What is the momentum of a proton traveling at v = 0.85c?
logistic_guy Senior Member Joined Apr 17, 2024 Messages 1,552 Jul 3, 2025 #1 What is the momentum of a proton traveling at v=0.85c\displaystyle v = 0.85cv=0.85c?
K khansaheb Senior Member Joined Apr 6, 2023 Messages 1,120 Jul 3, 2025 #2 logistic_guy said: What is the momentum of a proton traveling at v=0.85c\displaystyle v = 0.85cv=0.85c? Click to expand... Please show us what you have tried and exactly where you are stuck. Please follow the rules of posting in this forum, as enunciated at: READ BEFORE POSTING Please share your work/thoughts about this problem
logistic_guy said: What is the momentum of a proton traveling at v=0.85c\displaystyle v = 0.85cv=0.85c? Click to expand... Please show us what you have tried and exactly where you are stuck. Please follow the rules of posting in this forum, as enunciated at: READ BEFORE POSTING Please share your work/thoughts about this problem
logistic_guy Senior Member Joined Apr 17, 2024 Messages 1,552 Jul 7, 2025 #3 I have never seen beauty like this. p=mv1−v2c2=1.67×10−27(0.85)(299792458)1−(0.85)2=8.08×10−19 kg⋅m/s\displaystyle p = \frac{mv}{\sqrt{1-\frac{v^2}{c^2}}} = \frac{1.67 \times 10^{-27}(0.85)(299792458)}{\sqrt{1 - (0.85)^2}} = 8.08 \times 10^{-19} \ \text{kg} \cdot \text{m/s}p=1−c2v2mv=1−(0.85)21.67×10−27(0.85)(299792458)=8.08×10−19 kg⋅m/s
I have never seen beauty like this. p=mv1−v2c2=1.67×10−27(0.85)(299792458)1−(0.85)2=8.08×10−19 kg⋅m/s\displaystyle p = \frac{mv}{\sqrt{1-\frac{v^2}{c^2}}} = \frac{1.67 \times 10^{-27}(0.85)(299792458)}{\sqrt{1 - (0.85)^2}} = 8.08 \times 10^{-19} \ \text{kg} \cdot \text{m/s}p=1−c2v2mv=1−(0.85)21.67×10−27(0.85)(299792458)=8.08×10−19 kg⋅m/s