An unstable particle at rest breaks up into two fragments of unequal mass. The m

Question

An unstable particle at rest breaks up into two fragments of unequal mass. The mass of the

lighter fragment is 2.50 Å~ 10–28 kg, and that of the heavier fragment is 1.67 Å~ 10–27 kg. If the

lighter fragment has a speed of 0.893c after the breakup, what is the speed of the heavier

fragment?

Part B

what was the rest mass of the original unstable particle?

Question 4 options:

m=1.15 x 10-27 kg

m=2.3 x 10-27 kg

m=1.35 x 10-23 kg

m=2.7 x 10-23 kg

m=1.15 x 10-27 kg

m=2.3 x 10-27 kg

m=1.35 x 10-23 kg

m=2.7 x 10-23 kg

Explanation / Answer

Use conservation of momentum. Since there are no external forces, the total final momentum must equal the total initial momentum (which is zero). This means the two particles must have momenta that are equal in magnitude but opposite in direction.

P1 = P2
(m1)(v1)/(1-(v1/c)²) = (m2)(v2)/(1-(v2/c)²)

NOTE: This is the relativistic equation for momentum. The classical equation ((m1)(v1)=(m2)(v2)) WON'T WORK when the speeds are a significant fraction of "c".

They give you these numbers:
m1 = 2.5×10^-28 kg
v1 = 0.893c
m2 = 1.67×10^-27 kg

Therefore,

We will get equation,

C2/V22 =((m2/m1)2*( (C2/V12)-1)) +1

By putting all the value,

We will get,

V2 = 0.7647 *108 m/s

B) The rest mass of the original unstable particle is,

M = 2.3 *10-27 kg

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