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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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