Question

(a) An unstable particle with a mass equal to 3.34 ✕ 10⁻²⁷ kg is initially at rest. The particle decays into two fragments that fly off with velocities of 0.977c and −0.851c, respectively. Find the masses of the fragments. (Hint: Conserve both mass–energy and momentum.)

m(0.977c) =	kg
m(-0.851c) =	kg

(b) An unstable particle at rest breaks up into two fragments of unequal mass. The mass of the lighter fragment is equal to 4.00 ✕ 10⁻²⁸ kg and that of the heavier fragment is 1.63 ✕ 10⁻²⁷ kg. If the lighter fragment has a speed of 0.893c after the breakup, what is the speed of the heavier fragment?
c

Answer 1

A) here we have ,

step 1)

1=1/1-(0.977)²=4.6896

2=1/1-(0.851)²=1.9042

step 2) from conservation of energy we have

E1+E2=Etotal

1c²m1+​​​​​​2c²m2=mtotalc²

1.9042m2+4.6896m1=3.34*10^-27kg............1)

m2+2.4628m1=1.754*10^-27...........2)

now using eqn 1)

4.6896*0.977c*m1=1.9042*0.851c*m2

m1=0.3466m2............3)

now putting the value of m1 in eqn 2

m2+0.8536m2=1.754*10^-27kg

1.8536m2=1.754*10^-27kg

m2=9.46*10^-28kg

from eqn 3)

m1=0.3466*9.46*10^-28kg

m1=3.28*10^-28kg

so m(0.977c)=3.28*10^-28kg ( answer) or 3.35*10^-28 kg

and m(-0.851c)=9.46*10^-28kg( answer)

b) we know the formula

p=mv/(1-v²/c²)

from conservation of momentum we have

p1=p2=p

for lighter fragment

p=m1*0.893c/1-0.893²=1.9842m1c......................1)

for heavier fragment

p=m2v2/1-v2²/c²

p²(1-v²/c²)=m2²v2²

p²=v2²(m2²+p²/c²)

so now we have

v2=p²/m2²+p²/c².........2)

from eqn 1

p=1.9841cm1 , so now eqn 2 becomes

v2=(1.9841m1c)²/m2²+1.9841²m1²=1.9841*4*10^-28kg/(1.63*10^-27 )²+1.9841²*(4*10^-28)²=0.438c

so the answer is 0.438 c or 0.44 c