Example 1
A 250. N force is applied to a 30. kg object moving at 25 m to bring it to rest.
s
a. How long is the force applied?


Ft  p



Ft  p f  pi


Ft  mvi

mvi
t   
F
t  

30. kg 25 m
 250. N
t  3.0 s
s

Example 1
A 250. N force is applied to a 30. kg object moving at 25 m to bring it to rest.
s
b. How big of an average force would need to be applied if it lasted 6.0 s?
Notice: twice the time
requires only half the force!
 
 F t

t
F


Ft  p



Ft  p f  pi


Ft  mvi


mvi
F 
t

30. kg 25 m

s
F 
6.0 s

F  125 N

Example 2
A 0.45 kg baseball thrown horizontally at 40. m is hit by a bat such that the ball
s
leaves the bat traveling at 45 m at 45.
s
a. What impulse was given to the ball by the bat?


J  p
45 m
s
40. m
45
45 m
s
45

v fx

v fy
s




J y  p y
J x  px



J x  p fx  pix



J x  mv fx  mvix



J x  mv fx  vix 



J x  mv f cos 45  vi 

J x  0.45 kg 45 m cos 45   40. m
s
s

J x  32.3 Ns



Example 2
A 0.45 kg baseball thrown horizontally at 40. m is hit by a bat such that the ball
s
leaves the bat traveling at 45 m at 45.
s
a. What impulse was given to the ball by the bat?
45 m
s
40. m
45
45 m
s
45

v fx

v fy
s

Jy

Jy

Jy

Jy

Jy

Jy

 p y


 p fy  piy

 mv fy

 mv f sin 45


 0.45 kg 45 m sin 45
s
 14.3 Ns
Example 2
A 0.45 kg baseball thrown horizontally at 40. m is hit by a bat such that the ball
s
leaves the bat traveling at 45 m at 45.
s
a. What impulse was given to the ball by the bat?
45 m
s
40. m
45

J

32 Ns

J x  32 Ns
s

J y  14 Ns

J  32 Ns iˆ  14 Ns  ˆj

J  35 Ns at 24
14 Ns
Ns 
J  32.3 Ns   14.3 Ns    tan 

 32.3 Ns 
  24
J  35 Ns
2
2
1  14.3
Example 2
A 0.45 kg baseball thrown horizontally at 40. m is hit by a bat such that the ball
s
leaves the bat traveling at 45 m at 45.
s
b. What impulse was given to the bat by the ball?
45 m


Fbat
bat on
on ball
ball   Fball on bat
s
40. m
45
s
tbat on ball ttball
ballon
onbat
bat


 J bat on ball   J ball on bat

J   32 Ns iˆ   14 Ns  ˆj

204
J  35 Ns at 204
32 Ns
14 Ns

J

Newton’s
Third Law!