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Physics Tutorial 6.8 - Momentum and Impulse in Two Dimensions. Explosions. Revision Notes

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6.2Momentum and Impulse in Two Dimensions. Explosions.


In these revision notes for Momentum and Impulse in Two Dimensions. Explosions., we cover the following key points:

  • How to calculate momentum in two dimensions
  • The same for impulse in two dimensions
  • The meaning of explosions as the inverse process of inelastic collisions
  • How to deal with explosions in two dimensions

Momentum and Impulse in Two Dimensions. Explosions. Revision Notes

When objects are not moving according the line that connects their centres of mass but slightly diverted from it, the collision is not head-to-head. As a result, objects will move in more than one direction and the study of momentum and impulse for this kind of collision involves two directions: x- and y-direction as the movement of the two objects after the collision is a combination of these two basic directions.

The approach is the same as for momentum and impulse in one direction discussed earlier. The only difference is that we must write the known equations for each direction separately.

Thus, for elastic collisions we have:

x-direction Formula

m1 × v01(x) + m2 × v02(x) = m1 × v1(x) + m2 × v2(x)

y-direction Formula

m1 × v01(y) + m2 × v02(y) = m1 × v1(y) + m2 × v2(y)

and for inelastic collisions, we have

x-direction Formula

m1 × v01(x) + m2 × v02(x) = (m1 + m2) × vx

y-direction Formula

m1 × v01(y) + m2 × v02(y) = (m1 + m2) × vy

We can use the same approach as above for the impulse in two dimensions. Thus, since there is a single collision, there is also a single force exerted by an object to the other (and a single reaction force as well). However, this force extends in two directions and therefore, we must use two equations for the impulse - one for each direction.

We have:

J1 - 2 = F1 - 2 × ∆t = m1 × v01 - m1 × v1

and

J2-1 = F2-1 × ∆t = m2 × v02 - m2 × v2

Obviously, we need to calculate only one of the above impulses as J1 - 2 and J2-1 are equal and opposite. However, we must write it in components. For example, if we chose to work out the first impulse (J1 - 2), we must write

J1 - 2(x) = F1 - 2(x) × ∆t = m1 × v01(x) - m1 × v1(x)

and

J1 - 2(y) = F1 - 2(y) × ∆t = m1 × v01(y) - m1 × v1(y)

Explosions are the reverse phenomenon of inelastic collisions. Thus, in inelastic collisions the kinetic energy converts into other form after the impact, while in explosions, the energy stored in the object converts into kinetic energy, making it move faster than before. Therefore, the equation of explosions is the inverse of that used in inelastic collisions.

M × v0 = m1 × v1 + m2 × v2

where M is the mass of the original object, m1 and m2 are the masses of the respective pieces after the explosion, v0 is the initial velocity of the original object before the explosion, v1 and v2 are the velocities of the pieces after the explosion.

The simplest case of explosions is when the original object explodes and splits in two pieces, which move in opposite directions. In this case, we have only one dimension involved and therefore, a single equation to write (the equation of explosions written above).

In explosions in two dimensions, we use the same approach as for momentum and impulse in two dimensions, i.e. we write the equation of momentum conservation for each direction separately. We have

M × v0x = m1 × v1x + m2 × v2x + m3 × v3x + …
M × v0y = m1 × v1y + m2 × v2y + m3 × v3y + …

The total kinetic energy of a system after explosion is much gretaer than its initial kinetic energy when the object had not exploded yet. This means some of the stored (idle) energy of the object has been activated and it was converted into kinetic energy during the explosion. No break in the rules of energy conversion law occurs here.

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