Question

An unstable particle at rest spontaneously breaks into two fragments of unequal mass. The mass of the first fragment is 3.00 10-28 kg, and that of the other is 1.51 10-27 kg. If the lighter fragment has a speed of 0.834c after the breakup, what is the speed of the heavier fragment? (Assume the speeds are measured in a frame at rest with respect to the original particle.)

Answer 1

Answer:0.478 c

Step-by-step explanation:

Given

mass of lighter Particle
`(m_1)=3* 10^(-28) kg`

mass of heavier Particle
`(m_2)=1.51* 10^(-27) kg`

speed of lighter particle
`(v_1)=0.834 c`

Let speed of heavier particle
`=v_2`

and Momentum of the particle is given by

`P=\frac{mv}{\sqrt{1-((v)/(c))^2}}`

`P_1=\frac{m_1v_1}{\sqrt{1-((v_1)/(c))^2}}`

`P_1=\frac{3* 10^(-28)* 0.834 c}{\sqrt{1-((0.834 c)/(c))^2}}`

`P_1=8.219* 10^(-28) kg c`

`P_2=\frac{m_2v_2}{\sqrt{1-((v_2)/(c))^2}}`

as momentum is conserved therefore
`P_1=P_2`

`8.219* 10^(-28) kg c=\frac{1.51* 10^(-27)* v_2}{\sqrt{1-((v_2)/(c))^2}}`

`v_2=0.478 c`