Ball falls in a truck loaded with sand

  • Thread starter Thread starter 10bunny10
  • Start date Start date
  • Tags Tags
    Ball Truck
AI Thread Summary
The discussion centers on a ball falling into a truck filled with sand, examining the momentum changes during the collision. It is identified as a completely inelastic collision, with initial vertical and horizontal momentum components calculated as m * v * cos(α) and m * v * sin(α), respectively. Participants clarify the scenario, questioning whether the truck is stationary and addressing potential confusion over terminology. The conversation highlights the importance of understanding momentum conservation, particularly in the vertical direction, and whether friction plays a role in the collision dynamics. Overall, the complexities of momentum conservation in the context of inelastic collisions and friction are emphasized.
10bunny10
Messages
6
Reaction score
0

Homework Statement


a ball falls in a truck loaded with sand with velocity v having horizontal and vertical component. The angle under which it enters the truck is α, the mass of the ball is m of the track loaded with sand is M. What happens with the momentum after the collision?

The Attempt at a Solution


This should be considered as completely inelastic collision. In vertical direction the momentum before the collision is the momentum of the ball mcosv. After the collision the track is not moving in horizontal direction, but the momentum should be conserved. The ball sinks in the sand, so is this the explanation? In horizontal direction the momentum before the collision is msinv. What is the momentum after? There is friction during the collision. It should be (m+M)v', but what about the friction?
 
Physics news on Phys.org
hi bunny! welcome to physicsforums :)

I'm having some trouble understanding the question. Is it a truck (as in a vehicle, without the brakes on)? Because you said track a couple of times. And the truck is meant to be not moving before the collision? And when you said mcosv is the vertical momentum of the ball before collision, did you mean mvcosα? And so the angle is measured from the vertical, right? (Since the vertical momentum is mv when the angle is zero).

Also, you said: "After the collision the track is not moving in horizontal direction, but the momentum should be conserved. The ball sinks in the sand, so is this the explanation?" I'm guessing you meant to say vertical direction? In this case, why do you think vertical momentum is conserved? Even if you take away the complication of the truck, is the vertical momentum of a falling ball conserved?

You are right that the final momentum in the horizontal direction will be (m+M)v'. And what about the friction? Does the question mention that there is some dissipative friction?
 
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