Transcript Example - UGA Physics and Astronomy

Example
A 45-kg swimmer runs with a horizontal velocity of
+5.1 m/s off of a boat dock into a stationary 12-kg
rubber raft. Find the velocity that the swimmer and
raft would have after impact, if there were no
friction and resistance due to the water.
Solution:
Given: m1 = 45 kg, m2 = 12 kg,
Find:


m
vo1  5.1 s , vo2  0
 
v f 1, v f 2  ?
 Consider motion of boy and raft just before and
just after impact
 Boy and raft define the system
 Neglect friction and air resistance  no external
forces (which act in the direction of motion)
 Therefore, we can use the Conservation of
Linear Momentum

v o1
 

Po  Pf




po1  po 2  p f 1  p f 2




m1v o1  m2 v o 2  m1v f 1  m2 v f 2
 Since, the boy moves with the raft after the
impact
f1
f2
f


v v

v
m1v o1  (m1  m2 ) v f
m1
45
vf 
v o1 
5.1
(m1  m2 )
(45  12)
 4.0 ms



m
v f  4.0 s  v f 1  v f 2
 What if we have the case where vf1  vf2 ? We
then have two unknowns. So, we need another
equation.
 This is the situation discussed in example 7
 We can use Conservation of Mechanical Energy.
No non-conservative forces. No change in y – so
Eo  E f
only KE.
1
2
m v  m v  m2 v
2
1 o1
1
2
2
1 f1
1
2
2
f2
 From original conservation of momentum
equation, solve for vf2. Then substitute into
conservation of energy equation.
vf2
m1

( v o1  v f 1 )
m2
2

m1
2
2
2
1
1
1
 2 ( v o1  v f 1 ) 
2 m1 v o1  2 m1v f 1  2 m2 
 m2

2
1
m
2
m1 (v  v ) 
( v o1  v f 1 )
m2
1
Here is a trick!
2 m1 ( v o1  v f 1 )(v o1  v f 1 )
2
1 m1
2
( v o1  v f 1 )(v o1  v f 1 )
m2
m1
v o1  v f 1 
( v o1  v f 1 )
m2
m2 v o1  m2 v f 1  m1v o1  m1v f 1
v f 1 (m1  m2 )  v o1 (m1  m2 )
 m1  m2 
 v o1
v f 1  
Eq. (7.8a)
 m1  m2 
1
2
2
o1
2
f1
1
2
m1
vf2 
( v o1  v f 1 )
m2
m1
m1  m2
vf2 
( v o1 
v o1 )
m2
m1  m2
 m1 (m1  m2 )  m1 (m1  m2 ) 
 v o1
 
m2 (m1  m2 )


 m1m1  m1m2  m1m1  m1m2 
 v o1
 
m2 (m1  m2 )


 2m1m2 
2m1
 v o1 
 
v o1  v f 2
m1  m2
 m2 (m1  m2 ) 
Eq.
(7.8b)
Use numerical data from example
 45  12 
m
v f1  
5
.
1

2
.
95

s
 45  12 
 2 * 45 
m
vf2 
5
.
1

8
.
1

s
 45  12 
p f 1  m1v f 1  (45)(2.95)  133 kg ms
m
p f 2  m2 v f 2  (12)(8.1)  97 kg s
po1  m1v o1  (45)(5.1)  230 kg ms
Momentum is conserved!
Collisions in 2D
 Start with Conservation of Linear Momentum
vector equation
 Similar to Newton’s 2nd Law
 problems, break into
x- and y-components P  P
Pox  Pfx ,
o
f
Poy  Pfy
Example – Problem 7.34
Three guns are aimed at the center of a circle. They
are mounted on the circle, 120° apart. They fire in a
timed sequence, such that the three bullets collide at
the center and mash into a stationary lump.
Two of the bullets have identical masses of 4.50 g
each and speeds of v1 and v2. The third bullet has
a mass of 2.50 g and a speed of 575 m/s. Find the
unknown speeds.
Solution:
Given: m1 = m2 = 4.50 g, m3 = 2.50 g,
vo3 = 575 m/s, vf1 = vf2 = vf3 = 0
Find: vo1 and vo2
Method: If we neglect air resistance  then there
are no external forces (in the horizontal x-y plane;
gravity acts in the vertical direction)  we can use
Conservation of Linear Momentum



y
Po  Pf
m1 , vo1
Poy  Pfy

120
60
 m1v o1 sin 60
x


120
 m2 v o 2 sin 60  0
m3 , vo3
60
v o1  v o 2

m2 , vo 2
Pox  Pfx


m1v o1 cos60  m2 v o 2 cos60  m3 v o 3  0
1
1
m1v o1 ( 2 )  m1v o1 ( 2 )  m3 v o 3
m1v o1  m3 v o 3
m3
2 .5
v o1 
vo3 
575  319 ms  v o 2
m1
4 .5