Completely Inelastic Collisions

Click For Summary
SUMMARY

The discussion centers on calculating the maximum compression of a spring during a completely inelastic collision between two blocks. The blocks have masses m1=1.88 kg and m2=4.92 kg, with initial velocities of 10.3 m/s and 3.27 m/s, respectively. The correct approach involves using conservation of momentum to find the final velocity (vf) and then applying energy conservation to determine the spring compression (x). The user calculated x as 0.407 m, while the textbook answer is 0.359 m, indicating a potential error in the user's calculations or assumptions.

PREREQUISITES
  • Understanding of momentum conservation in inelastic collisions
  • Familiarity with energy conservation principles
  • Knowledge of spring force and Hooke's Law
  • Ability to solve quadratic equations
NEXT STEPS
  • Review the derivation of the inelastic collision formula: m1v1 + m2v2 = (m1 + m2)vf
  • Study energy conservation in mechanical systems, specifically involving springs
  • Practice similar problems involving spring compression after inelastic collisions
  • Examine potential sources of error in calculations, particularly in energy equations
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and collision problems, as well as educators seeking to clarify concepts related to inelastic collisions and energy conservation.

DeadFishFactory
Messages
25
Reaction score
0

Homework Statement


A block of mass m1=1.88 kg slides along a frictionless table with a speed of 10.3 m/s. Directly in front of it, and moving in the same direction, is a block m2=4.92 kg moving at 3.27 m/s. A massless spring with a force constant k=1120N/m is attached to the backside of m2. When the blocks collide, what is the maximum compression of the spring? (Hint: at the moment of maximum compression of the spring, the two blocks move as one, find the velocity by noting that the collision is compltely inelastic to this point).


Homework Equations


m1v1+m2v2=(m1+m2)vf
1/2(m1+m2)Vf^2=1/2Kx^2


The Attempt at a Solution



I solved for vf in the inelastic equation:

vf = (m1v1 + m2v2) / (m1 + m2)

and plugged it into the energy equation and solved for x:

[(m1v1 + m2v2)^2 / K(m1 + m2)]^1/2 = x

Plugged in numbers:

{[1.88(10.3) + 4.92(3.27)]^2 / 1120(1.88 + 4.92)}^1/2 = x

(1257/7620)^1/2 = x

x = .407 m = 40.7 cm

The book got 35.9 cm. Can anyone find my mistake, or if I did it completely wrong, can anyone tell me how to do it? Thanks in advance!
 
Physics news on Phys.org
I did a problem similar to this one from a different textbook with the values (everything is worded the same, except these values are in place):

m1: 2.00 kg
m2: 5.00 kg
v1: 10.0 m/s
v2: 3.00 m/s
K=1120 N/m

I approached this problem differenty when I drew the before and after reference frames of what it should look like. And I got:

m1v1+m2v2=(m1+m2)vf
2(10)+3(5)=7vf
vf=5 m/s

I went on to the energy conservation thing, but tweaked it according to my reference frame:

m1v1^2 + m2v2^2 + kx0^2 = (m1+m2)vf^2 + kx^2

Initially, the spring was compressed none, so:

2(10^2) + 5(3^2) = (7)(5^2) + 1120x^2

245 = 75 + 1120x^2

70 = 1120x^2

x= .25 m = 25 cm. I checked the answer for this problem from the different textbook and I got it right. The problem in my original post have values similar to this problem, but the answer is 35.9 cm. Any chance that they might've messed up? I proceeded to use these very steps on the problem above and wound up with an answer around 24.9 cm, or is it a HUGE coincidence that the way I did it for this problem is wrong, but wound up with the right answer anyways?
 

Similar threads

Replies
5
Views
3K
  • · Replies 10 ·
Replies
10
Views
4K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
8
Views
4K
  • · Replies 14 ·
Replies
14
Views
3K
Replies
4
Views
1K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 5 ·
Replies
5
Views
6K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K