Predicting Final Velocities for a Head-On Elastic Collision Problem

  • Thread starter Thread starter ~Fluffy
  • Start date Start date
  • Tags Tags
    Collision
AI Thread Summary
In a head-on elastic collision between two bodies of significantly different masses, the smaller mass (m) will bounce back with its initial speed if it collides with a larger mass (M) at rest. Conversely, if M approaches m with speed v_0 while m is at rest, the reference frame of M reveals that m approaches at speed v_0. In this frame, m will bounce back with speed v_0, while M remains stationary. Understanding this reference frame is crucial for predicting final velocities in elastic collisions. The discussion emphasizes the importance of analyzing the problem from different reference frames to derive accurate outcomes.
~Fluffy
Messages
1
Reaction score
0

Homework Statement



"Consider a head-on, elastic collision between two modies whose masses are m and M, with m << M. It is well known that if m has speed v_0 and M is initially t rest, m will bounce straight back with its speed unchanged, while M will remain at rest (to an excellent approximation). Use this fact to predict the final velocities if M approaches with speed v_0 and m is initially at rest. [Hint: Consider the reference frame attached to M.]

Homework Equations



1. p1 + p2' = p1 + p2'
2. m1*v1 + m2*v2 = m1*v1' + m2*v2' (elastic)
3. m1*v1 + m2*v2 = v' * (m1+m2) (inelastic)

The Attempt at a Solution



I came up with three possible scenarios for the answer. I'm not certain how to consider the reference frame hint into the problem, which is what I'm primarily interested in understanding.

My possible solutions:

1) M bounces off m, similar to how the example in the beginning of the problem.

ViM = V_0, Vim = 0.

VfM = V_0, Vfm = 0.

I don't think that's likely.

2) M stops at collision, all momentum is transferred to m.

ViM = V_0, Vim = 0.

VfM = 0, Vfm = (M / m) * V_0

3) Both m and M move at speed V_0 (roughly). I'm assuming that M is pushing m forward (so same speed).

ViM = V_0, Vim = 0.

VfM = Vfm = (M / (M+m)) * V_0 (rearrangement of equation 3)

My intuition tells me that solution 3 makes the most sense, but I would like to confirm its validity, as well as better understand the reference frame hint in relation to this problem.

Thank you! :)
 
Physics news on Phys.org
In the reference frame attached to M, what is the velocity of M? Of m?
 
Disclaimer: new poster to this forum so let me know if this is giving too much help

M is moving at velocity of v_0 in a static reference frame. But from the reference frame of M, it is not moving at all (almost by definition). Instead from M's point of view, it looks like m is approaching M at speed v_0 and M is standing still.

So if m was approaching M with a speed of v_0 the problem explained what would happen; it would bounce off and move away from M with a speed of v_0.

Does this give you a sense of how to use M's reference frame to answer the question?
 
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...

Similar threads

Back
Top