Again we must use the concept of conservation of momentum. The prob-
lem states that the first two fragments split into orthogonal directions, so we
are free to define the positive ˆ
x
direction as the direction of the fragment
of mass 3
m
and the positive ˆ
y
direction as the direction of the fragment of
mass 4
m
. Another key concept is the conservation of mass. The system had
a total mass of 8
m
prior to the explosion, thus it must have a total mass of
8
m
after the collision. Therefore, the mass of the third fragment must be
m
so that the three fragments have a total mass of 8
m
. The total momentum
of the system before the explosion,
~
p
0
, is 0.
~
p
0
8
m~v
8
m
0
0
.
This is a vector equation, so a 0 means 0 in each direction. So we must
have that the momentum of the fragments after the collision also sum to 0 in
each direction. It is often useful to break up problems about momenta into
components. Summing the momenta in the ˆ
x
direction:
p
f,system,x
0
p
f,
3
m,x
p
f,
4
m,x
p
f,m,x
3
mv
0
mv
f,m,x
.
This gives us the value of the third fragments velocity in the ˆ
x
direction
in terms of
v
,
v
f,m,x
3
v
. Similarly for the ˆ
y
component of the momentum:
p
f,system,y
0
p
f,
3
m,y
p
f,
4
m,y
p
f,m,y
0
4
mv
mv
f,m,y
.
This gives us the value of the third fragments velocity in the ˆ
y
direction
in terms of
v
,
v
f,m,y
4
v
. The speed of a particle is defined as the length
of the velocity vector, so using the equation for the magnitude of a vector:
~v
f,m
v
2
f,m,y
v
2
f,m,x
4
v
2
3
v
2
5
v.
This is choice a.
Remember, this was a special case where the total
momenta in each direction was 0. However, if for instance, the initial mo-
menta of the system were 3
mv
ˆ
x
, i.e.
~
p
0
3
mv
ˆ
x
, then we would have set
p
f,system,y
0, and
p
f,system,x
3
mv
.
8