# Particle Collisions (Momentum)

• goalieguy
In summary, when a proton with a mass m and initial velocity of 2.70*10^7 m/s collides elastically with a stationary nucleus of an unknown element, it rebounds with a velocity of 2.40*10^7 m/s. By applying the principles of conservation of linear momentum and kinetic energy, we can calculate the mass of the unknown nucleus to be 17m and its final velocity to be 0.30*10^7 m/s.
goalieguy
You are at the controls of a particle accelerator, sending beam of 2.70*10^7 m/s protons (mass m) at a gas target of an unknown element. Your detector tells you that some protons bounce straight back after a collision with one of the nuclei of the unknown element. All such protons rebound with a speed of 2.40*10^7 m/s. Assume that the initial speed of the target nucleus is negligible and that the collision is elastic.

(a) Find the mass of one of the nuclei of the unknown element. Express you answer in terms of the proton mass m.

(b) What is the speed of the unknown nucleus immediately after such a collision?

I really do not know how to start this problem. I know you somehow have to use momentum but i am not sure how to start it without the mass given?

Hint: conservation of linear momentum.

The mass is given if you have notice. It is m and you are suppose to leave it in such a way. Apply conservation of momentum and write out your equation for it then.

i don't get it... so the mass m is 2.10×10^7...?

i meant 2.70*10^7

Therefore, mass of the proton = m.
Assume, mass of the unknown nucleus = k.m, where 'k' is some constant. We need to find 'k'.
Assume, direction of proton before collision be +X.
Initial velocity of the proton, u = +2.70*10^7 m/s.
"Assume that the initial speed of the target nucleus is negligible."
Thus, initial velocity of the nucleus = 0 m/s.
Final velocity of the proton, u' = -2.40*10^7 m/s. [Note that it rebounds with this speed.]
Let, final velocity of the proton be +v.
Collision is elastic => |velocity of separation along the line of collision| = |velocity of approach along the line of collision|
=> |u' - v| = |u| => v + 2.40*10^7 m/s = 2.70*10^7 m/s.
=> v = +0.30*10^7 m/s.

Further, apply linear momentum balance along X-direction:
m.(u) + k.m.(0) = m.(u') + k.m.(v)
Solve it to get the value of 'k' (= 17).
Thus, mass of the unknown nucleus is 17m.

17m is not the right answer...

tmle04 said:
17m is not the right answer...

Not to beat a dead thread, but, perhaps this thread would be more useful if it contained a complete explanation for people who stumble on it in the future. Here is another way of looking at the problem:

It's an elastic collision where the target body is at rest.

Let the fired particle be A and the target nucleus be B.

The initial x-velocity of A is:

$$} v_{\texttt{A1x}} = 2.70\times10^7$$

The final x-velocity of A is:

$$v_{\texttt{A2x}} = -2.40\times10^7$$

Since B is initially at rest, it's x-velocity is:

$$v_{\texttt{B1x}} = 0$$

When the target body is at rest, the momentum conservation equation is:

$$m_{\texttt{A}}v_{\texttt{A1x}} = m_{\texttt{A}} v_{\texttt{A2x}} + m_{\texttt{B}} v_{\texttt{B2x}}$$

From this, we derive:

Eq. 1: $$m_{\texttt{B}} v_{\texttt{B2x}} = m_{\texttt{A}}(v_{\texttt{A1x}} - v_{\texttt{A2x}})$$

When the target body is at rest, the kinetic energy equation is:

Eq. 2: $$m_{\texttt{B}} v_{\texttt{B2x}}^2 = m_{\texttt{A}}(v_{\texttt{A1x}}^2 - v_{\texttt{A2x}}^2) = m_{\texttt{A}}(v_{\texttt{A1x}} - v_{\texttt{A2x}})(v_{\texttt{A1x}} + v_{\texttt{A2x}})$$

Dividing equation 2 by 1, we are left with:

$$v_{\texttt{B2x}} = v_{\texttt{A1x}} + v_{\texttt{A2x}}$$

Note that we were given the initial x-velocity and final x-velocity of A, so we can use these velocities to calculate the final x-velocity of B. This answers part (a) of the question.

Then, substitute your result for the final x-velocity of B into equation 1 and solve for the mass of B. This answers part (b) of the question.

## 1. What is a particle collision?

A particle collision is a phenomenon that occurs when two or more particles come into contact with each other and exchange energy and/or momentum. These collisions can happen at various speeds and can result in the creation of new particles or a change in the properties of the colliding particles.

## 2. How is momentum conserved in a particle collision?

Momentum is a fundamental quantity that is conserved in all physical interactions, including particle collisions. This means that the total momentum of the particles before and after the collision remains the same. The conservation of momentum is governed by the laws of physics and is a crucial concept in understanding the outcomes of particle collisions.

## 3. What are the different types of particle collisions?

There are three main types of particle collisions: elastic, inelastic, and completely inelastic. In an elastic collision, the total kinetic energy of the colliding particles is conserved. In an inelastic collision, some of the kinetic energy is lost, while in a completely inelastic collision, the colliding particles stick together and move as a single unit after the collision.

## 4. How are particle collisions studied?

Particle collisions are studied using particle accelerators, which are large machines that accelerate particles to very high speeds and then collide them with other particles or targets. Scientists use detectors to measure the properties of the colliding particles and analyze the data to understand the dynamics of the collision and the resulting particles.

## 5. What is the significance of studying particle collisions?

Studying particle collisions is essential for understanding the fundamental building blocks of matter and the laws of nature. By analyzing the outcomes of collisions, scientists can gain insights into the properties and interactions of particles, which can help in developing new technologies and advancing our understanding of the universe.

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