Momentum and Collisions - Mission MC8 Detailed Help

In a Physics lab, a 1.8-kg brick is dropped from rest upon a 4.6-kg cart moving east with a speed of 52 cm/s. The brick and cart move together after the collision. Assuming the system is isolated, fill in the momentum table and determine the final velocity of the brick and cart. Use the notation that east is the positive direction and west is the negative direction.
(Note: Your numbers are selected at random and likely different from the numbers listed here.)

The momentum (p) of an object can be calculated from knowledge of its mass (m) and velocity (v) using the formula:    
p = m • v

The ultimate goal of this analysis is to determine the post-collision velocity of the cart and the brick. The following strategy will prove effective.
  • Since the mass and the before-collision velocity of both the cart and the brick is given, the momentum values can be calculated (see Formula Frenzy section). Of course, brick is at rest, so its momentum is 0 units.
  • The total momentum of the system is simply the sum of the individual momentum values of the two objects. Once individual values have been calculated, the total momentum of the system before the collision can be determined.
  • The post-collision velocities of the cart and the brick are not known. The momentum of each individual object must be written in terms of the unknown velocity v. Since the brick is riding on top of the cart after the collision, they move with the same velocity; so the same variable v can be used for each object's velocity. For instance, for a 4.6-kg cart, the momentum can be represented as 4.6•v; and for a 1.8-kg brick, the momentum can be represented as 1.8•v. Together, the total system momentum after the collision would be 4.6•v + 1.8•v or 6.4•v.
  • Since the system is said to be isolated (see Define Help section), the post-collision momentum of the system is the same as the pre-collision momentum. The post-collision momentum is not known as an actual value; it is represented in terms of the unknown velocity v. Set the expression for momentum in terms of v equal to the pre-collision momentum. Then solve for the variable v. This is the post-collision velocity of both the cart and the brick.
  • Now that v is known, the post-collision momentum of each individual object can be calculated and entered into the cells of the table. The total momentum for the system can be calculated as well. This total system momentum should be the same as before the collision. If not, you will need to recheck your work.

Isolated System and External Forces:
A system of two colliding objects is considered to be an isolated system if the only momentum-changing forces exerted during the collision are the forces between the two objects themselves. If a third object exerts a force capable of changing one or both of the object's momentum during the collision, then this is considered an external force and the system is not isolated.


Tired of Ads?
Go ad-free for 1 year