What Happens to Two Blocks After an Elastic Collision on a Slope?

  • Thread starter Thread starter DavidHarmon
  • Start date Start date
  • Tags Tags
    Momentum Slope
AI Thread Summary
The discussion focuses on a physics problem involving two blocks in an elastic collision on a slope. The first block, with a mass of 2.80 kg, slides down a 30° incline and collides with a stationary block of 7.00 kg at the bottom. Participants suggest using energy conservation principles to determine the initial velocity of the lighter block before the collision. After finding the velocities post-collision, the next step is to calculate how far the smaller block travels back up the incline. The conversation emphasizes the need to establish the initial conditions and apply relevant equations for momentum and energy conservation.
DavidHarmon
Messages
5
Reaction score
0

Homework Statement


A block of mass m = 2.80 kg slides down a 30.0° incline which is 3.60 m high. At the bottom, it strikes a block of mass M = 7.00 kg which is at rest on a horizontal surface, Fig. 7-41. (Assume a smooth transition at the bottom of the incline, an elastic collision, and ignore friction.)
http://img141.imageshack.us/img141/4472/741ue2.gif


(a) Determine the speeds of the two blocks after the collision.
Lighter Block:
Heavier Block:

(b) Determine how far back up the incline the smaller mass will go.


Homework Equations



v=final velocity
u= Initial velocity
a=acceleration
s= Displacement
v^2=u^2+2as

m1u1+m2u2=m1v1+m2v2

The Attempt at a Solution



I resolved to find acceleration perpendicular to the plain of motion and also the displacement, but as i was not given an initial velocity i cannot find out the final velocity of the small block as it hits the large block

Does anyone have any ideas as to how i would go about solving this question?
 
Last edited by a moderator:
Physics news on Phys.org
i think you have to assume that the block starts at rest here.

to start, use energy conservation methods to find the velocity of the small block at the bottom of the incline.
 
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