Can someone please help me with this? I keep trying but I am not getting the rig
ID: 1374532 • Letter: C
Question
Can someone please help me with this? I keep trying but I am not getting the right answer. Thanks!
Two carbon atoms, each having mass 2.0x10^-26 Kg are joined together by a spring-like carbon-carbon bond. When the bond is at the equilibrium length, the 1st carbon atom moves toward the 2nd one with a speed of 500 m/s. The 2nd carbon atom remsins stationary all he time. When the speed of the 1st carbon atom becomes zero, the bond is stretched 0.038 nm. What is the bond's spring constant (N/m).
I keep getting 3.46, but others say it's 2.7, I don't know who is right. Thanks!
Explanation / Answer
m = 2 x 10-26 kg
V = 500 m/s
x = 0.038 nm = 3.8 x 10-11 m
k = spring constant.
using the formula
1/2 k X2 = 1/2 m v2
k = (m) v2/X2
inserting the above values
k = (2 x 10-26 ) 5002/(3.8 x 10-11 )2 = 3.46
the answer has to be 3.46