Conservation of Momentum - Collisions

AI Thread Summary
The discussion revolves around a collision problem involving a truck and a car, where the truck's initial and final velocities are provided, and the goal is to determine the car's velocity after the collision using the conservation of momentum. The initial calculations suggest a post-collision speed for the car of approximately 99 km/h, which raises concerns about violating energy conservation principles. Participants agree that while momentum is conserved in the collision, kinetic energy does not have to be conserved, but it cannot increase. The calculations presented indicate discrepancies in energy conservation, suggesting that the results may not be physically plausible. The conversation highlights the complexities of analyzing collisions and the importance of adhering to the laws of physics.
doug1
Messages
9
Reaction score
0

Homework Statement



A truck (4 000 kg) runs a red light and enters an intersection traveling at 81 km/h [E]. The truck collides with a car (2
000 kg) that was traveling at 54 km/h [N20oW]. Immediately after the collision the truck was traveling at 60 km/h [N
30o E]. Determine the velocity vector of the car immediately after the collision.


Homework Equations



I used the law of conservation of momentum. I used the concept that momentum is conserved in each component direction.

The Attempt at a Solution



The answer that I got was a velocity of approx 99 km/h [E32S]. Can anyone confirm this?
 
Physics news on Phys.org
This would violate energy conservation, unless the truck or the car have some active system to kick away other vehicles.
I don't understand the notation for the direction, but I would expect that momentum is not conserved either.
 
The notation [N30E] means 30 degrees east of north for example.

Momentum is conserved in this question, but energy is not
 
I agree that kinetic energy does not have to be conserved, but kinetic energy cannot increase in the process.

Working in units of 1000kg*km/h:
to east:
truck initial 4*81=364
car initial -2*54*sin(20°)
truck final 4*60*sin(30°)
car final ~207 or 104 km/h
Hmm... looks wrong.

to north:
truck initial 0
car initial 2*54*cos(20°)
truck final 4*60*cos(30°)
car final -106 or -53km/h

Total energy initially: 2*81^2+1*54^2=16000 (in 1000kg*(km/h)^2)
Total energy finally: 2*60^2+1*(104^2+53^2) = 20800
Clearly this cannot happen in a car accident. The same problem appears with your result of 99km/h total speed as well.
 
Last edited:
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