Question

three students each built a device to protect an egg when it was dropped to the ground from a height of 11.5 m. to keep the egg from breaking, each device needed to reduce the force applied to the egg to be less than 25 n. the students dropped the devices with the eggs inside and recorded some data for each device, as shown in the table below. for each student identify whether or not the egg inside the device broke when the device hit the ground, by typing either \egg broke\ or \egg did not break\ in the results column.

student	mass of egg (kg)	velocity of egg before impact (m/s)	time to stop egg (s)	result
2	0.05	14	0.05
3	0.05	14	0.10

Explanation:

Step1: Recall impulse - momentum theorem

The impulse - momentum theorem is $F\Delta t=\Delta p = m\Delta v$. Here, the initial velocity of the egg before impact is $v_i$ and the final velocity $v_f = 0$. So, $\Delta v=v_i - v_f=v_i$. The force exerted on the egg is $F=\frac{m\Delta v}{\Delta t}$.

Step2: Calculate the force for student 1

Given $m = 0.05\ kg$, $\Delta v=14\ m/s$, $\Delta t = 0.01\ s$. Then $F_1=\frac{m\Delta v}{\Delta t}=\frac{0.05\times14}{0.01}=\frac{0.7}{0.01}=70\ N$. Since $70\ N>25\ N$, the egg broke.

Step3: Calculate the force for student 2

Given $m = 0.05\ kg$, $\Delta v = 14\ m/s$, $\Delta t=0.05\ s$. Then $F_2=\frac{m\Delta v}{\Delta t}=\frac{0.05\times14}{0.05}=14\ N$. Since $14\ N<25\ N$, the egg did not break.

Step4: Calculate the force for student 3

Given $m = 0.05\ kg$, $\Delta v = 14\ m/s$, $\Delta t = 0.10\ s$. Then $F_3=\frac{m\Delta v}{\Delta t}=\frac{0.05\times14}{0.10}=7\ N$. Since $7\ N<25\ N$, the egg did not break.

Answer:

Student 1: Egg Broke
Student 2: Egg did not Break
Student 3: Egg did not Break