A rocket with a mass M moves along an x axis at the constant speed vi=40 m/s. A small explosion separates the rocket into a rear section (of mass m1) and a front section; both sections move along the x axis. The relative speed between the rear and front sections is 20 m/s. What are (a) the minimum possible value of final speed vf of the front section and (b) for what limiting value of m1 does it occur? (c) What is the maximum possible value of vf and (d) for what limiting value of m1 does it occur?

Question

A rocket with a mass M moves along an x axis at the constant speed vi=40 m/s. A small explosion separates the rocket into a rear section (of mass m1) and a front section; both sections move along the x axis. The relative speed between the rear and front sections is 20 m/s. What are (a) the minimum possible value of final speed vf of the front section and (b) for what limiting value of m1 does it occur? (c) What is the maximum possible value of vf and (d) for what limiting value of m1 does it occur?

i tried to use the conservation of momentum?

M(40m/s)=m1fv1f+m2fv2f

which simplifies to v2f=(40M+20mi)/M

Now there is three unknowns...i think its asking for a value though...

Answer 1

two more equations are:

m1+m2=M
v2-v1=20m/s

Answer 2

To find the minimum and maximum possible values of the final speed (vf) of the front section and the corresponding value of m1, you can use conservation of momentum and energy.

(a) To find the minimum possible value of vf, consider the scenario where the rear section (of mass m1) comes to a complete stop. In this case, the relative speed between the two sections is equal to the final speed of the front section. So, relative speed = vf = 20 m/s. Since the final speed of the rear section is 0 m/s, we can write the equation for conservation of momentum:

M * vi = m1 * vf + m2 * 0

Substituting the given values, we have:

M * 40 m/s = m1 * 20 m/s

Simplifying:

2M = m1

Therefore, the minimum possible value of m1 is 2M, and the minimum possible value of vf is 20 m/s.

(b) Now let's find the maximum possible value of vf by considering the scenario where all the mass is concentrated in the rear section (m1 = M). In this case, the rear section moves with the initial velocity (vi) of the rocket, and the front section comes to a complete stop. So, the relative speed is equal to the initial speed of the rocket. Therefore, relative speed = 40 m/s, and we can again use the conservation of momentum:

M * vi = M * vf + M * 0

Simplifying:

40M = M * vf

Dividing both sides by M:

vf = 40 m/s

Thus, the maximum possible value of vf is 40 m/s, and it occurs when m1 = M.

In summary:
(a) The minimum possible value of vf is 20 m/s, and it occurs when m1 = 2M.
(b) The maximum possible value of vf is 40 m/s, and it occurs when m1 = M.