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Review - Explain this using one of Newton's Laws

A lead brick has a lot of mass
and therefore a lot of inertia.

The greater the mass, the greater the tendency for an object to remain at rest, the smaller the acceleration that is produced by a given force.

Momentum Word Problems

 J = Ft = mΔv = Δp

 Ex 1)  A 5.0 kg mass has its velocity change from 8.0 m/s east to 2.0 m/s east.  Find the objects change in momentum.

Ex 1)  A 5.0 kg mass has its velocity change from 8.0 m/s east to 2.0 m/s east.  Find the objects change in momentum.

m = 5.0 kg
Vi = 8.0 m/s East
Vf = 2.0 m/s East
Δp = ?

 J = Ft = mΔv =Δp

Δp = mΔV

= (5.0 kg)(2.0 m/s - 8.0 m/s)

= -30. kgm/s East

or +30. kg m/s West

 Ex 2) A 5.0 kg mass moving with a velocity of 8.0 m/s east has an impulse applied to it which causes its velocity to change to 20. m/s East.   Find Impulse:

Ex 2) A 5.0 kg mass moving with a vector of 8.0 m/s east has an impulse applied to it which causes its velocity to change to 20. m/s East.

Find Impulse:

m = 5.0 kg
Vi = 8.0 m/s East
Vf = 20. m/s East
J = ?

 J = Ft = mΔv =Δp

J = mΔv = (5.0 kg)(12. m/s East)

= 60. kg m/s east

= 60. Ns East

Find the force if the impulse was applied for 3.0 sec.

F = ?

t = 3 seconds

m = 5.0 kg

Vi = 8.0 m/s East

Vf = 20. m/s East

J = 60. kg m/s east

 J = Ft = mΔv =Δp

J = Ft = 60. Ns East

F(3.0 sec) = 60. Ns East

F = 20. N East

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GolfBallhittingsteelsuperslomo.mp4

 Ex 3) How long would it take for a net upward force of 100. N, to increase the speed of a 50. kg object from 100. m/s to 150. m/s.

Ex 3) How long would it take for a net upward force of 100. N, to increase the speed of a 50. kg object from 100. m/s to 150. m/s.

F = 100. N

m = 50. kg

Vi = 100. m/s

Vf = 150. m/s

t = ?

 J = Ft = mΔv =Δp

FΔt  = mΔv

(100. N)t  =  50. kg(50. m/s)

t  =  25. secs

 Ex 4)  A 1.0 kg ball traveling @ 4.0 m/s strikes a wall and bounces straight back @ 2.0 m/s. Find Δp

Racquetball Slow-Mo - HiViz

Ex 4)  A 1.0 kg ball traveling @ 4.0  m/s strikes a wall and bounces straight back @ 2.0 m/s

Find Δp

m = 1.0 kg

Vi = 4.0 m/s

Vf = ?

Vf = -2.0 m/s

(opposite direction)

Δp = ?

 J = Ft = mΔv =Δp

(a) Δp = mΔv

= (1.0 kg)(-2.0 m/s - 4.0 m/s)

= - 6.0 kgm/s

(b) What is impulse applied to the ball?

J =  Δp = -6.0 kgm/s

(c) What is impulse applied to the wall?

J = +6.0 kgm/s

3rd Law, Action Reaction

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AP Physics

p. 188) 17

Dp = mDv

Momentum only changes in the x direction

toward wall +

Vi = Vsinq

Vf = -Vsinq

Dp = mDv

Dp = m(-2Vsinq)

= 2.1 kgm/s left