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

𝑚𝐴

𝑣𝐵=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

𝑚𝐴

𝑣𝐵=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.

0 0.05 0.1 0.15 0.2 0.25 0.3 0.350

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