CK-12-Physics - Intermediate

http://www.ck12.org Chapter 7. Momentum

and the Conservation of Energy. It will be easier to solve this problem in general terms rather than using the given
values at first.

FIGURE 7.14

Using Conservation of Momentum:

pi=mcvci+mb( 0 )

pf=mcvc f+mbvb f

pi=pf→mcvci+mb( 0 ) =mcvc f+mbvb f

EQUATION 1→mcvci=mcvc f+mbvb f

Using Conservation of Energy:

KEi=

1

2

mcvci^2 +

1

2

mb( 0 )^2

KEf=

1

2

mcvc f^2 +

1

2

mbvb f^2

KEi=KEf→

1

2

mcvci^2 =

1

2

mcvc f^2 +

1

2

mbvb f^2 →mcvci^2 =mcvc f^2 +mbvb f^2

EQUATION 2→mcvci^2 =mcvc f^2 +mbvb f^2

Equation 2 can be re-expressed as:mc(vci−vc f)(vci+vc f) =mbvb f^2

Equation 1 can be re-expressed as:mc(vci−vc f) =mbvb f

By dividing Equation 2 on the left bymc(vci−vc f)and on the right bymbvb fwe arrive at:

vci+vc f=vb f→EQUATION 3

The velocitiesvc fandvb fcan be found by using Equation 3 and Equation 1.

vc f=

mc−mb

mc+mb

vci=

10. 0 kg− 4. 0 kg

10. 0 kg+ 4. 0 kg

(

7. 00

m

s

)

= 3. 00

m

s

vb f=

2 mc

mc+mb

vci=

2 ( 10. 0 kg)

10. 0 kg+ 4. 0 kg

(

7. 00

m

s

)

= 10. 0

m

s

Check Your Understanding

1a. Verify the equations found forvc fandvb fof Example 7.4.1