Calculating the Compression Distance in an Inelastic Collision with a Spring

AI Thread Summary
In an inelastic collision involving a ball of clay and a block compressing a spring, the initial calculations using kinetic energy and spring work led to an incorrect distance of 4.5 m. The correct approach requires first applying conservation of momentum to determine the speed after the collision, as energy is not conserved in inelastic collisions. After finding the post-collision speed, energy conservation can be used to calculate the maximum compression of the spring. The final correct compression distance was found to be 0.266 m. This step-by-step method clarified the problem-solving process for the participants.
dancing123
Messages
4
Reaction score
0
A ball of clay of mass m = 0.5 kg strikes a block of mass M = 8.0 kg which slides on a frictionless table as it compresses a spring with spring constant k = 60 N/m. The initial speed of the ball of clay is v = 12 m/s. The spring is initially at its relaxed length.



W= delta K = 1/2mv^2
W = 1/2kx^2
These are the equations I tried using.




I tried setting delta K equal to the Work but the answer it gave me was too big. I got 4.5 m as my answer and it's wrong. Can someone show me how to do this problem step by step?
 
Physics news on Phys.org
welcome to pf!

hi dancing123! welcome to pf!

(have a delta : ∆ and try using the X2 icon just above the Reply box :wink:)
dancing123 said:
I tried setting delta K equal to the Work but the answer it gave me was too big. I got 4.5 m as my answer and it's wrong.

Show us your full calculations, and then we'll see what went wrong, and we'll know how to help! :smile:
 
Hi! Thank you for helping me. :)

This is what I tried to do:
W = 1/2kx^2
K = 1/2mv^2

1/2kx^2 = 1/2(m+M)v^2
x^2 = [(m+M)v^2]/k
x = sqrt{[(m+M)v^2]/k}
x = sqrt{[(0.5+8.0)12^2]/60}
x = 4.5166 m
 
ah, you have to split the problem into two parts …

the first is an inelastic collision so there's no conservation of energy, instead you use conservation of momentum to find the speed immediately after the collision …

from that time on, you can use energy :smile:
 
Sorry I got back to you so late but thank you so much. I was able to figure it out :) My answer turned out to be 0.266 m, which was the correct answer.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top