Question
Whether or not an object (such as a plate, glass, or bone) breaks upon impact depends on the average force exerted on that object by the surface. When a 1.2-kg glass figure hits the floor, it will break if it experiences an average force of 330 N. When it hits a tile floor, the glass comes to a stop in 0.015 s. From what minimum height must the glass fall to experience sufficient force to break? How would your answer change if the figure were falling to a padded or carpeted surface? Explain.
Final Answer
With a padded surface, the object could be dropped from a greater height before breaking. The padding increases the time of deceleration, which in turn decreases the net force the object experiences.
Solution video
OpenStax College Physics for AP® Courses, Chapter 8, Problem 6 (Test Prep for AP® Courses)
vote with a rating of
votes with an average rating of
.
Prev
Calculator Screenshots
Video Transcript
This is College Physics Answers with Shaun Dychko. A glass object is going to be dropped from some height that we have to figure out such that it will just begin to break and so what minimum height is could this thing be dropped from such that it would break is the question here and we are told that an average force of 330 newtons would break the glass object, it has a mass of 1.2 kilograms and we'll assume that it's in contact with the floor that it hits for 0.015 seconds. So to figure out this height that it will be dropped from, we are first going to figure out what is the minimum speed of contact with the floor such that the force on the object exceeds or equals 330 newtons? And so we are gonna use change in momentum over time equal to the net force formula to figure out what the speed should be because this v i multiplied by the mass will be the initial momentum and the final momentum will be zero because it will come to a stop on the floor. So we translate the information from this question into variables so we have t is 0.015 seconds, the force due to the floor— the normal force— is 330 newtons and the mass is 1.2 kilograms. So we know that net force is the rate of change in momentum and we also know that net force is the up forces minus the down forces so there's an up force F N and minus gravity, which is mg downwards and so we equate this and this since they are both equal to the net force so F N minus mg equals change in momentum over time. The change in momentum is mv f minus mv i— v f being zero so I can replace that term with 0— so we have negative mass times initial velocity and that is the velocity just before contacting the floor divided by Δt equals the normal force minus mg. So our job is to find v i here and then from that, we will figure out what height should this thing be dropped such that it would have a speed v i when it contacts the floor. So we multiply both sides by Δt, divided by m negative and so that means v i equals mg minus F N over m times Δt; this negative sign switched the minus between these two terms so F N is now minus and mg is now positive. Okay! So this speed then is 1.2 kilograms times 9.8 meters per second squared minus 330 newtons all divided by 1.2 kilograms times 0.015 seconds and that is negative 3.978 meters per second. So from what height could this object be dropped such that its speed here at the floor is negative 3.978 meters per second? We'll use this formula to figure that out. This is the kinematics formula and I am gonna replace these variables with things from our picture in a second. So we know that final speed squared equals initial speed squared plus 2 times acceleration times change in position but our final speed in this formula is v i in this picture— it's the speed just before hitting the floor— it's the initial speed in the context of talking about change in momentum so it's labeled v i. So I am replacing v f from our formula with v i v initial in this formula is v i prime, which is zero and I have written zero here and then plus 2 times the acceleration— the acceleration is negative g— and the change in position is the final y-position minus the initial y-position so this is minus 2g times 0— the final position being zero, we take the floor to be the reference level— and then minus the initial height that we are trying to find and so this works out to v initial squared equals 2g times initial height, divide both sides by 2g and now we have the initial height. So that's v initial squared over 2g so that's negative 3.978 meters per second squared divided by 2 times 9.8 meters per second squared and that is 0.81 meters. So this is the height beyond which the object will break. The next part of the question is asking what would happen if you had a padded surface? So a padded surface will increase the amount of time that the floor is in contact with the object while slowing it down, it will increase the time during which it is decelerating and as this time increases, the net force will decrease for a given change in momentum. So assuming the object is dropped from the same initial height, the change in momentum would be the same but the net force being that same change of momentum but now divided by new Δt and putting primes on these to say this is for a new scenario. So Δt prime is big now— it's increased due to this padded surface increasing the duration of deceleration— the net force will decrease since the net force is inversely proportional to this time. So this means a greater change in momentum would be tolerated before breaking so you could... drop from a higher height before breaking if you wanted to.