In the previous section, we completed a discussion on power. In this section we will see collisions
• In a previous chapter, we saw the Law of conservation of Momentum (Details here)
• In a previous section of the present chapter, we saw the law of conservation of Energy (Details here)
• We are now going to apply these two laws to a common phenomenon that we encounter in our day-to-day life, which is: collisions
1. Consider the row 'a', column '1' in fig.6.64 below:
• Two objects A and B of masses mA and mB move towards each other
• Their initial velocities are respectively vAi and vBi
♦ vAi is 'velocity of A (initial)'
♦ vBi is 'velocity of B (initial)'
2. Row 'a', column '2' shows the instant at which collision takes place
• We see that both the masses undergo deformation
♦ A is compressed due to the impact
♦ B is also compressed due to the impact
3. Assume that, both the masses are made of 'perfectly elastic' materials
• So they regain their original shapes after impact
• After impact, they move with velocities vAf and vBf
♦ vAf is 'velocity of A (final)'
♦ vBf is 'velocity of B (final)'
• This is shown in row 'a', column '3'
4. This situation is just like compressing and releasing a spring
• The energy used up for compressing the spring will be obtained back when the spring is released
• So there is no net change in energy
5. That means, in an elastic collision,
Total energy before collision = Total energy after collision
6. But the two masses are at the same level
• The levels do not change even after collision
• So they have the same gravitational potential energy before and after collision
• Thus, we need not consider the gravitational potential energy
7. So the result in (5) becomes:
■ In an elastic collision,
Total kinetic energy before collision = Total kinetic energy after collision
That is:
$\mathbf\small{\frac{1}{2}m_A v_{Ai}^2+\frac{1}{2}m_B v_{Bi}^2=\frac{1}{2}m_A v_{Af}^2+\frac{1}{2}m_B v_{Bf}^2}$
Thus we get:
$\mathbf\small{m_A v_{Ai}^2+m_B v_{Bi}^2=m_A v_{Af}^2+m_B v_{Bf}^2}$
8. If there is no external force acting on the system, the law of conservation of momentum is valid
So we have:
$\mathbf\small{m_A v_{Ai}+m_B v_{Bi}=m_A v_{Af}+m_B v_{Bf}}$ (Details here)
9. Equations in (7) and (8) can be used together, to solve problems in 'one-dimensional elastic collisions'
Solved example 6.33
Two objects undergo a one-dimensional elastic collision (Fig.6.65 below)
Calculate their velocities after collision. Given that: mA = 0.5 kg, mB = 3.5 kg, vAi = 4 ms-1, vBi = 0
Solution:
• We are required to find vAf and vBf
1. In an elastic collision, kinetic energy is conserved. So we have:
$\mathbf\small{m_A v_{Ai}^2+m_B v_{Bi}^2=m_A v_{Af}^2+m_B v_{Bf}^2}$
• Substituting the values, we get:
$\mathbf\small{0.5 \times 4^2+3.5 \times 0^2=0.5 v_{Af}^2+3.5 v_{Bf}^2}$
$\mathbf\small{\Rightarrow 16=v_{Af}^2+7 v_{Bf}^2}$
2. If there is no external force acting on the system, the law of conservation of momentum is valid
• So we have:
$\mathbf\small{m_A v_{Ai}+m_B v_{Bi}=m_A v_{Af}+m_B v_{Bf}}$
• Substituting the values, we get:
$\mathbf\small{0.5 \times 4+3.5 \times 0=0.5v_{Af}+3.5 v_{Bf}}$
$\mathbf\small{\Rightarrow 4=v_{Af}+7 v_{Bf}}$
$\mathbf\small{\Rightarrow v_{Af}=4-7 v_{Bf}}$
3. Substituting this value of vAf in (1), we get:
$\mathbf\small{16=(4-7 v_{Bf})^2+7 v_{Bf}^2}$
$\mathbf\small{\Rightarrow 16=16-56v_{Bf}+49v_{Bf}^2+7v_{Bf}^2}$
$\mathbf\small{\Rightarrow 0=-56v_{Bf}+56v_{Bf}^2}$
$\mathbf\small{\Rightarrow 0 =56v_{Bf}(-1+v_{Bf})}$
$\mathbf\small{\Rightarrow v_{Bf}=0\,\, \text{OR}\,\,v_{Bf}=1\,\, \text{ms}^{-1}}$
4. Substituting these values of vBf in (2), we get:
vAf = 4 ms-1 OR vAf = -3 ms-1.
5. Let us analyse the results:
• 'A' moved with a certain velocity and collided with 'B' (which was stationary)
• Since there is no external force, B will certainly move. So the result 'vBf = 0' should be discarded
• We get vBf = 1
• When vBf = 1, vAf is -3
• So the required result is: vAf = -3 vBf = 1
• This result is understandable
• A small mass (mA = 0.5 kg) moves towards the right (∵ vAi is positive)
• It collides with a larger mass (mB = 3.5 kg) which was stationary
• As a result of the collision,
♦ the larger mass move towards the right (∵ vBf is positive)
♦ the smaller mass rebounds and moves to the left (∵ vAf is negative)
1. Applying conservation of kinetic energy, we have:
$\mathbf\small{m_A v_{Ai}^2+m_B v_{Bi}^2=m_A v_{Af}^2+m_B v_{Bf}^2}$
• Where mA is the mass of A
• mB is the mass of B
• vAi is the initial velocity of A
• vBi is the initial velocity of B
• vAf is the final velocity of A
• vBf is the final velocity of B
2. But B is initially at rest. So vBi = 0
• So the equation becomes: $\mathbf\small{m_A v_{Ai}^2=m_A v_{Af}^2+m_B v_{Bf}^2}$
3. Applying the law of conservation of momentum, we have:
$\mathbf\small{m_A v_{Ai}+m_B v_{Bi}=m_A v_{Af}+m_B v_{Bf}}$
• But B is initially at rest. So vBi = 0
• So the equation becomes: $\mathbf\small{m_A v_{Ai}=m_A v_{Af}+m_B v_{Bf}}$
4. The result in (3) is: $\mathbf\small{m_A v_{Ai}=m_A v_{Af}+m_B v_{Bf}}$
• Multiply both sides by vBf. We get: $\mathbf\small{m_A v_{Ai}v_{Bf}=m_A v_{Af}v_{Bf}+m_B v_{Bf}^2}$
$\mathbf\small{\Rightarrow m_B v_{Bf}^2=m_A v_{Ai}v_{Bf}-m_A v_{Af}v_{Bf}}$
5. We can put this in the place of $\mathbf\small{m_B v_{Bf}^2}$ in (2). We get:
$\mathbf\small{m_A v_{Ai}^2=m_A v_{Af}^2+m_A v_{Ai}v_{Bf}-m_A v_{Af}v_{Bf}}$
$\mathbf\small{\Rightarrow v_{Ai}^2= v_{Af}^2+ v_{Ai}v_{Bf}- v_{Af}v_{Bf}}$
$\mathbf\small{\Rightarrow v_{Ai}^2-v_{Af}^2= v_{Bf}(v_{Ai}- v_{Af})}$
$\mathbf\small{\Rightarrow (v_{Ai}+ v_{Af})(v_{Ai}- v_{Af})= v_{Bf}(v_{Ai}- v_{Af})}$
$\mathbf\small{\Rightarrow (v_{Ai}+ v_{Af})= v_{Bf}}$
6. We can use this value of $\mathbf\small{v_{Bf}}$ in (3). We get:
$\mathbf\small{m_A v_{Ai}=m_A v_{Af}+m_B (v_{Ai}+ v_{Af})}$
$\mathbf\small{\Rightarrow m_A v_{Ai}=m_A v_{Af}+m_Bv_{Ai}+ m_Bv_{Af}}$
$\mathbf\small{\Rightarrow (m_A-m_B) v_{Ai}=(m_A + m_B)v_{Af}}$
$\mathbf\small{\Rightarrow v_{Af}=\frac{(m_A-m_B)v_{Ai}}{(m_A + m_B)}}$
• Thus we get the final velocity of A in terms of the three known quantities:
(i) Mass of A
(ii) Mass of B
(iii) Initial velocity of A
7. We can obtain the final velocity of B also
• For that, we put the above obtained value of $\mathbf\small{v_{Af}}$ in (5). We get:
$\mathbf\small{v_{Ai}+ \frac{(m_A-m_B)v_{Ai}}{(m_A + m_B)}= v_{Bf}}$
$\mathbf\small{\Rightarrow \frac{(m_A+m_B)v_{Ai}+(m_A-m_B)v_{Ai}}{(m_A + m_B)}= v_{Bf}}$
$\mathbf\small{\Rightarrow v_{Bf}=\frac{2m_Av_{Ai}}{(m_A + m_B)}}$
• Thus we get the final velocity of B in terms of the three known quantities:
(i) Mass of A
(ii) Mass of B
(iii) Initial velocity of A
8. Some interesting special cases can be analysed using the above expressions
Case 1: The two masses are equal
• When the two masses are equal, we can put: mA = mB = m
• Then from (6) we get: $\mathbf\small{\Rightarrow v_{Af}=\frac{(m-m)v_{Ai}}{(m + m)}=0}$
• From (7) we get: $\mathbf\small{v_{Bf}=\frac{2m\;v_{Ai}}{(m + m)}=v_{Ai}}$
• So 'case 1' can be written in the following 6 steps:
(i) Two bodies 'A' and 'B' collide elastically with each other
(ii) The two masses are equal. That is., mA = mB = m
(iii) Before the collision, 'A' was moving with a velocity vAi
(iv) Before the collision, 'B' was stationary. That is., vBi = 0
[Remember that, the equations in (6) and (7) are valid only if 'B' is initially at rest]
(v) After the collision, 'A' becomes stationary
(vi) After the collision, 'B' begins to move with vAi (the same velocity with which 'A' was moving)
Case 2: One of the two masses is very large when compared to the other
• Let us assume mB >> mA
• Then (mA-mB) will be very nearly equal to -mB
(That is., mA ≃ 0)
• Also (mA+mA) will be very nearly equal to mB
• Then from (6) we get: $\mathbf\small{v_{Af}=\frac{(-m_B)v_{Ai}}{( m_B)}=-v_{Ai}}$
• From (7) we get: $\mathbf\small{v_{Bf}=\frac{2\times 0\times v_{Ai}}{(0 + m_B)}=0}$
• So 'case 2' can be written in the following 6 steps:
(i) Two bodies 'A' and 'B' collide elastically with each other
(ii) mB >> mA
(iii) Before the collision, 'A' was moving with a velocity vAi
(iv) Before the collision, 'B' was stationary. That is., vBi = 0
[Remember that, the equations in (6) and (7) are valid only if 'B' is initially at rest]
(v) After the collision, 'A' begins to move with -vAi
(the same magnitude but opposite direction of the velocity with which 'A' was moving)
(vi) After the collision, 'B' remains stationary
1. Given that: mA = 0.5 kg, mB = 3.5 kg, vAi = 4 ms-1, vBi = 0
2. We have: $\mathbf\small{v_{Af}=\frac{(m_A-m_B)v_{Ai}}{(m_A + m_B)}}$
Substituting the known values, we get:
$\mathbf\small{v_{Af}=\frac{(0.5-3.5)\times 4}{(0.5 + 3.5)}=-3\,\,\text{ms}^{-1}}$ (Same as before)
3. We have: $\mathbf\small{v_{Bf}=\frac{2m_Av_{Ai}}{(m_A + m_B)}}$
Substituting the known values, we get:
$\mathbf\small{v_{Bf}=\frac{2\times 0.5\times 4}{(0.5 + 3.5)}=1\,\,\text{ms}^{-1}}$ (Same as before)
Solved example 6.34
In a nuclear reactor, a neutron of high speed (typically 107 ms-1) must be slowed to 103 ms-1. So that it can have a high probability of interacting with isotope $\mathbf\small{\overset{235}{\underset{92}{}}\text{U}}$ and causing it to fission. Show that a neutron can lose most of it's kinetic energy in an elastic collision with a light nuclei like deuterium which has only a few times the neutron mass. The material making up the light nuclei, usually heavy water (D2O) or graphite, is called a moderator
Solution:
1. We want to slow down a neutron
• We have to achieve this by making the neutron to collide with a deuterium nucleus
• This is shown in fig.6.66 below:
• It is an elastic collision
• We want to show that deuterium can be effectively used for slowing down a neutron
2. The deuterium nucleus is initially at rest. So we will use the formula:
$\mathbf\small{v_{Af}=\frac{(m_A-m_B)v_{Ai}}{(m_A + m_B)}}$
• Where mA is the mass of neutron
• mB is the mass of deuterium
• vAi is the initial velocity of neutron
• vAf is the final velocity of neutron
3. We have: $\mathbf\small{v_{Af}=\frac{(m_A-m_B)v_{Ai}}{(m_A + m_B)}}$
So final kinetic energy of the neutron (KAf)
= $\mathbf\small{\frac{1}{2}m_A\,v_{Af}^2=\frac{1}{2}m_A\,\left[\frac{m_A-m_B}{m_A + m_B}\right ]^2v_{Ai}^2}$
4. Initial kinetic energy of the neutron (KAi) = $\mathbf\small{\frac{1}{2}m_A\,v_{Ai}^2}$
5. Now take the ratio: $\mathbf\small{\frac{K_{Af}}{K_{Ai}}}$
• That is., ratio of the initial kinetic energy of neutron to it's final kinetic energy
• We will denote it as fA
• We have:
$\mathbf\small{f_A=\frac{K_{Af}}{K_{Ai}}=\frac{\frac{1}{2}m_A\,\left[\frac{m_A-m_B}{m_A + m_B}\right ]^2v_{Ai}^2}{\frac{1}{2}m_A\,v_{Ai}^2}=\left[\frac{m_A-m_B}{m_A + m_B}\right ]^2}$
6. So we can write:
Final kinetic energy of the neutron is $\mathbf\small{\left[\frac{m_A-m_B}{m_A + m_B}\right ]^2}$ times it's initial kinetic energy
• The numerator (mA-mB) will be less than the denominator (mA+mB)
• So the ratio $\mathbf\small{\left[\frac{m_A-m_B}{m_A + m_B}\right ]^2}$ is a fraction. It is not a whole number
• So final kinetic energy of the neutron will be certainly less than it's initial kinetic energy
• That means the final velocity of the neutron will be certainly less than the initial kinetic energy
• That means the neutron will be certainly slowed down
7. But knowing that the 'neutron will be slowed down' is not good enough
• We want to know 'how much slowing down' can be achieved
• If the neutron is not slowed down to a required value, the uranium nucleus will not be able to capture it
8. So we proceed as follows:
• Since it is an elastic collision,
Initial kinetic energy of the neutron-deuterium system = It's final kinetic energy
9. But initial velocity of deuterium = 0
• So initial kinetic energy of the system is contributed by the neutron alone
• That is., initial kinetic energy of the system = initial kinetic energy of the neutron
= $\mathbf\small{\frac{1}{2}m_A\,v_{Ai}^2}$
10. Combining (8) and (9), we get:
• Final kinetic energy of the system = Kf = KAi = $\mathbf\small{\frac{1}{2}m_A\,v_{Ai}^2}$
11. But the final kinetic energy of the system is made up by two items:
(i) Final kinetic energy of neutron KAf = $\mathbf\small{\frac{1}{2}m_A\,v_{Af}^2}$
(ii) Final kinetic energy of deuterium KBf = $\mathbf\small{\frac{1}{2}m_A\,v_{Bf}^2}$
• So we can write: Kf = KAi = KAf + KBf
12. Take out the second and third terms. We get: KAi = KAf + KBf
Divide both sides by KAi. We get: $\mathbf\small{1=\frac{K_{Af}}{K_{Ai}}+\frac{K_{Bf}}{K_{Ai}}}$
13. But we have already calculated $\mathbf\small{\frac{K_{Af}}{K_{Ai}}}$ in (5)
• We have: $\mathbf\small{\frac{K_{Af}}{K_{Ai}}=\left[\frac{m_A-m_B}{m_A + m_B}\right ]^2}$
• So the result in (12) becomes: $\mathbf\small{1=\left[\frac{m_A-m_B}{m_A + m_B}\right ]^2+\frac{K_{Bf}}{K_{Ai}}}$
$\mathbf\small{\Rightarrow \frac{K_{Bf}}{K_{Ai}}=1-\left[\frac{m_A-m_B}{m_A + m_B}\right ]^2}$
$\mathbf\small{\Rightarrow \frac{K_{Bf}}{K_{Ai}}=\frac{(m_A+m_B)^2-(m_A-m_B)^2}{(m_A + m_B)^2}}$
$\mathbf\small{\Rightarrow \frac{K_{Bf}}{K_{Ai}}=\frac{(m_A+m_B+m_A-m_B)(m_A+m_B-m_A+m_B)}{(m_A + m_B)^2}}$
$\mathbf\small{\left[\text{Using the identity}:(a^2-b^2)=(a+b)(a-b) \right]}$
$\mathbf\small{\Rightarrow \frac{K_{Bf}}{K_{Ai}}=\frac{(2m_A)(2m_B)}{(m_A + m_B)^2}}$
$\mathbf\small{\Rightarrow \frac{K_{Bf}}{K_{Ai}}=\frac{4m_A\;m_B}{(m_A + m_B)^2}}$
14. So we have two ratios:
(i) From (5) we have: $\mathbf\small{\frac{K_{Af}}{K_{Ai}}=\left[\frac{m_A-m_B}{m_A + m_B}\right ]^2}$
(ii) From (13) we have: $\mathbf\small{\frac{K_{Bf}}{K_{Ai}}=\frac{4m_A\;m_B}{(m_A + m_B)^2}}$
15. Given that: deuterium has only a few times the neutron mass
Assume that mB = 2 mA
(i) Then from 14(i), we get:
$\mathbf\small{\frac{K_{Af}}{K_{Ai}}=\left[\frac{m_A-m_B}{m_A + m_B}\right ]^2=\left[\frac{m_A-2m_A}{m_A + 2m_A}\right ]^2=\frac{1}{9}}$
(ii) From 14(ii) we get:
$\mathbf\small{\frac{K_{Bf}}{K_{Ai}}=\frac{4m_A\;m_B}{(m_A + m_B)^2}=\frac{4m_A\times 2m_A}{(m_A + 2m_A)^2}=\frac{8}{9}}$
16. So from 15(i) we get:
• Final kinetic energy of neutron is 1⁄9 of initial kinetic energy of neutron
⇒ Final kinetic energy of neutron is 1⁄9 of total kinetic energy of the system
• From 15(ii) we get:
Final kinetic energy of deuterium is 8⁄9 of initial kinetic energy of neutron
⇒ Final kinetic energy of deuterium is 8⁄9 of total kinetic energy of the system
• In percentage form:
1⁄9 = 11.1%
8⁄9 = 88.9%
17. So we can write:
• After the collision with deuterium, the neutron loses (100-11.1) = 88.9% of it's initial kinetic energy
♦ That means., the neutron will be left with only 11.1% of it's initial energy
• The lost 88.9% is acquired by the deuterium
18. Instead of deuterium, we can use carbon (graphite) as the moderator
• The formulas are the same that we wrote in (14)
• The only difference is this:
♦ When deuterium is used, we wrote: mB = 2 × mA
♦ When carbon is used, we can write: mB = k × mA
♦ We must put a suitable value for 'k'
• The results obtained are as follows:
After the collision with carbon, the neutron loses 71.6% of it's initial kinetic energy
♦ That means., the neutron will be left with only (100-71.6) = 28.4% of it's initial energy
• The lost 71.6% is acquired by the carbon
• In a previous chapter, we saw the Law of conservation of Momentum (Details here)
• In a previous section of the present chapter, we saw the law of conservation of Energy (Details here)
• We are now going to apply these two laws to a common phenomenon that we encounter in our day-to-day life, which is: collisions
1. Consider the row 'a', column '1' in fig.6.64 below:
 |
| Fig.6.64 |
• Their initial velocities are respectively vAi and vBi
♦ vAi is 'velocity of A (initial)'
♦ vBi is 'velocity of B (initial)'
2. Row 'a', column '2' shows the instant at which collision takes place
• We see that both the masses undergo deformation
♦ A is compressed due to the impact
♦ B is also compressed due to the impact
3. Assume that, both the masses are made of 'perfectly elastic' materials
• So they regain their original shapes after impact
• After impact, they move with velocities vAf and vBf
♦ vAf is 'velocity of A (final)'
♦ vBf is 'velocity of B (final)'
• This is shown in row 'a', column '3'
4. This situation is just like compressing and releasing a spring
• The energy used up for compressing the spring will be obtained back when the spring is released
• So there is no net change in energy
5. That means, in an elastic collision,
Total energy before collision = Total energy after collision
6. But the two masses are at the same level
• The levels do not change even after collision
• So they have the same gravitational potential energy before and after collision
• Thus, we need not consider the gravitational potential energy
7. So the result in (5) becomes:
■ In an elastic collision,
Total kinetic energy before collision = Total kinetic energy after collision
That is:
$\mathbf\small{\frac{1}{2}m_A v_{Ai}^2+\frac{1}{2}m_B v_{Bi}^2=\frac{1}{2}m_A v_{Af}^2+\frac{1}{2}m_B v_{Bf}^2}$
Thus we get:
$\mathbf\small{m_A v_{Ai}^2+m_B v_{Bi}^2=m_A v_{Af}^2+m_B v_{Bf}^2}$
8. If there is no external force acting on the system, the law of conservation of momentum is valid
So we have:
$\mathbf\small{m_A v_{Ai}+m_B v_{Bi}=m_A v_{Af}+m_B v_{Bf}}$ (Details here)
9. Equations in (7) and (8) can be used together, to solve problems in 'one-dimensional elastic collisions'
Solved example 6.33
Two objects undergo a one-dimensional elastic collision (Fig.6.65 below)
 |
| Fig.6.65 |
Solution:
• We are required to find vAf and vBf
1. In an elastic collision, kinetic energy is conserved. So we have:
$\mathbf\small{m_A v_{Ai}^2+m_B v_{Bi}^2=m_A v_{Af}^2+m_B v_{Bf}^2}$
• Substituting the values, we get:
$\mathbf\small{0.5 \times 4^2+3.5 \times 0^2=0.5 v_{Af}^2+3.5 v_{Bf}^2}$
$\mathbf\small{\Rightarrow 16=v_{Af}^2+7 v_{Bf}^2}$
2. If there is no external force acting on the system, the law of conservation of momentum is valid
• So we have:
$\mathbf\small{m_A v_{Ai}+m_B v_{Bi}=m_A v_{Af}+m_B v_{Bf}}$
• Substituting the values, we get:
$\mathbf\small{0.5 \times 4+3.5 \times 0=0.5v_{Af}+3.5 v_{Bf}}$
$\mathbf\small{\Rightarrow 4=v_{Af}+7 v_{Bf}}$
$\mathbf\small{\Rightarrow v_{Af}=4-7 v_{Bf}}$
3. Substituting this value of vAf in (1), we get:
$\mathbf\small{16=(4-7 v_{Bf})^2+7 v_{Bf}^2}$
$\mathbf\small{\Rightarrow 16=16-56v_{Bf}+49v_{Bf}^2+7v_{Bf}^2}$
$\mathbf\small{\Rightarrow 0=-56v_{Bf}+56v_{Bf}^2}$
$\mathbf\small{\Rightarrow 0 =56v_{Bf}(-1+v_{Bf})}$
$\mathbf\small{\Rightarrow v_{Bf}=0\,\, \text{OR}\,\,v_{Bf}=1\,\, \text{ms}^{-1}}$
4. Substituting these values of vBf in (2), we get:
vAf = 4 ms-1 OR vAf = -3 ms-1.
5. Let us analyse the results:
• 'A' moved with a certain velocity and collided with 'B' (which was stationary)
• Since there is no external force, B will certainly move. So the result 'vBf = 0' should be discarded
• We get vBf = 1
• When vBf = 1, vAf is -3
• So the required result is: vAf = -3 vBf = 1
• This result is understandable
• A small mass (mA = 0.5 kg) moves towards the right (∵ vAi is positive)
• It collides with a larger mass (mB = 3.5 kg) which was stationary
• As a result of the collision,
♦ the larger mass move towards the right (∵ vBf is positive)
♦ the smaller mass rebounds and moves to the left (∵ vAf is negative)
• If one of the masses is initially at rest, calculations become simpler. For such cases, we can derive some formulas. They will help us to solve problems quickly. We will write the steps:
$\mathbf\small{m_A v_{Ai}^2+m_B v_{Bi}^2=m_A v_{Af}^2+m_B v_{Bf}^2}$
• Where mA is the mass of A
• mB is the mass of B
• vAi is the initial velocity of A
• vBi is the initial velocity of B
• vAf is the final velocity of A
• vBf is the final velocity of B
2. But B is initially at rest. So vBi = 0
• So the equation becomes: $\mathbf\small{m_A v_{Ai}^2=m_A v_{Af}^2+m_B v_{Bf}^2}$
3. Applying the law of conservation of momentum, we have:
$\mathbf\small{m_A v_{Ai}+m_B v_{Bi}=m_A v_{Af}+m_B v_{Bf}}$
• But B is initially at rest. So vBi = 0
• So the equation becomes: $\mathbf\small{m_A v_{Ai}=m_A v_{Af}+m_B v_{Bf}}$
4. The result in (3) is: $\mathbf\small{m_A v_{Ai}=m_A v_{Af}+m_B v_{Bf}}$
• Multiply both sides by vBf. We get: $\mathbf\small{m_A v_{Ai}v_{Bf}=m_A v_{Af}v_{Bf}+m_B v_{Bf}^2}$
$\mathbf\small{\Rightarrow m_B v_{Bf}^2=m_A v_{Ai}v_{Bf}-m_A v_{Af}v_{Bf}}$
5. We can put this in the place of $\mathbf\small{m_B v_{Bf}^2}$ in (2). We get:
$\mathbf\small{m_A v_{Ai}^2=m_A v_{Af}^2+m_A v_{Ai}v_{Bf}-m_A v_{Af}v_{Bf}}$
$\mathbf\small{\Rightarrow v_{Ai}^2= v_{Af}^2+ v_{Ai}v_{Bf}- v_{Af}v_{Bf}}$
$\mathbf\small{\Rightarrow v_{Ai}^2-v_{Af}^2= v_{Bf}(v_{Ai}- v_{Af})}$
$\mathbf\small{\Rightarrow (v_{Ai}+ v_{Af})(v_{Ai}- v_{Af})= v_{Bf}(v_{Ai}- v_{Af})}$
$\mathbf\small{\Rightarrow (v_{Ai}+ v_{Af})= v_{Bf}}$
6. We can use this value of $\mathbf\small{v_{Bf}}$ in (3). We get:
$\mathbf\small{m_A v_{Ai}=m_A v_{Af}+m_B (v_{Ai}+ v_{Af})}$
$\mathbf\small{\Rightarrow m_A v_{Ai}=m_A v_{Af}+m_Bv_{Ai}+ m_Bv_{Af}}$
$\mathbf\small{\Rightarrow (m_A-m_B) v_{Ai}=(m_A + m_B)v_{Af}}$
$\mathbf\small{\Rightarrow v_{Af}=\frac{(m_A-m_B)v_{Ai}}{(m_A + m_B)}}$
• Thus we get the final velocity of A in terms of the three known quantities:
(i) Mass of A
(ii) Mass of B
(iii) Initial velocity of A
7. We can obtain the final velocity of B also
• For that, we put the above obtained value of $\mathbf\small{v_{Af}}$ in (5). We get:
$\mathbf\small{v_{Ai}+ \frac{(m_A-m_B)v_{Ai}}{(m_A + m_B)}= v_{Bf}}$
$\mathbf\small{\Rightarrow \frac{(m_A+m_B)v_{Ai}+(m_A-m_B)v_{Ai}}{(m_A + m_B)}= v_{Bf}}$
$\mathbf\small{\Rightarrow v_{Bf}=\frac{2m_Av_{Ai}}{(m_A + m_B)}}$
• Thus we get the final velocity of B in terms of the three known quantities:
(i) Mass of A
(ii) Mass of B
(iii) Initial velocity of A
8. Some interesting special cases can be analysed using the above expressions
Case 1: The two masses are equal
• When the two masses are equal, we can put: mA = mB = m
• Then from (6) we get: $\mathbf\small{\Rightarrow v_{Af}=\frac{(m-m)v_{Ai}}{(m + m)}=0}$
• From (7) we get: $\mathbf\small{v_{Bf}=\frac{2m\;v_{Ai}}{(m + m)}=v_{Ai}}$
• So 'case 1' can be written in the following 6 steps:
(i) Two bodies 'A' and 'B' collide elastically with each other
(ii) The two masses are equal. That is., mA = mB = m
(iii) Before the collision, 'A' was moving with a velocity vAi
(iv) Before the collision, 'B' was stationary. That is., vBi = 0
[Remember that, the equations in (6) and (7) are valid only if 'B' is initially at rest]
(v) After the collision, 'A' becomes stationary
(vi) After the collision, 'B' begins to move with vAi (the same velocity with which 'A' was moving)
Case 2: One of the two masses is very large when compared to the other
• Let us assume mB >> mA
• Then (mA-mB) will be very nearly equal to -mB
(That is., mA ≃ 0)
• Also (mA+mA) will be very nearly equal to mB
• Then from (6) we get: $\mathbf\small{v_{Af}=\frac{(-m_B)v_{Ai}}{( m_B)}=-v_{Ai}}$
• From (7) we get: $\mathbf\small{v_{Bf}=\frac{2\times 0\times v_{Ai}}{(0 + m_B)}=0}$
• So 'case 2' can be written in the following 6 steps:
(i) Two bodies 'A' and 'B' collide elastically with each other
(ii) mB >> mA
(iii) Before the collision, 'A' was moving with a velocity vAi
(iv) Before the collision, 'B' was stationary. That is., vBi = 0
[Remember that, the equations in (6) and (7) are valid only if 'B' is initially at rest]
(v) After the collision, 'A' begins to move with -vAi
(the same magnitude but opposite direction of the velocity with which 'A' was moving)
(vi) After the collision, 'B' remains stationary
In the solved example 6.33, one of the two objects is at rest. So the formulas should be applicable. Let us check:
2. We have: $\mathbf\small{v_{Af}=\frac{(m_A-m_B)v_{Ai}}{(m_A + m_B)}}$
Substituting the known values, we get:
$\mathbf\small{v_{Af}=\frac{(0.5-3.5)\times 4}{(0.5 + 3.5)}=-3\,\,\text{ms}^{-1}}$ (Same as before)
3. We have: $\mathbf\small{v_{Bf}=\frac{2m_Av_{Ai}}{(m_A + m_B)}}$
Substituting the known values, we get:
$\mathbf\small{v_{Bf}=\frac{2\times 0.5\times 4}{(0.5 + 3.5)}=1\,\,\text{ms}^{-1}}$ (Same as before)
In a nuclear reactor, a neutron of high speed (typically 107 ms-1) must be slowed to 103 ms-1. So that it can have a high probability of interacting with isotope $\mathbf\small{\overset{235}{\underset{92}{}}\text{U}}$ and causing it to fission. Show that a neutron can lose most of it's kinetic energy in an elastic collision with a light nuclei like deuterium which has only a few times the neutron mass. The material making up the light nuclei, usually heavy water (D2O) or graphite, is called a moderator
Solution:
1. We want to slow down a neutron
• We have to achieve this by making the neutron to collide with a deuterium nucleus
• This is shown in fig.6.66 below:
 |
| Fig.6.66 |
• We want to show that deuterium can be effectively used for slowing down a neutron
2. The deuterium nucleus is initially at rest. So we will use the formula:
$\mathbf\small{v_{Af}=\frac{(m_A-m_B)v_{Ai}}{(m_A + m_B)}}$
• Where mA is the mass of neutron
• mB is the mass of deuterium
• vAi is the initial velocity of neutron
• vAf is the final velocity of neutron
3. We have: $\mathbf\small{v_{Af}=\frac{(m_A-m_B)v_{Ai}}{(m_A + m_B)}}$
So final kinetic energy of the neutron (KAf)
= $\mathbf\small{\frac{1}{2}m_A\,v_{Af}^2=\frac{1}{2}m_A\,\left[\frac{m_A-m_B}{m_A + m_B}\right ]^2v_{Ai}^2}$
4. Initial kinetic energy of the neutron (KAi) = $\mathbf\small{\frac{1}{2}m_A\,v_{Ai}^2}$
5. Now take the ratio: $\mathbf\small{\frac{K_{Af}}{K_{Ai}}}$
• That is., ratio of the initial kinetic energy of neutron to it's final kinetic energy
• We will denote it as fA
• We have:
$\mathbf\small{f_A=\frac{K_{Af}}{K_{Ai}}=\frac{\frac{1}{2}m_A\,\left[\frac{m_A-m_B}{m_A + m_B}\right ]^2v_{Ai}^2}{\frac{1}{2}m_A\,v_{Ai}^2}=\left[\frac{m_A-m_B}{m_A + m_B}\right ]^2}$
6. So we can write:
Final kinetic energy of the neutron is $\mathbf\small{\left[\frac{m_A-m_B}{m_A + m_B}\right ]^2}$ times it's initial kinetic energy
• The numerator (mA-mB) will be less than the denominator (mA+mB)
• So the ratio $\mathbf\small{\left[\frac{m_A-m_B}{m_A + m_B}\right ]^2}$ is a fraction. It is not a whole number
• So final kinetic energy of the neutron will be certainly less than it's initial kinetic energy
• That means the final velocity of the neutron will be certainly less than the initial kinetic energy
• That means the neutron will be certainly slowed down
7. But knowing that the 'neutron will be slowed down' is not good enough
• We want to know 'how much slowing down' can be achieved
• If the neutron is not slowed down to a required value, the uranium nucleus will not be able to capture it
8. So we proceed as follows:
• Since it is an elastic collision,
Initial kinetic energy of the neutron-deuterium system = It's final kinetic energy
9. But initial velocity of deuterium = 0
• So initial kinetic energy of the system is contributed by the neutron alone
• That is., initial kinetic energy of the system = initial kinetic energy of the neutron
= $\mathbf\small{\frac{1}{2}m_A\,v_{Ai}^2}$
10. Combining (8) and (9), we get:
• Final kinetic energy of the system = Kf = KAi = $\mathbf\small{\frac{1}{2}m_A\,v_{Ai}^2}$
11. But the final kinetic energy of the system is made up by two items:
(i) Final kinetic energy of neutron KAf = $\mathbf\small{\frac{1}{2}m_A\,v_{Af}^2}$
(ii) Final kinetic energy of deuterium KBf = $\mathbf\small{\frac{1}{2}m_A\,v_{Bf}^2}$
• So we can write: Kf = KAi = KAf + KBf
12. Take out the second and third terms. We get: KAi = KAf + KBf
Divide both sides by KAi. We get: $\mathbf\small{1=\frac{K_{Af}}{K_{Ai}}+\frac{K_{Bf}}{K_{Ai}}}$
13. But we have already calculated $\mathbf\small{\frac{K_{Af}}{K_{Ai}}}$ in (5)
• We have: $\mathbf\small{\frac{K_{Af}}{K_{Ai}}=\left[\frac{m_A-m_B}{m_A + m_B}\right ]^2}$
• So the result in (12) becomes: $\mathbf\small{1=\left[\frac{m_A-m_B}{m_A + m_B}\right ]^2+\frac{K_{Bf}}{K_{Ai}}}$
$\mathbf\small{\Rightarrow \frac{K_{Bf}}{K_{Ai}}=1-\left[\frac{m_A-m_B}{m_A + m_B}\right ]^2}$
$\mathbf\small{\Rightarrow \frac{K_{Bf}}{K_{Ai}}=\frac{(m_A+m_B)^2-(m_A-m_B)^2}{(m_A + m_B)^2}}$
$\mathbf\small{\Rightarrow \frac{K_{Bf}}{K_{Ai}}=\frac{(m_A+m_B+m_A-m_B)(m_A+m_B-m_A+m_B)}{(m_A + m_B)^2}}$
$\mathbf\small{\left[\text{Using the identity}:(a^2-b^2)=(a+b)(a-b) \right]}$
$\mathbf\small{\Rightarrow \frac{K_{Bf}}{K_{Ai}}=\frac{(2m_A)(2m_B)}{(m_A + m_B)^2}}$
$\mathbf\small{\Rightarrow \frac{K_{Bf}}{K_{Ai}}=\frac{4m_A\;m_B}{(m_A + m_B)^2}}$
14. So we have two ratios:
(i) From (5) we have: $\mathbf\small{\frac{K_{Af}}{K_{Ai}}=\left[\frac{m_A-m_B}{m_A + m_B}\right ]^2}$
(ii) From (13) we have: $\mathbf\small{\frac{K_{Bf}}{K_{Ai}}=\frac{4m_A\;m_B}{(m_A + m_B)^2}}$
15. Given that: deuterium has only a few times the neutron mass
Assume that mB = 2 mA
(i) Then from 14(i), we get:
$\mathbf\small{\frac{K_{Af}}{K_{Ai}}=\left[\frac{m_A-m_B}{m_A + m_B}\right ]^2=\left[\frac{m_A-2m_A}{m_A + 2m_A}\right ]^2=\frac{1}{9}}$
(ii) From 14(ii) we get:
$\mathbf\small{\frac{K_{Bf}}{K_{Ai}}=\frac{4m_A\;m_B}{(m_A + m_B)^2}=\frac{4m_A\times 2m_A}{(m_A + 2m_A)^2}=\frac{8}{9}}$
16. So from 15(i) we get:
• Final kinetic energy of neutron is 1⁄9 of initial kinetic energy of neutron
⇒ Final kinetic energy of neutron is 1⁄9 of total kinetic energy of the system
• From 15(ii) we get:
Final kinetic energy of deuterium is 8⁄9 of initial kinetic energy of neutron
⇒ Final kinetic energy of deuterium is 8⁄9 of total kinetic energy of the system
• In percentage form:
1⁄9 = 11.1%
8⁄9 = 88.9%
17. So we can write:
• After the collision with deuterium, the neutron loses (100-11.1) = 88.9% of it's initial kinetic energy
♦ That means., the neutron will be left with only 11.1% of it's initial energy
• The lost 88.9% is acquired by the deuterium
18. Instead of deuterium, we can use carbon (graphite) as the moderator
• The formulas are the same that we wrote in (14)
• The only difference is this:
♦ When deuterium is used, we wrote: mB = 2 × mA
♦ When carbon is used, we can write: mB = k × mA
♦ We must put a suitable value for 'k'
• The results obtained are as follows:
After the collision with carbon, the neutron loses 71.6% of it's initial kinetic energy
♦ That means., the neutron will be left with only (100-71.6) = 28.4% of it's initial energy
• The lost 71.6% is acquired by the carbon
• So deuterium and carbon are very effective moderators
• However high losses like 88.9% and 71.6% can be achieved only if the neutron does a head-on collision with deuterium or carbon
• We will see the details of head-on collisions in a later section
• However high losses like 88.9% and 71.6% can be achieved only if the neutron does a head-on collision with deuterium or carbon
• We will see the details of head-on collisions in a later section
In the next section we will see inelastic collisions
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