# Solving Collisions Problem: Velocity of Third Fragment

• bondgirl007
In summary: Please check your work.In summary, a 600 kg canon initially at rest exploded upon firing and broke into three fragments. One fragment of mass of 200 kg traveled east at 40m/s and a second fragment of mass 300 kg traveled due south at 20 m/s. Using the principles of momentum conservation, the missing fragment of mass 100 kg was found to have a velocity of 100 m/s at an angle of 37 degrees north of west.
bondgirl007

## Homework Statement

a 600 kg canon, initially at rest, exploded upon firing and broke into three fragments. One fragment of mass of 200 kg traveled east at 40m/s and a second fragment of mass 300 kg traveled due south at 20 m/s. What was the velocity of the third fragment?

## Homework Equations

sum of P = sum of P'
0= P1 + P2 + P3

## The Attempt at a Solution

I'm not sure whether I'm supposed to use trig or something to find the missing velocity.
Any help would be GREATLY appreciated.

bondgirl007 said:

## Homework Statement

a 600 kg canon, initially at rest, exploded upon firing and broke into three fragments. One fragment of mass of 200 kg traveled east at 40m/s and a second fragment of mass 300 kg traveled due south at 20 m/s. What was the velocity of the third fragment?

## Homework Equations

sum of P = sum of P'
0= P1 + P2 + P3

## The Attempt at a Solution

I'm not sure whether I'm supposed to use trig or something to find the missing velocity.
Any help would be GREATLY appreciated.
Momentum is a vector quantity, and as such, the sum of the momenta must obey the laws of vector additions. Are you familiar with resultants and vector sums?

This is what I have so far. Am I on the right track?
Will the resultant be the hypotenuse?

http://img408.imageshack.us/my.php?image=physicsay5.jpg

#### Attachments

• physics.JPG
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I did Pythragoras to find the missing vector.

(8000)^2 + (6000)^2 = c^2
c=100 m/s

tan$$\theta$$ = 6000/8000
$$\theta$$ = 37

I got 100 m/s for my velocity and 37 degrees N of W. Is this right?

You know anythin about momentum conservation?

bondgirl007 said:
I did Pythragoras to find the missing vector.

(8000)^2 + (6000)^2 = c^2
c=100 m/s

tan$$\theta$$ = 6000/8000
$$\theta$$ = 37

I got 100 m/s for my velocity and 37 degrees N of W. Is this right?
Yes, that is correct, but your diagram and equation are not consistent with this answer. Looks like you combined a step when solving for 'c' , and the direction of the diagonal is wrong.

## 1. What is a collision problem?

A collision problem is a scenario where two or more objects come into contact with each other and transfer energy and momentum between them.

## 2. What is the velocity of the third fragment in a collision problem?

The velocity of the third fragment in a collision problem is the speed and direction at which the third object or fragment moves after the collision has occurred.

## 3. How is the velocity of the third fragment calculated?

The velocity of the third fragment is calculated using the conservation of momentum and energy equations, taking into account the masses and velocities of all objects involved in the collision.

## 4. What factors can affect the velocity of the third fragment in a collision problem?

The velocity of the third fragment can be affected by the masses and velocities of the objects involved in the collision, as well as the type of collision (elastic or inelastic) and any external forces acting on the objects.

## 5. How can the velocity of the third fragment be determined experimentally?

The velocity of the third fragment can be determined experimentally by measuring the masses and initial velocities of the objects involved in the collision and using equations of conservation of momentum and energy to calculate the final velocity of the third fragment.

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