# Collision and deviation angle

• Saitama
In summary, the conversation discusses an elastic collision between two particles - one stationary and one with a greater mass. The maximum angle through which the striking particle may be deviated as a result of the collision is found to be ##\sin(\theta_{max})=m_2/m_1##. The solution involves using the conservation of momentum and energy equations, and manipulating them to get a quadratic equation in order to find the maximum angle. The use of vectors is suggested to simplify the problem.

## Homework Statement

A particle of mass ##m_1## collides elastically with a stationary particle of mass ##m_2## (##m_1>m_2##). Find the maximum angle through which the striking particle may be deviated as a result of the collision.

(Ans. ##\sin(\theta_{max})=m_2/m_1##)

## The Attempt at a Solution

I don't understand the problem. After the collision, both the particles continue the motion in the initial direction of motion. I don't see why the particles deviate. No geometry about the particles is specified.

Any help is appreciated. Thanks!

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I think you're supposed to assume a glancing collision, rather than head-on.

TSny said:
I think you're supposed to assume a glancing collision, rather than head-on.

Please have a look at the attachment. I have shown the situations before and after the collision.

Conserving momentum in x-direction,
$$m_1v=m_1v_1\cos(\theta_1)+m_2v_2\cos(\theta_2) (*)$$

Conserving momentum in y-direction,
$$m_1v_1\sin(\theta_1)=m_2v_2\sin(\theta_2) (**)$$

Since collision is elastic, energy conservation holds,
$$\frac{1}{2}m_1v^2=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2 (***)$$

How do I solve this? :uhh:

EDIT:
From (**)
$$v_2=\frac{m_1v_1\sin(\theta_1)}{m_2\sin(\theta_2)}$$

Substituting this in (*)
$$m_1v=m_1v_1\cos(\theta_1)+m_2\cos(\theta_2)\frac{m_1v_1\sin(\theta_1)}{m_2\sin(\theta_2)}$$
$$\Rightarrow v=v_1\cos(\theta_1)+v\frac{v_1\sin(\theta_1)\cos(\theta_2)}{\sin(\theta_2)}=v_1\frac{\sin(\theta_1+\theta_2)}{\sin(\theta_2)}$$
$$\Rightarrow v_1=\frac{v\sin(\theta_2)}{\sin(\theta_1+\theta_2)}$$

Using this expression for ##v_1##,
$$v_2=\frac{m_1}{m_2}\frac{v\sin(\theta_1)}{\sin(\theta_1+\theta_2)}$$

Substituting ##v_1## and ##v_2## in the energy equation and simplifying,
$$\sin^2(\theta_1+\theta_2)=\sin^2(\theta_2)+\frac{m_1}{m_2}\sin^2( \theta _1)$$
How do I solve after this? :(

#### Attachments

• collision.png
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Ok initially the momentum has only x coordinates .
So you have to have zero y-component of momentum after the collision

Can you try using sin(a+b) = Sin(a)cos(b)...

yands said:
Can you try using sin(a+b) = Sin(a)cos(b)...

I would rather refrain from using that here. It would lead to a very ugly expression.

And, the correct expansion is sin(a+b)=sin(a)cos(b)+cos(a)sin(b).

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I would approach this differently. Consider the change in momentum of the first particle. What is the square of its magnitude?

1 person
voko said:
I would approach this differently. Consider the change in momentum of the first particle. What is the square of its magnitude?

Change in momentum is ##(m_1v_1\cos(\theta_1)-m_1v)\hat{i}-m_1v_1\sin(\theta_1)\hat{j}##. Square of the magnitude is,
$$m_1^2(v_1^2+v^2-2vv_1\cos(\theta_1))$$

How does this help?

What is that equal to?

voko said:
What is that equal to?

Magnitude of impulse?

What is that square equal to, given the conservation of momentum?

voko said:
What is that square equal to, given the conservation of momentum?

I still don't understand what you ask me here. It is equal to ##m_1^2|\vec{v_1}-\vec{v}|^2##. I can't think of anything else.

Conservation of momentum is ## m_1 \vec {v} = m_1 \vec{v}_1 + m_2 \vec{v}_2 ##. You got ## m_1^2 (\vec{v} - \vec{v}_1)^2 ##. It obviously is equal to ## m_2^2 v_2^2 ##. Now use conservation of energy to get rid of ## v_2 ##.

voko said:
Conservation of momentum is ## m_1 \vec {v} = m_1 \vec{v}_1 + m_2 \vec{v}_2 ##. You got ## m_1^2 (\vec{v} - \vec{v}_1)^2 ##. It obviously is equal to ## m_2^2 v_2^2 ##. Now use conservation of energy to get rid of ## v_2 ##.

I followed your suggestion and found ##\vec{v_1}## in terms of ##\vec{v}##. What should I do next?

This is what I get:
$$\vec{v_1}=\vec{v}\frac{(m_1/m_2-1)}{(m_1/m_2+1)}$$

Really? How did you do that?

After eliminating ## v_2 ##, you were supposed to get a quadratic equation for ## v_1 ##. Its analysis would solve the problem.

voko said:
Really? How did you do that?

After eliminating ## v_2 ##, you were supposed to get a quadratic equation for ## v_1 ##. Its analysis would solve the problem.

No, I did not get a quadratic.

Here's what I did:
$$v_2^2=\frac{m_1^2(\vec{v}-\vec{v_1})^2}{m_2^2}$$
From the energy equation:
$$\frac{1}{2}m_1v^2=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2\frac{m_1^2(\vec{v}-\vec{v_1})^2}{m_2^2}$$
$$\Rightarrow v^2=v_1^2+\frac{m_1}{m_2}(\vec{v}-\vec{v_1})^2$$
$$\Rightarrow (\vec{v}-\vec{v_1})(\vec{v}+\vec{v_1})=\frac{m_1}{m_2}(\vec{v}-\vec{v_1})^2$$
$$\Rightarrow \vec{v}+\vec{v_1}=\frac{m_1}{m_2}(\vec{v}-\vec{v_1})$$
Solving this gives me the relation I posted. Is there a problem with cancelling out ##(\vec{v}-\vec{v_1})## from both the sides?

Pranav-Arora said:
Solving this gives me the relation I posted. Is there a problem with cancelling out ##(\vec{v}-\vec{v_1})## from both the sides?

You bet. You cannot cancel the square - that would mean going from scalars to vectors, and this cannot be done unambiguously. Instead, you need to expand that square, just like you did previously, and go from there.

voko said:
You bet. You cannot cancel the square - that would mean going from scalars to vectors, and this cannot be done unambiguously. Instead, you need to expand that square, just like you did previously, and go from there.

This is the quadratic I get:
$$v_1^2\left(\frac{m_1}{m_2}+1\right)-2\frac{m_1}{m_2}vv_1\cos(\theta_1)+v^2\left(\frac{m_1}{m_2}-1\right)=0$$

Since the quadratic has real roots, discriminant must be greater than or equal to zero. (is this a correct way to say it?)

Solving the inequality, I end up with
$$\sin(\theta_1) \leq \frac{m_2}{m_1}$$

Thanks voko! :)

Also, how did you come up with this method? I mean, dealing with vectors here really made the job a lot easier. How did you know that we should go with vectors? Usually in collision questions, I conserve momentum in x and y direction, dealing in vectors never hit me before.

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I actually prefer staying with vectors for as long as I can. In this case, additionally, you demonstrated that going component-wise results in some very nasty trigonometry. It is actually possible to continue with your original approach and massage the equation you got into something manageable, but that's too much algebra and very little physics.

Re the method I proposed, it occurred to me that the difference between the initial and final momenta will have only one angle involved, so one of the angles is eliminated early on. I cannot quite explain my train of thoughts, it somehow just dawned upon me :)

voko said:
I actually prefer staying with vectors for as long as I can. In this case, additionally, you demonstrated that going component-wise results in some very nasty trigonometry. It is actually possible to continue with your original approach and massage the equation you got into something manageable, but that's too much algebra and very little physics.

Re the method I proposed, it occurred to me that the difference between the initial and final momenta will have only one angle involved, so one of the angles is eliminated early on. I cannot quite explain my train of thoughts, it somehow just dawned upon me :)

Using vectors is a great idea, I too will try to use them as much as possible. Thanks a lot voko!

## 1. What is a collision angle?

A collision angle is the angle at which two objects collide with each other. It is measured from the direction of motion of the first object to the direction of motion of the second object at the point of impact.

## 2. What factors affect the collision angle?

The factors that affect the collision angle include the mass, velocity, and direction of motion of the objects, as well as the shape and surface properties of the objects.

## 3. How is the deviation angle calculated?

The deviation angle is calculated by measuring the angle between the original direction of motion of an object and its resulting direction of motion after a collision. This can be determined using trigonometric principles.

## 4. What is the relationship between collision angle and deviation angle?

The collision angle and deviation angle are directly related. As the collision angle increases, the deviation angle also increases. This means that a more oblique collision results in a larger deviation angle.

## 5. How do collisions and deviation angles impact real-world scenarios?

Understanding collision and deviation angles is important in predicting the outcomes of real-world scenarios such as car crashes, sports collisions, and particle collisions in physics experiments. It allows scientists to calculate and analyze the energy, force, and direction of motion involved in these collisions.