Collision Problem: Masses, Speed, Force

  • Thread starter Thread starter ges9503
  • Start date Start date
  • Tags Tags
    Collision
AI Thread Summary
The discussion focuses on a physics problem involving a box dropped onto a skateboard, analyzing the final speed after collision and the average force exerted during the impact. The final speed of the skateboard and box combination is calculated to be 4.06 m/s. For the average force during the collision, the participant suggests considering impulse in both the vertical and horizontal directions due to the box's change in momentum. They recognize the need to combine these components to determine the total impulse vector and solve for the average force. The conversation emphasizes understanding the principles of momentum and impulse in collision scenarios.
ges9503
Messages
3
Reaction score
0

Homework Statement



A box of mass 6.30 kg is dropped from a height of 2.10 m onto a super-sized skateboard of mass 8.70 kg that rolls on a horizontal surface where no friction acts. Initially the skateboard moves at 7.00 m/s.
(a) Assume that the box stays on the skateboard. What is the final speed?
(b) The collision of the box and the skateboard lasts a time of 8.00 ms. What is the average force exerted by the skateboard on the box during the time of the collision?

Homework Equations



J= Favg * t

The Attempt at a Solution


I already have found part a to equal 4.06 m/s. I just need help figuring out what to do with part b, which is to be reported in i-hat j-hat notation.
Thanks
 
Physics news on Phys.org
I'll admit that I am not completely sure of this myself, but I'm thinking that you have to find the impulse in both the x and y directions. Because there will be a change in momentum vertically as the box hits the board, but also because the box ends up moving at the same speed as the board then the board must also be applying a force (or impulse) to the box in the horizontal direction.
Then recombine the components to find the total impulse vector and solve for t.
 
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