Conservation of momentum and skaters
A very good example of the
conservation of momentum in an 'explosion' is given by a pair of skaters. They stand facing
each other and push themselves apart, the same force acts on each skater. If their masses
are equal they both recoil with the same (but opposite) velocity.
Let the mass of the girl and her twin sister be m
and their velocities after separation be v
1 and v
2.
Momentum before
separation = 0
Momentum after impact = mv
1 + mv
2By the law of
conservation of momentum:
Momentum before they push apart = 0 = momentum after
they push apart = mv
1 + mv
2Therefore: 0 = mv
1 + mv
2
and so mv
1 = - mv
2 v
1 = - v
2However if they are
of unequal mass the person with the greater mass will move off much more slowly after the
initial push.
Let the mass
of the girl be m, the mass of the man M and their velocities after separation be v and
V.
Momentum before separation = 0
Momentum after impact = mv
1 +
MV
By the law of conservation of momentum:
Momentum before they push apart = 0
= momentum after they push apart = mv + MV
Therefore: 0 = mv + MV and so mv = -
MV v = - [M/m]V
If the mass of the man is double that of the girl she will 'recoil' at
twice his speed.
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