Help understanding this collision question

  • Thread starter Thread starter urbano
  • Start date Start date
  • Tags Tags
    Collision
AI Thread Summary
In a head-on collision between a truck and a small car, both vehicles exert equal forces on each other due to Newton's third law, but their accelerations differ because of their mass. The impulse experienced by both vehicles is equal if they are in contact for the same duration and exert the same force. However, the change in momentum is not equal, as the velocities of the vehicles before and after the collision differ significantly. The smaller car will experience greater deceleration due to its lower mass compared to the truck. Understanding these dynamics clarifies the differences in force, impulse, momentum, and deceleration in collision scenarios.
urbano
Messages
36
Reaction score
0

Homework Statement


I've been checking out the ol' truck vs small car in a head on collision problem. The question asks which about which vehicle will experience the greatest force of impact ? The greater impulse? The greater change in momentum ? The greater deceleration.

A number of aspects of it have me second guessing my understanding of what is going on...and even when I feel I may know the answer I don't actually know WHY it is the way it is



Homework Equations



Momentum = m * v
Impulse = F * t , which means a change in momentum = change in m * v

Newtons 3rd law

The Attempt at a Solution



So in terms of the force on impact, I believe if Newton's third law says for each action there is an equal and opposite reaction, then therefore they should impart the same amount of force on each other...although this confuses me. If F= m*a, how can they exert the same force on each other given the truck is likely to have a greater force given its mass ?

The greater impulse ...I believe this would be equal. If they are exerting the same amount of force and in contact for the same amount of time then they both are creating an equal impulse. However, the change in momentum confuses me...if it = m * v, the mass of each vehicle is likely to stay the same and the velocity would stay the same. For example if the truck hits the car and pushes it east, the truck and the car will both travel east and their masses will have remained the same...therefore making the impulse equal for both ?

The change in momentum...as per above wouldn't they both change by equal amounts?

The greater deceleration. .for this I believe it to be the mini (or both) as A= d/t and I'm presuming the mini would end up going backwards.
 
Physics news on Phys.org
So in terms of the force on impact, I believe if Newton's third law says for each action there is an equal and opposite reaction, then therefore they should impart the same amount of force on each other
Right

If F= m*a, how can they exert the same force on each other given the truck is likely to have a greater force given its mass ?
The truck will have a different acceleration.

The greater impulse ...I believe this would be equal. If they are exerting the same amount of force and in contact for the same amount of time then they both are creating an equal impulse.
Right.

However, the change in momentum confuses me...if it = m * v, the mass of each vehicle is likely to stay the same and the velocity would stay the same.
The velocity will not stay the same. In addition, the velocity differences will be different, too.

For example if the truck hits the car and pushes it east, the truck and the car will both travel east and their masses will have remained the same...therefore making the impulse equal for both ?
Their velocity, not their momentum. In addition, the question asks for momentum transfer (change in momentum), not momentum.

The change in momentum...as per above wouldn't they both change by equal amounts?
Right.

The greater deceleration. .for this I believe it to be the mini (or both) as A= d/t and I'm presuming the mini would end up going backwards.
What is d?
You can use F=ma and the result for the force to answer this question.
 
mfb said:
Right
The velocity will not stay the same. In addition, the velocity differences will be different, too.

.

I don't quite get this...at the moment they collide aren't they both for a very small instant of time starting at 0 velocity, i.e they are not moving ? Then if the truck is in contact with the mini and pushes it backwards aren't they moving in the same direction for the same amount of time ? i.e both have the same velocity ? the mini will keep being pushed backwards so long as the truck keeps moving forwards. so the truck pushes forward 5 meters over 5 seconds, the mini moves 5 meters over 5 seconds as well.

Obviously I'm wrong but I don't know why.

**Update** I just watched an animation on this exact scenario, an depending on if it's elastic or non-elastic then the results vary ...ARGHH!
 
Last edited:
at the moment they collide aren't they both for a very small instant of time starting at 0 velocity
Could happen during the collision (but here you get the additional problem that not all parts of the cars are moving at the same speed), but it does not matter.

Then if the truck is in contact with the mini and pushes it backwards aren't they moving in the same direction for the same amount of time ? i.e both have the same velocity ?
After the collision, if the collision is perfectly inelastic, they move with the same velocity. Before the collision, they do not (otherwise they would not collide).
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Back
Top