1.3b Explosions and Newton’s third law
Explosions
Explosions are treated in the same way as collisions, in that total momentum is
conserved. For example, in the case of a bullet being fired from a gun, the total
momentum before firing is zero, since nothing is moving.
After firing, the bullet has momentum in the forward direction. The gun must
therefore have the same magnitude of momentum in the opposite direction so the
two momenta cancel each other out, leaving the total momentum still equal to
zero. For this reason the gun must have a recoil velocity after the explosion (i.e.
the gun ‘jumps’ backwards).
It should be obvious that in an explosion kinetic energy is not conserved.
Think about this. If a bomb explodes leading to an overall gain in kinet ic energy.
What kind of energy does the bomb have before the explosion?
Worked example
A gun of mass 1 kg fires a bullet of mass 5 g at a speed of 100 m s –1 . Calculate
the recoil velocity of the gun.
Before
After
0 m s–1
1 kg
0.005 kg
?
100 m s–1
0.005 kg
1 kg
Take motion  as +
momentum = 0
m1v1 = 1 × v1
m 2 v 2 = 0.005 × –100
= –0.5
total momentum before = total momentum after
OUR DYNAMIC UNIVERSE (H, PHYSICS)
© Learning and Teaching Scotland 2011
1
0 = (1 × v 1 ) – 0.5
v 1 = 0.5
ie v 1 = 0.5 m s –1 in the opposite direction to the
bullet
2
OUR DYNAMIC UNIVERSE (H, PHYSICS)
© Learning and Teaching Scotland 2011
Momentum and Newton’s third law
It can be shown that conservation of momentum and Newton’s third law mean
the same thing.
Starting from conservation of momentum:
total momentum before
Rearrange:
or:
ie
= total momentum after
m1u1 + m2u2
= m1v1 + m2v2
m2u2 – m2v2
= m1v1 – m1u1
m 2 (u 2 – v 2 )
= m 1 (v 1 – u 1 )
–m 2 (v 2 – u 2 )
= m 1 (v 1 – u 1 )
–(change in momentum of
object 2)
= (change in momentum of
object 1)
In other words, in any particular example involving two objects colliding, if the
momentum of one object increases by, for example, 6, then the momentum of the
other object must decrease by 6.
Consider again the example given above. We had concluded that:
–m 2 (v 2 – u 2 ) = m 1 (v 1 – u 1 )
Applying the impulse relationship:
–F 2 t = F 1 t
so
–F 2 = F 1
Newton’s third law states that if one body exerts a force on a second body, the
second body exerts a force on the first body that is equal in size and opposite in
direction.
Note: These forces operate on different bodies so they do not cancel each other
out.
OUR DYNAMIC UNIVERSE (H, PHYSICS)
© Learning and Teaching Scotland 2011
3