Average force in totally inelastic collision

In summary, the average force on mass A is -14400i+2400j N, and the average force on mass B is -2400i+14400j N.
  • #1
uk9999
11
0

Homework Statement



Two bodies A and B collide in a totally inelastic collision. Using the relevant equations and given that the mass of body A is 1200kg and the collision lasts for 0.2s, determine the average force vectors acting on each body during the collision.

Homework Equations



vA=5i+3j m/s
vB=-i+4j m/s
mA=(3/2)mB
Common velocity after collision v=2.6i+3.4j m/s


The Attempt at a Solution


attempt 1
Ft = (mA+mB)v - (mAvA+mBvB)
calculation checked and rechecked... leads to wrong answer
attempt 2
Ft = (mA+mB)v - (mAvA-mBvB)
calculation checked and rechecked... leads to wrong answer
attempt 3
FAt=0.5(mA+mB)v - mAvA
FBt=0.5(mA+mB)v - mBvB
calculation checked and rechecked... leads to wrong answer

Answer is stated as "FA= -FB= (-14400i+2400j) N in the mark scheme
 
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  • #2
Hi there uk9999. Well i think you need to rethink the way you looking at the question. First let's have a look back at how we define impulse and momentum:

[tex]\textbf{I} = \textbf{F}t = m\textbf{v} - m\textbf{u}[/tex]

now I, F, v and u can all be vectors as we have in this question, but in order to solve a problem involving vectors we should brake it down into its components, in this scenario in 2D so well equate the x comonets and the y comonents seperatly.

Now in actual fact the question gives you far more information than you actually need, and it is not nessesary to consider both bodies, as the question tells us the initial and final velocity of both, and Impulse is described for a single body, so you can take you pick as to which one you want to use. so looking back at the first equation we need to modify this so that we can consider componets:

[tex]I_x = F_x t = mv_x - mu_x[/tex]
[tex]I_y = F_y t = mv_y - mu_y[/tex]

now i decided to use body A, no reason just was the first one I came to :D, so if wee input those values into you componentised Impluse equations we get:

[tex]I_x = F_x t = 1200(2.6) - 1200(5)[/tex]
[tex]I_y = F_y t = 1200(3.4) - 1200(3)[/tex]

now have a go from there, I think there is enough info there for you to finish off the question :D have fun
 
  • #3
uk9999 said:

The Attempt at a Solution


attempt 1
Ft = (mA+mB)v - (mAvA+mBvB)
calculation checked and rechecked... leads to wrong answer
attempt 2
Ft = (mA+mB)v - (mAvA-mBvB)
calculation checked and rechecked... leads to wrong answer
attempt 3
FAt=0.5(mA+mB)v - mAvA
FBt=0.5(mA+mB)v - mBvB
calculation checked and rechecked... leads to wrong answer
In all of your attempts you have tried to use the combined momentum of both masses. Don't!

To find the average force on mass A, you need the change in momentum of mass A. Similarly, to find the average force on mass B, you need the change in momentum of mass B. (Of course, you don't have to calculate the force twice. Use Newton's 3rd law.)
 
  • #4
Ah thank you knew I was doing something wrong :blushing:
 

Related to Average force in totally inelastic collision

1. What is the definition of average force in a totally inelastic collision?

The average force in a totally inelastic collision is the average amount of force exerted during the collision between two objects. It is a measure of the impact of the collision and is calculated by dividing the change in momentum by the time duration of the collision.

2. How does the average force in a totally inelastic collision differ from other types of collisions?

In a totally inelastic collision, the objects involved stick together and move with a common final velocity. This means that the average force is generally higher compared to other types of collisions where the objects bounce off each other and have different final velocities.

3. What factors affect the average force in a totally inelastic collision?

The average force in a totally inelastic collision is affected by the masses of the objects involved, their initial velocities, and the duration of the collision. The greater the masses and initial velocities, the higher the average force will be.

4. Can the average force in a totally inelastic collision be negative?

Yes, the average force in a totally inelastic collision can be negative. This occurs when the objects involved have opposite initial velocities and the resulting force during the collision is in the opposite direction, causing a decrease in momentum.

5. How is the concept of average force in totally inelastic collisions used in real-world applications?

The concept of average force in totally inelastic collisions is used in various fields such as engineering, physics, and sports. It is used to calculate the impact and safety of car crashes, design protective gear for athletes, and analyze the effects of collisions in machinery and structures.

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