#### To determine

**To show:** The relation between the force *F* acting on a body and the mass of the body *m*, acceleration a, speed of light c, and velocity v as F=m0a1−v2c23.

#### Answer

Thus, the relationship between force acting on the body and initial mass, velocity and acceleration is shown as F=m0a1−v2c23.

#### Explanation

**Given:**

The given equation for the mass of particle from the theory of relativity is as below.

m=m01−v2c2 (1)

The given equation for the rate of change of momentum is as below.

F=ddt(mv) (2)

Simplify the equation (2) and differentiate it with respect to variable t.

F=vdmdt+mdvdt

Substitute m01−v2c2 for m, and *a* for dvdt.

F=vddt(m01−v2c2)+ma=vm0ddt((1−v2c2)−1/2)+ma=-vm012((1−v2c2)−3/2)(0−2vc2dvdt)+ma=-vm012((1−v2c2)−3/2)(0−2vc2a)+ma

F=vm0((1−v2c2)−3/2)(vc2a)+maF=vm0((1−v2c2)−3/2)(vc2a)+ma

Substitute m01−v2c2 for m.

F=vm0((1−v2c2)−3/2)(vc2a)+m01−v2c2a=v2m0a(1−v2c2)3/2(1c2)+m01−v2c2a=m0a(1−v2c2)[v2c2(1−v2c2)+1]=m0a(1−v2c2)[v2+c2(1−v2c2)c2(1−v2c2)]

F=m0a(1−v2c2)[c2c2(1−v2c2)]=m0a(1−v2c2)[1(1−v2c2)]F=m0a(1−v2c2)3/2

Thus, the relationship between force acting on the body and initial mass, velocity and acceleration is shown as F=m0a1−v2c23.