1 / 6

Conservation of Momentum Momentum before a collision must Equal

Conservation of Momentum Momentum before a collision must Equal Momentum after a collision. Before = After m 1 v i + m 2 v i = m 1 v f + m 2 v f. Remember Momentum (p) is a vector so “+” & “-” for direction need to be used before & after the collision

howe
Télécharger la présentation

Conservation of Momentum Momentum before a collision must Equal

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Conservation of Momentum • Momentum before a collision • must Equal • Momentum after a collision. Before = After m1vi + m2vi = m1vf + m2vf

  2. Remember Momentum (p) is a vector so “+” & “-” for direction need to be used before & after the collision (A negative “-” means opposite direction) Sample 6D pg 218 *Elastic Collision*  Figure 6-8 Elastic collision m1vf + m2vf (After) m1vi + m2vi (Before) *m2vf decreases while m1vf increases

  3. Inelastic Collisions m1vi + m2vi = (m1 + m2) Vf Sample Problem 6E pg 223

  4. Kinetic Energy is not conserved in inelastic collisions (1/2 mv2) Why? Sample 6F, pg 225 * KE Calculations K.E. is constant in an Elastic Collision. [Figure 6-12] pg 227 Sample 6G pg 228 * Notice KE Calculations Table 6-2 pg 230

  5. HMWK #1Test #7 Chp. 6 D-G BK Chp. 6 D-F Pg. 219 6D 1,3 Pg. 224 6E 3,5 Pg. 226 6F 2,3 #2 a.6.2 m/s S b. 3 J

  6. HMWK #2Test #7 Chp. 6 D-G WKBK: 6-D: 3. Vi= .4 m/s forward 4. Vi= .54 m/s backward 6-E: 1. m2= 270 kg 2. m2= 4.97 kg 8. Vf= .69 m/s Up 9. Vf = .456 m/s left 6F: 2. m2= 4.4 x 102kg ΔKE= -3280 J 3. Vf= 2 m/s right ΔKE = -5450 J 6G: 2. Vi= 5m/s right KE = KE 932 J = 933 J 3. Vi= 2 m/s left 29m/s2 = 29m/s2

More Related