Car 1 & Car 2: Inertia & Velocity at the Edge of a Cliff

AI Thread Summary
Car 1, having twice the mass of Car 2, experiences greater inertia and thus achieves a lower velocity when both cars reach the cliff edge. Despite equal forces applied to both cars, Car 2 accelerates faster due to its lesser mass. As a result, Car 2 has a greater horizontal velocity when leaving the cliff, allowing it to travel farther than Car 1. The discussion emphasizes the relationship between mass, inertia, and acceleration in determining the final velocities of the cars. Overall, Car 2's lighter mass contributes to its superior performance in this scenario.
deanine3
Messages
24
Reaction score
0

Homework Statement


Car 1 and Car 2 are initially at rest on a horizontal parking lot at the edge of a steep cliff. Car 1 has twice the mass as car 2. Equal and constant forces are applied to each car and they accelerate across equal distance to the cliff. We ignore the effects of friction. When they reach the far end of the lot, the force is suddenly removed, whereupon they sail through the air and crash to the ground below.


Homework Equations


Velocity = speed x direction
inertia
acceleration = change in velocity/time

The Attempt at a Solution


Car 1 has twice the mass it has twice the inertia and less velocity then car 2. Since car 2 has 1/2 the mass, it has greater acceleration leaving the cliff it therefore "shoots out" farther than car 1. Would it be correct to say it has greater "horizontal velocity" leaving the cliff?
 
Physics news on Phys.org


Yep.

You were an easy one! :cool:
 


Hoorah!
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging 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