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Wize University Physics Textbook (Master) > Momentum, Impulse and Collisions

Shell Explosion

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Two-Dimensional Linear Momentum Conservation Next Section
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Momentum Conservation in the Center of Mass Frame


When looking at more complicated situations it is useful to consider the center of mass of the system.


Wize Concept
No matter how many objects or pieces we have before and after a collision (or explosion), the motion of the center of mass depends only on the net external force acting on the entire system.


  • The center of mass behaves as though it were a single particle of combined mass M\bcth M located at an imaginary point, where that net external force would be applied.
  • The location of the center of mass is where the mass of the entire system would be concentrated if you wanted to replace all other masses by just one big mass.

Exam Tip
You need to determine whether there is an external force acting on the system, and in which direction.

  • If there is no external force, momentum is conserved in that direction. The center of mass will be either at rest or it will move with a constant velocity.

  • If there is an external force, there will be an impulse on the center of mass in that direction.

Newton's Second Law applies as usual:
 ∑F=Macm \boxed{ \ \sum F=Ma_{cm}\ }

where acma_{cm} is the acceleration of the center of mass.

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Shell explosion


The shell explosion problem is a special case where going to the center of mass frame is really helpful. A shell is traveling through the air (projectile motion) and suddenly splits up into a few pieces which continue to fly, but along different trajectories than the original piece.

Let's look at the forces that act on the objects in our system:

  • In the horizontal direction there is no force at all, which means that the center of mass will move with a constant velocity.

  • In the vertical direction there is the force of gravity, which means that we will have an impulse acting on the center of mass.
Extra Practice
Practice: Fireworks
A firework explodes and forms three fragments of equal mass. One fragment travels directly upward at 1414 m/s and the second fragment moves at 2525 m/s at 25°25\degree below the horizontal to the right. Determine the velocity of the third fragment immediately after the explosion.
Practice: Firework
A fireworks explodes and forms three fragments of equal mass. One fragment travels directly upward at 14 m/s and the second fragment moves at 25m/s at 25o below the horizontal to the right. Determine the velocity of the third fragment immediately after the explosion.
Kinematics: Momentum in shell explosion
A fireworks explodes and forms three fragments of equal mass. One fragment travels directly upward at 14 m/s and the second fragment moves at 25m/s at 25o below the horizontal to the right. Determine the speed of the third fragment immediately after the explosion. Enter the answer in m/s.
A 1.0kg explosive is at rest when it suddenly goes off. It breakes into 2 pieces. A 0.30kg piece goes north at 10.0m/s. What is the magnitude of the momentum of the other piece?
A fireworks explodes and forms three fragments of equal mass. One fragment travels directly upward at 14 m/s and the second fragment moves at 25m/s at 25o below the horizontal to the right. Determine the velocity of the third fragment immediately after the explosion.
A radioactive atom at rest with a mass of 100×10−27 kg100\times 10^{-27}~kg is decaying to three particles with masses of 25×10−27 kg25\times 10^{-27}~kg , 40×10−27 kg40\times 10^{-27}~kg and 35×10−27 kg35\times 10^{-27}~kg respectively. Assume that all three particles remain in x-y plane. If the first particle is moving in the direction of positive y axis at speed of 20m/s and the second particle moves in the direction of 25 degree clockwise respect to positive x-axis at speed of 15 m/s, what is the angle and speed of the of the third particle?