Average force exerted by momentum

AI Thread Summary
The discussion revolves around calculating the average force exerted on a car after a collision, where a 10 kg car traveling at 10 m/s hits another car, causing both to come to rest. The initial momentum is calculated as 100 kg*m/s, but the time taken for the collision is incorrectly assumed to be 1 second. It is clarified that to find the correct time, uniform deceleration must be assumed, and the mass of the second vehicle is unknown, complicating the calculation. There is a suggestion that the problem may require using energy principles instead of momentum to find the average force over distance. The need for accurate assumptions and calculations is emphasized to arrive at the correct average force.
PhizKid
Messages
477
Reaction score
2

Homework Statement


A car of mass 10 kg hits another car at 10 m/s and they both come to rest at the same time. In the process, the car that was hit moved back 10 meters. What is the average force exerted on it?


Homework Equations


Average force = change in momentum divided by time


The Attempt at a Solution


I'm taking it so everything is going to the right, so positive rightwards.
Final momentum is 0. Initial is 10kg * 10m/s so: Average force = -100kg*m/s (because final - initial) divided by the time which is 1 second I think. 10m = 10m/s * t = 1 second.
So -100kg*m/s divided by 1 second = -100N.

But this is wrong. I can't find my error but I'm told I calculated the time incorrectly. How do I get the correct time?
 
Physics news on Phys.org
If it only took 1 sec, the speed of the first car would have to be 10m/s for the whole second (then suddenly 0). To get the time, you have to assume uniform deceleration after the impact. That's reasonable, but worse is that you are not told the mass of the second vehicle, so you don't know what the speed is just after impact. So to get an answer you'll have to assume they're the same mass.
I've a nasty feeling that the questioner intends you to use energy = force * distance. That will give you the force averaged over distance, but that is not the same as the force averaged over time.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top