# Two Knights Collide: Solving Homework Problem

In summary, the knights collide when Sir Alfred's acceleration has a magnitude of 0.364 m/s2 relative to Sir George's starting point, where they have a distance of 60.69 meters.

## Homework Statement

In an historical movie, two knights on horseback start from rest 80.5 m apart and ride directly toward each other to do battle. Sir George's acceleration has a magnitude of 0.377 m/s2, while Sir Alfred's has a magnitude of 0.364 m/s2. Relative to Sir George's starting point, where do the knights collide?

## Homework Equations

a kinematic equation?
V^22nd-(Vo^2,1st + 2a1st *x1st) / 2a2nd

## The Attempt at a Solution

a1st=.377
vo1st=0 m/s
x1st=80.5

a2nd=-.364
Vo2nd= 0m/s
x2nd=80.5

im so lost this is my first physics class ever and this is homework from chapter 2...

Welcome to PF!

The first thing to do is choose and appropriate kinematic equation by looking at the variables you have and the one that you want.

https://www.physicsforums.com/showpost.php?p=905663&postcount=2

Secondly set up the equations correctly remembering that St George's position is the reference point, and thirdly solve them.

i do not know what to do with two accelerations? also am i have had one class and this is my homework, the professor just posted the equations on the board. I'm teaching myself from square one...

so wold i use
velocity & displacement:
v^2 = v_0^2 + 2 a Delta x

also wouldn't sir alfreds be negative cause he is moving backwards(Left) on the x plane

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You will have two equations one for each knight. Then you can solve them simultaneously.

o-i-c

so
(0)^2+2(.377)(80.5)=60.69
(0_^2+2(-.364)(80.5)=-58.6
neither are correct,i apologize i hate not even having common sense,but I am so lost in this stuff..

First off you're trying to find the distance they travel when they hit each other so you can't use the distance between them. You may want to find a different kinematic equation to use.

x = x_0 + v_0 t + (1/2) a t^2 ?

Yes that's the one. Now remember the reference position is for one of the knights so be careful setting up the other knights equation.

so i use that formula twice and do what with the two numbers? subtract or something
what is X_0 and the t?
do i convert the .377 and .364 into seconds

You need to set up the equation for each knight. That is put in the variables yu know for each particular knight. One of them starts at the origin so what will x_0 be then? The other starts 80.5 meters away from the origin so what will x_0 for that knight be and what sign will his acceleration have? Don't worry about the uknonws for now. If you need a bit more help understanding why you're doing this have a look for simultaneous equations.

http://www.tech.plym.ac.uk/maths/resources/PDFLaTeX/simultaneous.pdf

Last edited by a moderator:
80.5+V_0t+(1/2)(.377)T^2

AND

-80.5+V_0t(1/2)(-.364)t^2

but what do i do with the t and V_0t...
i really think this class is above me...

Don't give up! It should become clearer once you've solved one example and try another.

V0 is the initial speed of each knight, what are their initial speeds.

One more thing the first knight starts at the origin (i.e. x0 = 0), and the other is 80.5 metres away not -80.5. once you have that sorted you can solve the equations by eliminating one of the 2 unknowns.

0+V_0t+(1/2)(.377)T^2

AND

80.5+V_0t(1/2)(-.364)t^2

That looks ok except there is a + sign missing from the second one. So their initial speeds are zero so you can get rid of the terms with V0 in there. Then you'll be ready to solve simultaneously.

man i need more help I am not asking for answers but something has to give...i have spent over four hours on this problem and nothing i have not examples,no hard copy practice its just lecture and online homework this is bs...

idk what I'm doing I'm questioning my basic algebraic skills now.
ihave:

0+T+(1/2)(.377)T^2::: T^2=.1885
80.5+T+(1/2)(-.364)T^2:::T^2=-.182

Now you have:

$$s = \frac{1}{2} a_1t^2$$
$$s = 80.5 - \frac{1}{2}a_2t^2$$

Now eliminate the variable you don't need (i.e. the time). Check out the link I gave you about simultaneous equations for more info on that.

im sorry but i don't what to do, all this is just confusing me more.i need answers
i have forced my self to go this far. if i plug it all in it will make sense

I'm afraid we can't give out answers only hints. I've told you what you need to do, so do it. You want the distance s and you want to get rid of t. Make t the subject for one of the equations and substitute it into the other, then solve for s.

1/2(.377)t^2=80.5-1/2(.364)t^2

## 1. How do I know which knight will win in a collision?

The winning knight can be determined by calculating their momentum. The knight with the greater momentum will win the collision.

## 2. What is the formula for calculating momentum in a collision?

The formula for momentum is: momentum = mass x velocity. In the case of two knights colliding, you would need to calculate the momentum for each knight and compare them to determine the winner.

## 3. Can the collision between two knights be perfectly elastic?

Yes, it is possible for the collision between two knights to be perfectly elastic. This means that both knights will bounce off each other with no energy lost during the collision.

## 4. How can I solve a homework problem involving two knights colliding?

To solve a homework problem involving two knights colliding, you will need to apply the principles of conservation of momentum and energy. Set up the equations and solve for the unknown variables.

## 5. Are there any real-world applications for understanding the collision between two knights?

While the concept of two knights colliding may seem abstract, it can actually be applied to real-world situations such as car crashes or sports collisions. Understanding the principles of momentum and energy can help us better understand and predict the outcomes of these events.

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