In the previous section, we saw elastic collision. It was illustrated in the first row 'a' in fig.6.64 of the previous section. Now we take up the second row 'b'. For convenience, the fig.6.64 is shown again below:
1. Consider the row 'b', column '1' in fig.6.64 above.
• 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 'b', 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 inelastic' materials
• So they do not even try to regain their original shapes after impact
• That means, whatever 'deformation caused' remains as such
• Since they do not even try to recover, they do not get separated
• After impact, they move together with velocity vf
♦ vf is 'velocity of (A+B) (final)'
• This is shown in row 'b', column '3'
■ This is called a completely inelastic collision or a perfectly inelastic collision
• A portion of the initial energy will be lost as heat, sound, vibrations etc.,
4. This situation is just like compressing and not releasing a spring
• The energy used up for compressing the spring will not be obtained back if the spring is not released
• So there is a reduction in total energy
5. That means, in a completely inelastic collision,
Total energy before collision > Total energy after collision
6. 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:
$\mathbf\small{\frac{1}{2}m_A v_{Ai}^2+\frac{1}{2}m_B v_{Bi}^2>\frac{1}{2}(m_A+m_B) v_f^2}$
• So we have the '>' sign instead of the '=' sign
8. We can replace the '>' sign by the '=' sign by making a small modification
• The modification is: Bring in the 'lost energy' also
• If we add the 'lost energy' on the right side, the energies will balance. So we get:
$\mathbf\small{\frac{1}{2}m_A v_{Ai}^2+\frac{1}{2}m_B v_{Bi}^2=\frac{1}{2}(m_A+m_B) v_f^2+\text{Energy lost due to collision}}$
9. On many occasions, we will want to know the 'lost energy'
• For example, in a collision between two cars, the 'lost energy' is the energy which causes the damage to the cars and the passengers
• For finding the 'lost energy', we can use the equation in (8)
• All the quantities in that equation are known except 'vf' and the 'energy lost'
• To find two unknowns, we need two equations
10. To write the second equation, we use the law of conservation of momentum
• If there is no external force acting on the system, the law of conservation of momentum is valid
• So we have:
$\mathbf\small{m_A v_{Ai}+m_B v_{Bi}=(m_A+m_B) v_f}$ (Details here)
11. Equations in (8) and (10) can be used together, to solve problems in 'one-dimensional perfectly inelastic collisions'
Solved example 6.35
Two objects undergo a one-dimensional, completely inelastic collision (Fig.6.67 below)
(i) Calculate the velocity of (A+B) after collision
(ii) Also calculate the amount of energy lost
Given that: mA = 0.150 kg, mB = 70.0 kg, vAi = 35 ms-1, vBi = 0
Solution:
We are required to find vf
1. For a completely inelastic collision, we have:
$\mathbf\small{\frac{1}{2}m_A v_{Ai}^2+\frac{1}{2}m_B v_{Bi}^2=\frac{1}{2}(m_A+m_B) v_f^2+\text{Energy lost due to collision}}$
• Substituting the values, we get:
$\mathbf\small{\frac{1}{2}\times 0.150 \times 35^2+\frac{1}{2}\times 70 \times 0^2=\frac{1}{2}\times (70.15) v_f^2+\text{Energy lost}}$
$\mathbf\small{\Rightarrow 91.875+0=35.075\;v_f^2+\text{Energy lost}}$
$\mathbf\small{\Rightarrow 91.875=35.075\;v_f^2+\text{Energy lost}}$
2. If there is no external force acting on the system, the law of conservation of momentum is valid
So we have:
$\mathbf\small{m_A v_{Ai}+m_B v_{Bi}=(m_A+m_B) v_f}$
• Substituting the values, we get:
$\mathbf\small{0.150 \times 35+70 \times 0=70.15\;v_f}$
$\mathbf\small{\Rightarrow 5.25=70.15\;v_f}$
$\mathbf\small{\Rightarrow v_f=0.0748\;\text{ms}^{-1}}$
• This is the answer for part (i)
3. Substituting this value of vf in (1), we get:
$\mathbf\small{91.875=35.075\times 0.0748^2+\text{Energy lost}}$
⇒ Energy lost = 91.68 J
• This is the answer for part (ii)
Solved example 6.36
A bullet of mass 0.012 kg and horizontal speed 70 ms-1, strikes a block of wood of mass 0.4 kg and instantly comes to rest with respect to the block. The block is suspended from the ceiling by means of thin wires. Calculate the height to which the block rises. Also estimate the amount of heat produced in the block
Solution:
1. In the fig.6.68(a) below, the wooden block of mass 0.4 kg is suspended from point 'O'
• 'O' is a point in the ceiling
2. When the bullet strikes the block, it gets embedded in the block
• Due to the initial velocity of the bullet, the block (with the bullet embedded) will tend to move horizontally
3. But since the block is suspended from the ceiling, horizontal motion is not possible. Instead, it will move in a circular path
• This circular path UVWX is shown in green color. U is the initial position of the block
• The block cannot complete the full circle because the ceiling is present
• If the length of the suspending wires is sufficiently large, the block will not reach even up to the ceiling
4. Let 'P' be the 'maximum possible point'. This is shown in fig.b
• Draw a horizontal through P. This horizontal meets the vertical (through O) at Q
• Then UQ = 'h', the maximum possible height
• We want to know this height 'h'
5. When the bullet strikes the block, a 'completely inelastic collision' takes place
• Because, after the collision, the bullet and the block move together
6. For a completely inelastic collision, we have:
$\mathbf\small{\frac{1}{2}m_A v_{Ai}^2+\frac{1}{2}m_B v_{Bi}^2=\frac{1}{2}(m_A+m_B) v_f^2+\text{Energy lost due to collision}}$
• Substituting the values, we get:
$\mathbf\small{\frac{1}{2}\times 0.012 \times 70^2+\frac{1}{2}\times 0.4 \times 0^2=\frac{1}{2}\times (0.412) v_f^2+\text{Energy lost}}$
$\mathbf\small{\Rightarrow 29.4=0.206\;v_f^2+\text{Energy lost}}$
7. If there is no external force acting on the system, the law of conservation of momentum is valid
So we have:
$\mathbf\small{m_A v_{Ai}+m_B v_{Bi}=(m_A+m_B) v_f}$
• Substituting the values, we get:
$\mathbf\small{0.012 \times 70+0.4 \times 0=0.412\;v_f}$
$\mathbf\small{\Rightarrow 0.84=0.412\;v_f}$
$\mathbf\small{\Rightarrow v_f=2.039\;\text{ms}^{-1}}$
8. Substituting this value of vf in (6), we get:
$\mathbf\small{29.4=0.206\times 2.039^2+\text{Energy lost}}$
⇒ Energy lost = 28.54 J
• This is the answer for part (ii)
9. Next, let us apply the law of conservation of energy to the points U and P
■ By the Law of conservation of energy, the total energy must be the same at those two points
• We can write EU = EP
12. Let us write the various energies:
(i) Total energy EU:
• Kinetic energy = $\mathbf\small{\frac{m_{(A+B)}\,v_f^2}{2}}$
• Gravitational potential energy = 0
■ So total energy EA = $\mathbf\small{\frac{m_{(A+B)}\,v_f^2}{2}}$
(ii) Total energy EP
• Kinetic energy = 0
• Gravitational potential energy = $\mathbf\small{m_{(A+B)}gh}$
■ So total energy EP = $\mathbf\small{m_{(A+B)}gh}$
13. Equating the two energies we get:
$\mathbf\small{\frac{m_{(A+B)}\,v_f^2}{2}=m_{(A+B)}gh}$
$\mathbf\small{\Rightarrow \frac{v_f^2}{2}=gh}$
$\mathbf\small{\Rightarrow h=\frac{v_f^2}{2g}}$
• Substituting the known values, we get:
$\mathbf\small{h=\frac{2.039^2}{2 \times 9.8}=0.212\;\text{m}}$
• This is the answer for part (i)
So we have seen the discussion on 'completely inelastic collision'. It was illustrated in the second row 'b' in fig.6.64 above. Now we take up the third row 'c'.
1. Consider the row 'c', column '1' in fig.6.64 above.
• 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 'b', 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, the masses are neither 'perfectly elastic' nor 'perfectly inelastic'
• Since they are not 'perfectly elastic', they will not 'fully regain' their original shapes
• Since they are not 'perfectly inelastic', they will not stay in the 'deformed shape'
■ So this is an intermediate state between 'elastic collision' and 'completely inelastic collision'
4. After collision, they will move separately with velocities vAf and vBf
• Also there will be some 'lost energy'. We can be sure about this because, the original shapes are not regained
• In other words, some energy is lost for 'causing the deformations'
5. So there are 3 unknowns: vAf , vBf and 'lost energy'
• But we have only two equations:
(i) $\mathbf\small{m_A v_{Ai}+m_B v_{Bi}=m_A v_{Af}+m_B v_{Bf}}$
(ii) $\mathbf\small{\frac{1}{2}m_A v_{Ai}^2+\frac{1}{2}m_B v_{Bi}^2=\frac{1}{2}m_A v_{Af}^2+\frac{1}{2}m_B v_{Bf}^2+\text{Energy lost due to collision}}$
• We will see the methods to solve such problems in higher classes
 |
| Fig.6.64 |
• 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 'b', 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 inelastic' materials
• So they do not even try to regain their original shapes after impact
• That means, whatever 'deformation caused' remains as such
• Since they do not even try to recover, they do not get separated
• After impact, they move together with velocity vf
♦ vf is 'velocity of (A+B) (final)'
• This is shown in row 'b', column '3'
■ This is called a completely inelastic collision or a perfectly inelastic collision
• A portion of the initial energy will be lost as heat, sound, vibrations etc.,
4. This situation is just like compressing and not releasing a spring
• The energy used up for compressing the spring will not be obtained back if the spring is not released
• So there is a reduction in total energy
5. That means, in a completely inelastic collision,
Total energy before collision > Total energy after collision
6. 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:
$\mathbf\small{\frac{1}{2}m_A v_{Ai}^2+\frac{1}{2}m_B v_{Bi}^2>\frac{1}{2}(m_A+m_B) v_f^2}$
• So we have the '>' sign instead of the '=' sign
8. We can replace the '>' sign by the '=' sign by making a small modification
• The modification is: Bring in the 'lost energy' also
• If we add the 'lost energy' on the right side, the energies will balance. So we get:
$\mathbf\small{\frac{1}{2}m_A v_{Ai}^2+\frac{1}{2}m_B v_{Bi}^2=\frac{1}{2}(m_A+m_B) v_f^2+\text{Energy lost due to collision}}$
9. On many occasions, we will want to know the 'lost energy'
• For example, in a collision between two cars, the 'lost energy' is the energy which causes the damage to the cars and the passengers
• For finding the 'lost energy', we can use the equation in (8)
• All the quantities in that equation are known except 'vf' and the 'energy lost'
• To find two unknowns, we need two equations
10. To write the second equation, we use the law of conservation of momentum
• If there is no external force acting on the system, the law of conservation of momentum is valid
• So we have:
$\mathbf\small{m_A v_{Ai}+m_B v_{Bi}=(m_A+m_B) v_f}$ (Details here)
11. Equations in (8) and (10) can be used together, to solve problems in 'one-dimensional perfectly inelastic collisions'
Solved example 6.35
Two objects undergo a one-dimensional, completely inelastic collision (Fig.6.67 below)
 |
| Fig.6.67 |
(ii) Also calculate the amount of energy lost
Given that: mA = 0.150 kg, mB = 70.0 kg, vAi = 35 ms-1, vBi = 0
Solution:
We are required to find vf
1. For a completely inelastic collision, we have:
$\mathbf\small{\frac{1}{2}m_A v_{Ai}^2+\frac{1}{2}m_B v_{Bi}^2=\frac{1}{2}(m_A+m_B) v_f^2+\text{Energy lost due to collision}}$
• Substituting the values, we get:
$\mathbf\small{\frac{1}{2}\times 0.150 \times 35^2+\frac{1}{2}\times 70 \times 0^2=\frac{1}{2}\times (70.15) v_f^2+\text{Energy lost}}$
$\mathbf\small{\Rightarrow 91.875+0=35.075\;v_f^2+\text{Energy lost}}$
$\mathbf\small{\Rightarrow 91.875=35.075\;v_f^2+\text{Energy lost}}$
2. If there is no external force acting on the system, the law of conservation of momentum is valid
So we have:
$\mathbf\small{m_A v_{Ai}+m_B v_{Bi}=(m_A+m_B) v_f}$
• Substituting the values, we get:
$\mathbf\small{0.150 \times 35+70 \times 0=70.15\;v_f}$
$\mathbf\small{\Rightarrow 5.25=70.15\;v_f}$
$\mathbf\small{\Rightarrow v_f=0.0748\;\text{ms}^{-1}}$
• This is the answer for part (i)
3. Substituting this value of vf in (1), we get:
$\mathbf\small{91.875=35.075\times 0.0748^2+\text{Energy lost}}$
⇒ Energy lost = 91.68 J
• This is the answer for part (ii)
Solved example 6.36
A bullet of mass 0.012 kg and horizontal speed 70 ms-1, strikes a block of wood of mass 0.4 kg and instantly comes to rest with respect to the block. The block is suspended from the ceiling by means of thin wires. Calculate the height to which the block rises. Also estimate the amount of heat produced in the block
Solution:
1. In the fig.6.68(a) below, the wooden block of mass 0.4 kg is suspended from point 'O'
 |
| Fig.6.68 |
2. When the bullet strikes the block, it gets embedded in the block
• Due to the initial velocity of the bullet, the block (with the bullet embedded) will tend to move horizontally
3. But since the block is suspended from the ceiling, horizontal motion is not possible. Instead, it will move in a circular path
• This circular path UVWX is shown in green color. U is the initial position of the block
• The block cannot complete the full circle because the ceiling is present
• If the length of the suspending wires is sufficiently large, the block will not reach even up to the ceiling
4. Let 'P' be the 'maximum possible point'. This is shown in fig.b
• Draw a horizontal through P. This horizontal meets the vertical (through O) at Q
• Then UQ = 'h', the maximum possible height
• We want to know this height 'h'
5. When the bullet strikes the block, a 'completely inelastic collision' takes place
• Because, after the collision, the bullet and the block move together
6. For a completely inelastic collision, we have:
$\mathbf\small{\frac{1}{2}m_A v_{Ai}^2+\frac{1}{2}m_B v_{Bi}^2=\frac{1}{2}(m_A+m_B) v_f^2+\text{Energy lost due to collision}}$
• Substituting the values, we get:
$\mathbf\small{\frac{1}{2}\times 0.012 \times 70^2+\frac{1}{2}\times 0.4 \times 0^2=\frac{1}{2}\times (0.412) v_f^2+\text{Energy lost}}$
$\mathbf\small{\Rightarrow 29.4=0.206\;v_f^2+\text{Energy lost}}$
7. If there is no external force acting on the system, the law of conservation of momentum is valid
So we have:
$\mathbf\small{m_A v_{Ai}+m_B v_{Bi}=(m_A+m_B) v_f}$
• Substituting the values, we get:
$\mathbf\small{0.012 \times 70+0.4 \times 0=0.412\;v_f}$
$\mathbf\small{\Rightarrow 0.84=0.412\;v_f}$
$\mathbf\small{\Rightarrow v_f=2.039\;\text{ms}^{-1}}$
8. Substituting this value of vf in (6), we get:
$\mathbf\small{29.4=0.206\times 2.039^2+\text{Energy lost}}$
⇒ Energy lost = 28.54 J
• This is the answer for part (ii)
9. Next, let us apply the law of conservation of energy to the points U and P
■ By the Law of conservation of energy, the total energy must be the same at those two points
• We can write EU = EP
12. Let us write the various energies:
(i) Total energy EU:
• Kinetic energy = $\mathbf\small{\frac{m_{(A+B)}\,v_f^2}{2}}$
• Gravitational potential energy = 0
■ So total energy EA = $\mathbf\small{\frac{m_{(A+B)}\,v_f^2}{2}}$
(ii) Total energy EP
• Kinetic energy = 0
• Gravitational potential energy = $\mathbf\small{m_{(A+B)}gh}$
■ So total energy EP = $\mathbf\small{m_{(A+B)}gh}$
13. Equating the two energies we get:
$\mathbf\small{\frac{m_{(A+B)}\,v_f^2}{2}=m_{(A+B)}gh}$
$\mathbf\small{\Rightarrow \frac{v_f^2}{2}=gh}$
$\mathbf\small{\Rightarrow h=\frac{v_f^2}{2g}}$
• Substituting the known values, we get:
$\mathbf\small{h=\frac{2.039^2}{2 \times 9.8}=0.212\;\text{m}}$
• This is the answer for part (i)
So we have seen the discussion on 'completely inelastic collision'. It was illustrated in the second row 'b' in fig.6.64 above. Now we take up the third row 'c'.
1. Consider the row 'c', column '1' in fig.6.64 above.
• 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 'b', 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, the masses are neither 'perfectly elastic' nor 'perfectly inelastic'
• Since they are not 'perfectly elastic', they will not 'fully regain' their original shapes
• Since they are not 'perfectly inelastic', they will not stay in the 'deformed shape'
■ So this is an intermediate state between 'elastic collision' and 'completely inelastic collision'
4. After collision, they will move separately with velocities vAf and vBf
• Also there will be some 'lost energy'. We can be sure about this because, the original shapes are not regained
• In other words, some energy is lost for 'causing the deformations'
5. So there are 3 unknowns: vAf , vBf and 'lost energy'
• But we have only two equations:
(i) $\mathbf\small{m_A v_{Ai}+m_B v_{Bi}=m_A v_{Af}+m_B v_{Bf}}$
(ii) $\mathbf\small{\frac{1}{2}m_A v_{Ai}^2+\frac{1}{2}m_B v_{Bi}^2=\frac{1}{2}m_A v_{Af}^2+\frac{1}{2}m_B v_{Bf}^2+\text{Energy lost due to collision}}$
• We will see the methods to solve such problems in higher classes
• The collisions that we saw in this section and in the previous section, were all one-dimensional collisions. They are also called head-on collisions
• We will see more details about such collisions in the next section. After that, we will discuss about two-dimensional collisions
• We will see more details about such collisions in the next section. After that, we will discuss about two-dimensional collisions
