Explosion And Collision

[Isabella Kells]

Its worth pointing out how many equations and unknown variables we are dealing with when it comes to collision problems, because it is quite predictable and can give us some insight into how hard a particular problem might be before we get started on it. As discussed above, momentum is conserved in every collision, so

Since this is a vector equation, it actually contains a number of linear independent equations equal to the dimension of the problem (typically 1 or 2 for us, but generally 3). Since these linear equations can only be solved if there are an equal number of unknown variables and equations, we can only solve problems that have the same number of unknowns as dimensions (for example, a 1D problem can only ask one question - "what is the final velocity?" or "what was the mass of the first object?" - never "what was the final velocity AND the mass of the first object?"). This is the complete story for inelastic collisions - the number of unknowns has to match the dimension.

This is a scalar equation, and represents one further constraint on our system. However, that extra relationship means we can leave one further quantity unspecified - it is no longer free to be set, but must satisfy the extra equation from conservation of energy. This makes elastic collisions generally more complicated than inelastic problems, because we have an extra equation and unknown to deal with. To take the example from above, in 1D we have now two equations that govern our collision:

The first step to solving this is recognizing that there is an \(m\) in every term, so we can divide by that. Physically, that means the mass does not contribute to the physics at all - the solution will be the same no matter what the mass is. Solving this for the final velocity gets us

So, the first cart moves at one quarter the speed, no matter what it's initial mass is. (Note that although the mass does not matter, the relative sizes of the masses do matter. If their ratio was anything besides \(m_2/m_1=3\), we would get a different answer here. The same goes for the speeds - if we picked a final speed of something besides \(v_0/4\), we would get a different final answer.)

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A proton (mass [latex] 1.67\,\,10^-27\,\textkg [/latex]) collides with a neutron (with essentially the same mass as the proton) to form a particle called a deuteron. What is the velocity of the deuteron if it is formed from a proton moving with velocity [latex] 7.0\,\,10^6\,\textm/s [/latex] to the left and a neutron moving with velocity [latex] 4.0\,\,10^6\,\textm/s [/latex] to the right?

Define the system to be the two particles. This is a collision, so we should first identify what kind. Since we are told the two particles form a single particle after the collision, this means that the collision is perfectly inelastic. Thus, kinetic energy is not conserved, but momentum is. Thus, we use conservation of energy to determine the final velocity of the system.

This is an example of the type of analysis done by investigators of major car accidents. A great deal of legal and financial consequences depend on an accurate analysis and calculation of momentum and energy.

As discussed in a previous part of Lesson 2, total system momentum is conserved for collisions between objects in an isolated system. For collisions occurring in isolated systems, there are no exceptions to this law. This same principle of momentum conservation can be applied to explosions. In an explosion, an internal impulse acts in order to propel the parts of a system (often a single object) into a variety of directions. After the explosion, the individual parts of the system (that is often a collection of fragments from the original object) have momentum. If the vector sum of all individual parts of the system could be added together to determine the total momentum after the explosion, then it should be the same as the total momentum before the explosion. Just like in collisions, total system momentum is conserved.

Momentum conservation is often demonstrated in a Physics class with a homemade cannon demonstration. A homemade cannon is placed upon a cart and loaded with a tennis ball. The cannon is equipped with a reaction chamber into which a small amount of fuel is inserted. The fuel is ignited, setting off an explosion that propels the tennis ball through the muzzle of the cannon. The impulse of the explosion changes the momentum of the tennis ball as it exits the muzzle at high speed. The cannon experienced the same impulse, changing its momentum from zero to a final value as it recoils backwards. Due to the relatively larger mass of the cannon, its backwards recoil speed is considerably less than the forward speed of the tennis ball.

In the exploding cannon demonstration, total system momentum is conserved. The system consists of two objects - a cannon and a tennis ball. Before the explosion, the total momentum of the system is zero since the cannon and the tennis ball located inside of it are both at rest. After the explosion, the total momentum of the system must still be zero. If the ball acquires 50 units of forward momentum, then the cannon acquires 50 units of backwards momentum. The vector sum of the individual momenta of the two objects is 0. Total system momentum is conserved.

As another demonstration of momentum conservation, consider two low-friction carts at rest on a track. The system consists of the two individual carts initially at rest. The total momentum of the system is zero before the explosion. One of the carts is equipped with a spring-loaded plunger that can be released by tapping on a small pin. The spring is compressed and the carts are placed next to each other. The pin is tapped, the plunger is released, and an explosion-like impulse sets both carts in motion along the track in opposite directions. One cart acquires a rightward momentum while the other cart acquires a leftward momentum. If 20 units of forward momentum are acquired by the rightward-moving cart, then 20 units of backwards momentum is acquired by the leftward-moving cart. The vector sum of the momentum of the individual carts is 0 units. Total system momentum is conserved.

Just like in collisions, the two objects involved encounter the same force for the same amount of time directed in opposite directions. This results in impulses that are equal in magnitude and opposite in direction. And since an impulse causes and is equal to a change in momentum, both carts encounter momentum changes that are equal in magnitude and opposite in direction. If the exploding system includes two objects or two parts, this principle can be stated in the form of an equation as:

The diagram below depicts a variety of situations involving explosion-like impulses acting between two carts on a low-friction track. The mass of the carts is different in each situation. In each situation, total system momentum is conserved as the momentum change of one cart is equal and opposite the momentum change of the other cart.

Since total system momentum is conserved in an explosion occurring in an isolated system, momentum principles can be used to make predictions about the resulting velocity of an object. Problem solving for explosion situations is a common part of most high school physics experiences. Consider for instance the following problem:

The strategy for solving for the speed of the cannon is to recognize that the cannon travels 6.1 cm at a constant speed in the 0.0218 seconds. The speed can be assumed constant since the problem states that it was measured after the impulse of the explosion when the acceleration had ceased. Since the cannon was moving at constant speed during this time, the distance/time ratio will provide a post-explosion speed value.

The strategy for solving for the post-explosion speed of the tennis ball involves using momentum conservation principles. If momentum is to be conserved, then the after-explosion momentum of the system must be zero (since the pre-explosion momentum was zero). For this to be true, then the post-explosion momentum of the tennis ball must be equal in magnitude (and opposite in direction) of that of the cannon. That is,

The negative sign in the above equation serves the purpose of making the momenta of the two objects opposite in direction. Now values of mass and velocity can be substituted into the above equation to determine the post-explosion velocity of the tennis ball. (Note that a negative velocity has been inserted for the cannon's velocity.)

It's worth noting that another method of solving for the ball's velocity would be to use a momentum table similar to the one used previously in Lesson 2 for collision problems. In the table, the pre- and post-explosion momentum of the cannon and the tennis ball. This is illustrated below.

The variable v is used for the post-explosion velocity of the tennis ball. Using the table, one would state that the sum of the cannon and the tennis ball's momentum after the explosion must sum to the total system momentum of 0 as listed in the last row of the table. Thus,

Using the table means that you can use the same problem solving strategy for both collisions and explosions. After all, it is the same momentum conservation principle that governs both situations. Whether it is a collision or an explosion, if it occurs in an isolated system, then each object involved encounters the same impulse to cause the same momentum change. The impulse and momentum change on each object are equal in magnitude and opposite in direction. Thus, the total system momentum is conserved.

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