Completely Inelastic Collision: A block and a block with a spring

Click For Summary
The discussion revolves around solving a physics problem involving a completely inelastic collision between two blocks, one of which is attached to a spring. The user attempts to calculate the maximum compression of the spring after the collision by applying conservation of momentum and energy equations. Despite repeated attempts, they struggle with determining the correct initial and final kinetic energies, which are critical for finding the spring compression. The main issue identified is an error in calculating these energies, which leads to incorrect results. Accurate energy calculations are essential to solve the problem correctly.
brendan3eb
Messages
53
Reaction score
0
I actually have two problems that I have been struggling with that are very similar, so I believe that I am making the same incorrect assumption for both problems, but I am at that point where I have been re-trying the problems for so long that I do not think I am going to find my mistake.

Homework Statement


A block of mass m1=2.0 kg slides along a frictionless table with a speed of 10 m/s. Directly in front of it, and moving in the same direction, is a block of mass m2=5.0 kg moving at 3.0 m/s. A massless spring with spring constant k=1120 N/m is attached to the near side of m2, as shown in Fig. 10-35. 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 completely inelastic at this point.)


Homework Equations


m1v1i+m2v2i=(m1+m2)v
K2-K1=W
W=(1/2)kx^2

The Attempt at a Solution


m1=2.0 kg
v1i=10 m/s
m2=5.0 kg
v2i=3 m/s
k=1120 N/m
Pi=Mv
m1v1i+m2v2i=(m1+m2)v1
v1=(m1v1i+m2v2i)/(m1+m2)
v1=(2*10+5*3)/(2+5)=5 m/s
K2-K1=W
(1/2)m1(v1)^2-(1/2)m1(v1i)^2=-(1/2)kx^2
(1/2)s cancel out
(2)(5)^2-(2)(10)^2=-(1120)(x)^2
x^2=.134
x=.37 m

It looks as though all the units cancel out right and I do not have to convert any the units.
 
Physics news on Phys.org
re-tried it a few more times to no avail
 
Your problem is with computing the energies.

Before the collision, what is the total kinetic energy of the two blocks?
After the collisiion, what is the total kinetic energy of the two blocks?

The difference equals (1/2)kx^2, but both your initial and final KE's are wrong.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 9 ·
Replies
9
Views
2K
Replies
5
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
17
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 7 ·
Replies
7
Views
5K
Replies
21
Views
842
Replies
3
Views
2K
Replies
2
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K