What is the duration of impact for a steel ball hitting a plastic box?

  • Thread starter Thread starter MEriksson
  • Start date Start date
  • Tags Tags
    Shock
Click For Summary
SUMMARY

The impact duration of a steel ball with a diameter of 0.51mm and mass of 0.54kg, dropped from a height of 0.74m onto a plastic box, is estimated to be approximately 0.8 seconds based on calculations involving average impact force and elastic collision assumptions. The impact speed is recorded at 3.86m/s, resulting in a force of 4J. The analysis incorporates the principles outlined in UL standard 60068-2-27 and IEC standard 60950-1, emphasizing the importance of understanding the stiffness of the materials involved and the behavior of the plastic during impact.

PREREQUISITES
  • Understanding of Newton's Second Law (F=ma)
  • Familiarity with elastic and inelastic collisions
  • Knowledge of impact force calculations
  • Basic principles of wave propagation in solids
NEXT STEPS
  • Research the specifics of UL standard 60068-2-27 and IEC standard 60950-1
  • Learn about calculating impact forces in elastic collisions
  • Explore the properties of materials, particularly plastics, under impact conditions
  • Study wave propagation in solids and its relevance to impact analysis
USEFUL FOR

Engineers, material scientists, and anyone involved in impact testing or analyzing the behavior of materials under stress will benefit from this discussion.

MEriksson
Messages
1
Reaction score
0
Hello, I'm trying to estimate how long the impact of a steel ball hitting a plastic box would be.

The actual case: you drop a steal ball d=0.51mm, m=0.54kg from h=0.74m onto a box made of plastic, the impact speed is 3.86m/s and force 4J. I want to know the deceleration speed during impact or how long the impact lasts so i can use F=ma, to get the other. Anyone have a estimate for these numbers?

The reason: I'm comparing UL standard 60068-2-27 with IEC standard 60950-1 :P
 
Engineering news on Phys.org
impulse = Favg * t = m * v
Favg = avg. impact force
t = time of contact or collision
m = mass of falling ball
v = change in vel of ball after and before collision

it must be assumed if it is an elastic or inelastic collision to find out velocity of ball after collision using conservation of momentum considering losses.

impact force has to be found out somehow. it has to be greater than m* g . it is usually taken to be twice of static force due to mass mg.

from the above data you can find out collision time.

i considered elastic collision.
v = 3.86ms-1 for both before and after collision. hence net change is 7.72ms-1.
Favg = 2*0.54*9.81

gives t = 0.8sec
 
It depends only on the stiffness of both objects if the collision is elastic. The answer is in the low ms range, or lower with your small objects, and the shock often in the 10,000G range. In the plastic range, it depends on plastic behaviour, sure.

Not obvious: the "elastic behaviour" involves generally the propagation of waves in the solids. Seriously difficult to determine with objects that were not designed for their shock behaviour.

I doubt about d=0.51mm, m=0.54kg.
 

Similar threads

  • · Replies 8 ·
Replies
8
Views
21K
  • · Replies 2 ·
Replies
2
Views
3K