We left off with a set of problems to do with applications of the rocket principle and Newton's 3rd law. To recap, the problems were as follows:
1) A liquid fuel rocket traveling in space has a speed of 3, 000 m/s. Its engines are turned on and fuel is ejected in a direction opposite to the rocket's motion at a speed of 5,000 m/s relative to the rocket. Find the speed of the rocket once its mass is reduced to one half its mass before ignition.
2) A toy balloon is found to move at a speed of 10 m/s after all its air is expelled. Assuming its initial speed = 0.001 m/s estimate the mass of the balloon alone if its initial mass is 0.1g and the velocity of ejected air relative to a stationary frame is found to be 4 m/s.
3)A rifle of mass 5 kg fires a bullet of mass 0.002 kg which attains a velocity of 200 m/s. Find the velocity of recoil of the rifle.
4)A toy trolley of 1 kg mass has a wooden support affixed to one end and rests on a frictionless track. An arrow of mass 100 g is fired at the wood support with a velocity of 20 m/s. Find the velocity with which the trolley moves off with the arrow embedded in it after impact.
5) A radionuclide of mass 210u (u denotes atomic mass units) undergoes fission with the release of an alpha-particle of mass 4u. Find the kinetic energy of the residual nucleus.
Now, the Solutions:
1)v(f) – v(i) = v’ ln (Mi)/M(f))
and we need v(f) with v(i) = 3 000 m/s and v’ = 5,000 m/s. The mass reduction condition means ‘M(i)/M(f) = 1/ (½) = 2 so ln(2)
Therefore: v(f) = v’ ln (Mi)/M(f)) + v(i) = (5,000 m/s) ln(2) + 3 000 m/s
Or: v(f) = 3470 m/s + 3,000 m/s = 6, 470 m/s
2) Here, M(i) = 0.1 g = 0.0001 kg and we are to find M(f) the actual balloon mass once all the air has vacated. We are giving an estimate so let v(i) ~ 0 m/s and therefore,
v(f) – v(i) = 10 m/s while v’ = 4 m/s.
Then: ln (Mi)/M(f)) = [v(f) – v(i)] / v’ = 10 m/s/ 4 m/s = 2.5
Using anti-logarithms, this mean in order to obtain 2.5 from the log of the mass ratio, one needs to have M(i)/M(f)) = 12.2. So we only need to solve this for M(f):
M(f) = M(i)/ 12.2 = 0.0001 kg/ 12.2 = 8.2 x 10^-6 kg
Or, M(f) = mass (balloon) = 0.0082 g
2) See diagram showing situation:
By conservation of linear momentum:
mu1 + Mu2 = mv1 + Mv2
where m = bullet mass (0.002 kg) and M = rifle mass (5 kg)
Before firing: u1 = u2 = 0 (bullet and rifle at rest)
After firing: bullet has v1 = 200 m/s, and we must find v2.
Total momentum before firing = total momentum after firing, so:
0 = mv1 + Mv2, or:
v2 = -(mv1)/M = - (0.002 kg)(200 m/s)/ 5 kg = - 0.08 m/s
and the negative sign means the rifle recoils in the opposite direction to the bullet’s
4)See diagram to show situation after arrow embeds in toy trolley.
Momentum of the system before is just the momentum of the arrow since the trolley is at rest, thus:
Momentum before collision = m(A) v(A) = 0.1 kg (20 m/s) = 2 kg m/s
Momentum after collision = [m(A) + m(T)] V
Where V denotes the common velocity of the coupled masses, m(A) and m(T)
Therefore total momentum before = total momentum after so:
m(A) v(A) = 2 kg m/s = [m(A) + m(T)] V = (1.1 kg) V
Then: V = 2 kg m/s/ 1.1 kg = 1.8 m/s
5) Original mass of radionuclide = 210 u, and m(He) = 4 u (mass of alpha particle)
The fission of the alpha particle can be treated as a collision with recoil, so that before the fission occurs the original momentum of the system is:
m(He) v(He) + m(R)v(R) = 0
where ‘R’ denotes the residual nucleus. We can write this as a proportion:
v(He)/ v(R) = - m(R)/ m(He)
Squaring both sides we get:
{v(He)/ v(R)}^2 = {m(R)/ m(He)}^2
Now, the ratio of kinetic energies can be posed as:
E(R)/ E = [½m(R) v(R)^2]/ [½m(He) v(He)^2] = m(R) v(R)^2/ m(He) v(He)^2
But from earlier: v(R)^2/ v(He)^2 = m(He) ^2/ m(R)^2
So we can write the new proportion:
E(R)/ E = m(He)/ m(R) = (4u/ 206u)E = 2E/ 103
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