Promise to rate you if you can help me with this by 12:00 AMPST. my answers are
ID: 1679161 • Letter: P
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Promise to rate you if you can help me with this by 12:00 AMPST. my answers are 1.79e-28 and 1.10e-27 but it says I'm off by10%. Down to my last submission!An unstable particle with a mass equal to 3.34 x10-27 kg is initially at rest. The particle decays intotwo fragments that fly off with velocities of 0.987c and -0.870c, respectively. Find the masses of thefragments. (Hint: Conserve both mass-energy andmomentum.)
m0.987c = kg m-0.870c = kg Promise to rate you if you can help me with this by 12:00 AMPST. my answers are 1.79e-28 and 1.10e-27 but it says I'm off by10%. Down to my last submission!
An unstable particle with a mass equal to 3.34 x10-27 kg is initially at rest. The particle decays intotwo fragments that fly off with velocities of 0.987c and -0.870c, respectively. Find the masses of thefragments. (Hint: Conserve both mass-energy andmomentum.)
m0.987c = kg m-0.870c = kg
Explanation / Answer
Conservation of linear momentum : m1*v1/(1-(v1/c)^2) = m2*v2/(1-(v2/c)^2) (1) => m2/(1-(v2/c)^2) = m1/(1-(v1/c)^2) *v1/v2 Conservation of mass-energy Mo.c^2 = m1/(1-(v1/c)^2) * c^2 + m2/(1-(v2/c)^2) *c^2= m1.c^2/(1-(v1/c)^2) *(1 + v1/v2) => m1 = Mo*(1-(v1/c)^2) /(1 + v1/v2) = 2.56x10^-28 kg (1)=> m2 = 8.91 x10^-28 kg
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