Linear Momentum: Collisions

Initially, since both balls are moving on the x-axis, the y component of the momentum is zero. Since momentum is conserved, we can state that the momentum of each ball must be equal and opposite after the collision, as they move along the y-axis. Since both masses are equal, the velocity of each ball must be equal and opposite.

No, the second ball must also leave the collision at an angle. The first ball has a component of linear momentum in the y direction after the collision, given by v _1fsinθ . Since both balls were traveling in the x direction before the collision, there was no initial momentum in the y direction. Thus, for momentum to be conserved, the second ball must travel in the negative y direction, to counteract the momentum of the first ball. If the second ball remained stationary, momentum would not be conserved.

The collision is completely inelastic, and we have two variables, v _f and θ , and the two equations of conservation of linear momentum. We start by relating momentum before and after the collision in the x direction:

implying that

Now equating the y components,

Implying that

We have two independent equations for v _f and θ If we divide the second one by the first, v _f will cancel, and we will be left with an expression for θ only:

Thus

And θ = 71.6 ^o . Substituting this in to find v _f , we find that:

Thus the two stuck together objects have a final velocity of 2.11 m/s directed 71.6 ^o above horizontal.

To solve this problem we begin with our familiar momentum equations for both x and y components. Since we only have two variables ( v ₁ and θ ) we need not generate a third equation from conservation of kinetic energy. Thus we equate the x and y components of linear momentum before and after the collision:

Here we have two equations relating θ and v ₁ . To solve we can simply substitute our expression for v ₁ in terms of θ into our first equation:

Thus the pool cue will be deflected about 30 degrees from horizontal.