3 ms·
Coincidently I'm reading Ernst Mach's "The Science of Mechanics" at this very moment. It includes a lot of analysis of setups like this. As for this problem:
by phaedrus 16y ago
Coincidently I'm reading Ernst Mach's "The Science of Mechanics" at this very moment. It includes a lot of analysis of setups like this.
As for this problem: consider first the weights. Each one's downward force is 1 kg, due to gravity. There must also be exactly 1 kg's force's worth of tension in the string, because if it were less the weights would be moving down and if it were more the weights would be pulled up. Because the system (is assumed to be) not moving, we may replace the right side weight and pulley wheel with a fixed anchor for the string, in which case there would be no doubt that the spring balance displays 1 kg. Argument from symmetry says the scale would also read 1 kg if the left side were anchored and the right side free. Because the system in unmoving, it really makes no difference whether the opposite side of the scale is fixed to an anchor or balanced by a free weight: it will read 1 kg in all cases.
This is very similar to a Car Talk puzzler that their physicist consultant clarified: is it worse to be in a car going 60 mph that hits an unmovable highway overpass wall, or to be in a car going 60 mph that hits an identical car perfectly head-on that is also going 60 mph? The physicist's answer was that there is no difference: in both cases you experience an identical impulse going from 60 mph to a stop in the same time.
You may be inclined disbelieve this on the idea that the crumple zones in the car-to-car collision make for a "softer" collision, but consider this: define the point at which the bumpers of the two cars as "the origin". Assuming the cars and the collision are perfectly symmetrical, each car is going to crumple up on its respective side of the origin identically. Suppose the force starts small and ramps up gradually as the crumpling occurs; each car will be pushing the other away with the same force. An unmoving highway overpass barrier would resist with the same force.
- Ralith 16y agoPedantry: The kg is a unit of mass, not force. You want Newtons.
- dedward 16y agoMoving some vectors around to make it look more similar - would I like to hit a nice solid wall at 60MPH or a parked car head-on at 120MPH. The head-on collision has twice as much energy to dissipate, and introduces all kinds of projectiles and other stuff heading in my direction. I'll take the wall over the car.
- modeless 16y agoYour first paragraph is wrong. You can't switch frames like that because in a collision both cars will quickly come to rest relative to the ground, so kinetic energy in the ground frame is what matters. A car at 120 MPH has 4 times the kinetic energy of a car at 60 MPH, not 2 times, because kinetic energy increases as the square of velocity. I'll agree with the projectile thing though.
- RiderOfGiraffes 16y agoNot entirely true. After colliding with a stationary car the previously moving car will retain some of its velocity. That will then dissipate over time and be less violent. So to some extent - when simplified, to a first approximation - you really can switch frames of reference like that.
- modeless 16y agoWhen simplified, yes, but if you start talking about projectiles then you are explicitly not simplifying things, and should also account for the effect of the ground because it's probably a larger concern than projectiles in a collision like this.