Chapter 7 Page

 Mass in motion  Inertia in motion  It is a vector quantity 2

 Mass ( kg)  Velocity (m/s) 3

p = mv  p = momentum (kg∙m/s)  m = mass (kg)  v = velocity (m/s)  mass and velocity are directly related to momentum 4

7) m = 40. kg v = 2.5 m/s a.p = mv = (40. kg) (2.5 m/s) = 100 kg∙m/s = 1.0 x 10 2 kg∙m/s b.P = mv = (80. kg) (2.5 m/s) = 200 kg∙m/s = 2.0 x 10 2 kg∙m/s 5

8) m = 3.5 kg v = 1.5 m/s a.p = mv = (3.5 kg) (1.5 m/s) = 5.3 kg∙m/s b.p = mv = (3.5 kg) (0.75 m/s) = 2.7 kg∙m/s 6

9 ) All momenta are the same. a.p = mv = (60. kg) (4.0 m/s) = 240 kg∙m/s b.p = mv 240 kg∙m/s = (55 kg) v v = (240 kg∙m/s) / (55 kg) = 4.4 m/s c. p = mv 240 kg∙m/s = m (2.0 m/s) m = (240 kg∙m/s) / (2.0 m/s) = 120 kg 7

 The mosquito because it has a velocity. If the velocity is zero (tractor trailer), the momentum will be zero. 8

 Yes, if the elephant’s velocity is much less than the cheetah or if the elephant is not moving at all. 9

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 The amount of force applied at a certain amount of time.  Impulse is applied in the opposite direction of the motion to slow/stop an object  Impulse is applied in the same direction of the motion 11

I = ∆p  I = impulse (N∙s)  ∆p = change in momentum (kg∙m/s) Ft = m∆v  F = force (N)  t = time (s)  ∆v = change in velocity (m/s) 12

 In a collision, an object experiences a force for a specific amount of time that results in a change in momentum 13

 A tennis racket is moving east to apply an impulse of 120 N∙s on a tennis ball that is moving west. What would be the change in momentum of the tennis ball?  120 kg∙m/s east 14

 The velocity change is greatest in case B. The velocity changes from +30 m/s to -28 m/s. This is a change of 58 m/s (-) and is greater than in case A (-15 m/s). 15

 Change in momentum depends on change in velocity. Case B has a greater change in velocity. 16

 Because B has a greater change in velocity. 17

 Impulse is equal to the change in momentum. Case B has a greater change in momentum. 18

a. I = Ft = (50. N)(2.5 s) = 130 N∙s b. I = Ft = (25 N)(0.75 s) = 19 N∙s I = ∆p = 19 kg∙m/s 19

c. ∆p = m∆v = (15 kg) (-10. m/s) = -150 kg∙m/s ∆p = I = -150 N∙s I = Ft -150 N∙s = F (3.2 s) F = -150 N∙s / 3.2s = -47 N 20

d. I = Ft = (200. N)(12 s) = 2400 N∙s = ∆p = 2400 kg∙m/s ∆p = m∆v 2400 kg∙m/s = (25 kg) ∆v ∆v = (2400 kg∙m/s)/(25 kg) = 96 m/s 21

21) When you increase the time of collision, you decrease the force of collision. 22) The airbag increases the time of collision resulting in the decrease of the force of collision. 23) When you bend your knees, you increase the time of impact which decreases the force of impact. 22

24) Rebounding is when colliding objects bounce off each other. 25) The change in velocity is greater when a car rebounds compared to it crumpling upon impact. It is more beneficial for the car to crumple because less change in velocity means less change in momentum, which means less impulse on the car. 23