is defined as product of an average force () and time (), or the area underneath the force time graph. The units for impulse are N s∙ .
Impulse
𝐽=𝐹 ∆ 𝑡
𝐽=∫𝑡 0
𝑡 1
𝐹 𝑑𝑡
Compute the Impulse (J) for the force shown below with an average force () 95.6 N and time () of 0.217 s.
Computing Impulse using Average Force
Impulse-MomentumThe impulse – momentum relationship is derived from Newton’s law of acceleration.
𝐹=𝑚𝑎 𝑎=𝑉 𝑓 −𝑉 𝑖
Δ𝑡
𝐹=𝑚𝑉 𝑓 −𝑉 𝑖
Δ𝑡
𝐹 (∆ 𝑡 )=𝑚𝑉 𝑓 −𝑚𝑉 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.
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