Question
Elastic Collisions Momentum (4 PTS) Collision M red (kg) M blue (kg) v redinitial (m/s) v blueinitial (m/s) v redfinal (m/s) v bluefinal (m/s) P
Elastic Collisions Momentum (4 PTS)
| Collision | Mred (kg) | Mblue (kg) | vredinitial (m/s) | vblueinitial (m/s) | vredfinal (m/s) | vbluefinal (m/s) | Ptotal initial (kg/m/s) P=m.v | Ptotal final (kg/m/s) P=m.v |
| 1a | 1 | 2 | 10 | 2 | -0.7 | 7.30 | Pi=(1x10+2x2)=(10+4)=14 | Pf=(1.x-.7)+(2x7.3) =(-.7+14.6)=14 |
| 2a | 1 | 3 | 7 | -2 | -6.5 | 2.5 | Pf=(1x7)+(3x-2)=(7+-6)=1 | Pf=(1x-6.5)+(3x2.5) =(-6.5+7.5)=1 |
| 3a | 1 | 2 | 10 | 3 | 0.7 | 7.7 | 1*10+2*3=10+6=16 | 1*-.7+2*7.7=-.7+14.6 |
| 4a | 2 | 3 | 10 | -5 | -8 | 7 | 2*10+3*-5=20+(-15)=5 | 2*-8+3*7=-16+21=5 |
Elastic Collisions Kinetic Energy- (4 PTS)
| Collision | KEredinitial(J) (KE)=12mv2(J) | KEblueinitial(kg) | KEtotalinitial(J) | KEredfinal(J)
| KEbluefinal(J)
| KEtotalfinal(kg) |
| 1b | =1/2x1x 10^2 =50 | =1/2x2x2^2=4
| 50+4=54 | =1/2x1x(-.7)^2 =.245 | =1/2x2x(7.3)^2 =53.29 | =.245+53.29=53.54 |
| 2b | =1/2x1x7^2 =24.5 | =1/2x3x(-2)^2 =6 | =24.5+6 =30.5 | =1/2x1x(-6.5)^2 =21.125 | =1/2x3x(2.5)^2 =9.375 | =21.125+9.375 =30.5 |
| 3b | =*1*10^2 =50 | =*2*3^2 =9 | 50+9=59 | =*1*0.7^2 =0.245 | =*2*7.7^2 =59.29 | .245+59.29= 59.535 |
| 4b | =*2*10^2 =100 | =*3*(-5)^2 =-37.5 | 100+(-37.5)=62.5
| =*2*-8^2 =73.5 | =*3*7^2 =73.5 | 73.5+73.5 =147 |
Q1 Was momentum conserved in this collision? How do you know? (1 PTs)
Q2 Was Kinetic Energy conserved in this collision? How do you know? (1 PTs)
Inelastic Collisions Momentum (4 PTS)
| Collision | Mred (kg) | Mblue (kg) | vredinitial (m/s) | vblueinitial (m/s) | vredfinal (m/s) | vbluefinal (m/s) | Ptotal initial (kg/m/s) P=m.v | Ptotal final (kg/m/s) P=m.v |
| 5a | 1 | 1 | 10 | 0 | 5 | 5 | (1x10)+(1x0)=10+0 =10 | =(1+1)x5 2x5=10 |
| 6a | 1 | 3 | 8 | -5 | -1.7 | -1.7 | (1x8)+(3x-5) =8-15=-7 | (1+3)x-1.7 = -6.8 |
| 7a | 2 | 3 | 6 | 3 | 4.2 | 4.2 | (2*6)+(3*3)=12+9=21 | 8.4+12.6=21 |
| 8a | 2 | 3 | 10 | -5 | 1 | 1 | (2*10)+(3+(-5))=20-15=5 | 2+3=5 |
Inelastic Collisions Kinetic Energy (4 PTS)
| Collision | KEredinitial(J) (KE)=12mv2 | KEblueinitial(kg) | KEtotalinitial(kg) | KEtotalfinal(kg) |
| 5b | =1/2(1)x(10)^2=50 | =1/2x1x(0)^2=0 | 50+0=50 | =1/2 (1+1)x(5)^2 =25 |
| 6b | =1/2(1)x(8)^2=32 | . (3)x(-5)^2=37.5 | 32+37.5=69.5 | =1/2 x(1+3)x (-1.7)^2 =5.78 |
| 7b | (2)(6)^2=36 | (3)(3)^2+13.5 | (3)(4.2)^2=17.64 | (3)(4.2)^2=26.46 |
| 8b | (2)(10)^2=100 | (3)(-5)^2=37.5 | (2)(1)^2=1 | (3)(1)^2=1.5 |
Q4 . Was momentum conserved in this collision? How do you know? (1 PTs)
Q5 . Was Kinetic Energy conserved in this collision? How do you know? (1 PTs)
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Principles And Practice Of Physics
Authors: Eric Mazur
2nd Global Edition
1292364793, 978-1292364797
