Problem: If an object of mass 9 kg is moving at 3 m/s, on a collision course with a 6 kg object moving at 6 m/s in the opposite direction, and if the objects stick together after collision, then will they go off in the direction of the first (enter 1) or the second (enter 2)?
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
Solution: The collision effectiveness of an object is proportional to its mass, and also to its velocity. It is therefore proportional to the product of its mass and velocity. So we define the momentum of an object to be its collision effectiveness, which is the product of its mass and its velocity. Object 1 thus has momentum ( 9 kg)( 3 m/s) = 27 kg m/s, while object 2 has momentum ( 6 kg)( 6 m/s) = 36 kg m/s in the opposite direction. Thus object 2 has the dominant momentum, or collision effectiveness, and motion after collision will be in its direction.
The sum of the momenta will be
total momentum = pTot = m1 v1 + m2 v2.
This sum will be either positive or negative, depending on whether m1 v1 has greater or lesser magnitude than m2 v2. The total momentum will not change during collision, since the objects exert equal and opposite forces on one another and therefore by the impulse-momentum theorem experience equal and opposite momentum changes.
If the sum of the momenta is positive, then the total momentum is positive and will continue to be so after the collision, so the pair will go off in the direction of m1. If the sum is negative, the total momentum is and will remain negative and be pair will go off in the negative direction--that of m2.
.
.
.
.
.
.
.
.
.
.