In the previous section we saw how to find the apparent weight in a lift. In this section we will see another situation where there is a combined application of the Second and Third laws of motion.
1. A bullet is fired from a gun. See fig.5.24 below:
We are given the following information:
• mG is the mass of the gun
• mB is the mass of the bullet
• $\mathbf\small{\vec{v_B}}$ is the velocity with which the bullet is fired
• $\mathbf\small{\vec{v_G}}$ is the recoil velocity of the gun
2.Let us assume that, the motion of the bullet is along the x axis
• We can begin the calculations:
• First we do the calculations for the bullet:
x direction:
(i) Initial momentum of the bullet in the x direction = $\mathbf\small{\vec{p_{Bx1}}=m_B \times \vec{v_{Bx1}}=m_B \times 0=0}$
(ii) Final momentum of the bullet in the x direction = $\mathbf\small{\vec{p_{Bx2}}=m_B \times \vec{v_{Bx2}}=m_B \times \vec{v_{B}}}$
■ So change in momentum in the x direction = $\mathbf\small{\Delta \vec{p_{Bx}}=(\vec{p_{Bx2}}-\vec{p_{Bx1}})=(m_B \vec{v_{B}}-0)=m_B \vec{v_{B}}}$
y direction:
• We do not need to worry about the y direction because, there is no motion along that direction
• We can straight away write: $\mathbf\small{\Delta \vec{p_{By}}}$ = 0
8. The bullet suffers a change in momentum because, a force acted on it
Obviously, this force was exerted by the gun
• Let us denote this force (which is applied on the bullet by the gun) as $\mathbf\small{\vec{F_{BG}}}$
Conservation of momentum
1. A bullet is fired from a gun. See fig.5.24 below:
 |
| Fig.5.24 |
• mG is the mass of the gun
• mB is the mass of the bullet
• $\mathbf\small{\vec{v_B}}$ is the velocity with which the bullet is fired
• $\mathbf\small{\vec{v_G}}$ is the recoil velocity of the gun
2.Let us assume that, the motion of the bullet is along the x axis
• We can begin the calculations:
• First we do the calculations for the bullet:
x direction:
(i) Initial momentum of the bullet in the x direction = $\mathbf\small{\vec{p_{Bx1}}=m_B \times \vec{v_{Bx1}}=m_B \times 0=0}$
(ii) Final momentum of the bullet in the x direction = $\mathbf\small{\vec{p_{Bx2}}=m_B \times \vec{v_{Bx2}}=m_B \times \vec{v_{B}}}$
■ So change in momentum in the x direction = $\mathbf\small{\Delta \vec{p_{Bx}}=(\vec{p_{Bx2}}-\vec{p_{Bx1}})=(m_B \vec{v_{B}}-0)=m_B \vec{v_{B}}}$
y direction:
• We do not need to worry about the y direction because, there is no motion along that direction
• We can straight away write: $\mathbf\small{\Delta \vec{p_{By}}}$ = 0
3. Resultant change in momentum $\mathbf\small{\Delta \vec{p_B}}$ is the resultant of $\mathbf\small{\Delta \vec{p_{Bx}}}$ and $\mathbf\small{\Delta \vec{p_{By}}}$
• But $\mathbf\small{\Delta \vec{p_{By}}}$ = 0
■ So we get: $\mathbf\small{\Delta \vec{p_B}=\Delta \vec{p_{Bx}}=m_B \vec{v_{B}}}$
4. We can write the following four points:
(i) The bullet suffers a change in momentum of $\mathbf\small{m_B \vec{v_{B}}}$
(ii) This 'change in momentum' is a vector quantity
(iii) It's magnitude is $\mathbf\small{m_B\, |\vec{v_B}|}$
(iv) It's direction is towards the positive side of the x axis
5. Now we do the calculations for the gun:
x direction:
(i) Initial momentum of the gun in the x direction = $\mathbf\small{\vec{p_{Gx1}}=m_G \times \vec{v_{Gx1}}=m_G \times 0=0}$
(ii) Final momentum of the gun in the x direction = $\mathbf\small{\vec{p_{Gx2}}=m_G \times \vec{v_{Gx2}}=-m_G \times \vec{v_{G}}}$
■ So change in momentum in the x direction = $\mathbf\small{\Delta \vec{p_{Gx}}=(\vec{p_{Gx2}}-\vec{p_{Gx1}})=(-m_G \vec{v_{G}}-0)=-m_G \vec{v_{G}}}$
y direction:
• We do not need to worry about the y direction because, there is no motion along that direction
• We can straight away write: $\mathbf\small{\Delta \vec{p_{Gy}}}$ = 0
6. Resultant change in momentum $\mathbf\small{\Delta \vec{p_G}}$ is the resultant of $\mathbf\small{\Delta \vec{p_{Gx}}}$ and $\mathbf\small{\Delta \vec{p_{Gy}}}$
• But $\mathbf\small{\Delta \vec{p_{By}}}$ = 0
■ So we get: $\mathbf\small{\Delta \vec{p_B}=\Delta \vec{p_{Bx}}=m_B \vec{v_{B}}}$
4. We can write the following four points:
(i) The bullet suffers a change in momentum of $\mathbf\small{m_B \vec{v_{B}}}$
(ii) This 'change in momentum' is a vector quantity
(iii) It's magnitude is $\mathbf\small{m_B\, |\vec{v_B}|}$
(iv) It's direction is towards the positive side of the x axis
5. Now we do the calculations for the gun:
x direction:
(i) Initial momentum of the gun in the x direction = $\mathbf\small{\vec{p_{Gx1}}=m_G \times \vec{v_{Gx1}}=m_G \times 0=0}$
(ii) Final momentum of the gun in the x direction = $\mathbf\small{\vec{p_{Gx2}}=m_G \times \vec{v_{Gx2}}=-m_G \times \vec{v_{G}}}$
■ So change in momentum in the x direction = $\mathbf\small{\Delta \vec{p_{Gx}}=(\vec{p_{Gx2}}-\vec{p_{Gx1}})=(-m_G \vec{v_{G}}-0)=-m_G \vec{v_{G}}}$
y direction:
• We do not need to worry about the y direction because, there is no motion along that direction
• We can straight away write: $\mathbf\small{\Delta \vec{p_{Gy}}}$ = 0
6. Resultant change in momentum $\mathbf\small{\Delta \vec{p_G}}$ is the resultant of $\mathbf\small{\Delta \vec{p_{Gx}}}$ and $\mathbf\small{\Delta \vec{p_{Gy}}}$
• But $\mathbf\small{\Delta \vec{p_{Gy}}}$ = 0
■ So we get: $\mathbf\small{\Delta \vec{p_G}=\Delta \vec{p_{Gx}}=-m_B \vec{v_{G}}}$
7. We can write the following four points:
(i) The gun suffers a change in momentum of $\mathbf\small{-m_G \vec{v_{G}}}$
(ii) This 'change in momentum' is a vector quantity
(iii) It's magnitude is $\mathbf\small{m_G\, |\vec{v_G}|}$
(iv) It's direction is towards the negative side of the x axis
■ So we get: $\mathbf\small{\Delta \vec{p_G}=\Delta \vec{p_{Gx}}=-m_B \vec{v_{G}}}$
7. We can write the following four points:
(i) The gun suffers a change in momentum of $\mathbf\small{-m_G \vec{v_{G}}}$
(ii) This 'change in momentum' is a vector quantity
(iii) It's magnitude is $\mathbf\small{m_G\, |\vec{v_G}|}$
(iv) It's direction is towards the negative side of the x axis
Obviously, this force was exerted by the gun
• Let us denote this force (which is applied on the bullet by the gun) as $\mathbf\small{\vec{F_{BG}}}$
• Let the time duration for which the force acts be Δt
• Consider the product: $\mathbf\small{\vec{F_{BG}}\times\Delta t}$
9. We know that this (product of force and time for which the force acts) is equal to the ‘change in momentum’
• So we can write $\mathbf\small{\vec{F_{BG}}\times\Delta t=m_B \vec{v_{B}}}$
10. The gun suffers a change in momentum because, a force acted on it
Obviously, this force was exerted by the bullet
• Let us denote this force (which is applied on the gun by the bullet) as $\mathbf\small{\vec{F_{GB}}}$
9. We know that this (product of force and time for which the force acts) is equal to the ‘change in momentum’
• So we can write $\mathbf\small{\vec{F_{BG}}\times\Delta t=m_B \vec{v_{B}}}$
10. The gun suffers a change in momentum because, a force acted on it
Obviously, this force was exerted by the bullet
• Let us denote this force (which is applied on the gun by the bullet) as $\mathbf\small{\vec{F_{GB}}}$
• The time duration for which the force acts will be the same Δt
• Consider the product: $\mathbf\small{\vec{F_{GB}}\times\Delta t}$
11. We know that this (product of force and time for which the force acts) is equal to the ‘change in momentum’
• So we can write $\mathbf\small{\vec{F_{GB}}\times\Delta t=-m_G \vec{v_{G}}}$
12. So we have the following two information:
(i) The bullet suffers a change in momentum $\mathbf\small{\vec{\Delta p_B}}$:
$\mathbf\small{\vec{\Delta p_B}=\vec{F_{BG}}\times\Delta t=m_B \vec{v_{B}}}$
(ii) The gun suffers a change in momentum $\mathbf\small{\vec{\Delta p_G}}$:
$\mathbf\small{\vec{\Delta p_G}=\vec{F_{GB}}\times\Delta t=-m_G \vec{v_{G}}}$
13. Take out the first and second part from 12(i) and 12(ii):
(i) $\mathbf\small{\vec{\Delta p_B}=\vec{F_{BG}}\times\Delta t}$
(ii) $\mathbf\small{\vec{\Delta p_G}=\vec{F_{GB}}\times\Delta t}$
14. Applying Newton's third law, we have:
$\mathbf\small{\vec{F_{BG}}}$ = -$\mathbf\small{\vec{F_{GB}}}$
15. So we can replace $\mathbf\small{\vec{F_{BG}}}$ in 13(i). We get:
$\mathbf\small{\vec{\Delta p_B}=-\vec{F_{GB}}\times\Delta t}$
• But from 13(ii), $\mathbf\small{-\vec{F_{GB}}\times\Delta t}$ = $\mathbf\small{-\vec{\Delta p_G}}$
• So we get: $\mathbf\small{\vec{\Delta p_B}=-\vec{\Delta p_G}}$
■ That is., 'change in momenta' are numerically equal, but have opposite signs
16. Now, the change in momentum for the bullet can be written as:
• Change in momentum of bullet = Final momentum of bullet - Initial momentum of bullet
• That is., $\mathbf\small{\vec{\Delta p_B}=[\vec{p_{B1}}-\vec{p_{B2}}]}$
17. Similarly, the change in momentum for the gun can be written as:
• Change in momentum of gun = Final momentum of gun - Initial momentum of gun
• That is., $\mathbf\small{\vec{\Delta p_G}=[\vec{p_{G1}}-\vec{p_{G2}}]}$
18. But according to the result in (15), the two 'change in momenta' are numerically equal, but have opposite signs
• So we can change the sign of (17) and equate it to (16). We get:
$\mathbf\small{[\vec{p_{B1}}-\vec{p_{B2}}]}= {-[\vec{p_{G1}}-\vec{p_{G2}}]}$
$\mathbf\small{\Rightarrow \vec{p_{B1}}-\vec{p_{B2}}}= {-\vec{p_{G1}}-\vec{p_{G2}}}$
$\mathbf\small{\Rightarrow \vec{p_{B1}}+\vec{p_{G1}}}= {\vec{p_{B2}}+\vec{p_{G2}}}$
19. On the left side, we have:
• Sum of momenta before the firing of the gun
On the right side, we have:
• Sum of momenta after the firing of the gun
• So we can write: The total momentum of the (bullet + gun) system is conserved
20. The Law of Conservation of momentum states that:
■ The total momentum of an isolated system of interacting particles is conserved
Solved example 5.10
A shell of mass 0.020 kg is fired by a gun of mass 100 kg. If the muzzle speed of the shell is 80 ms-1, what is the recoil speed of the gun?
Solution:
• For this problem, there is one-dimensional motion only. So we do not need to use vector notations.
1. Applying the law of conservation of momentum, we have:
pS1 + pG1 = pS2 + pG2
2. But pS1 = pG1 = 0
• So we get: pS2 = -pG2
⟹ mS × vS2 = mG × vG2
3. Substituting the values, we get:
0.020 × 80 = 100 × vG2
⟹ vG2 = 0.016 ms-1.
Solved example 5.11
In a quarry, a stationary rock explodes into 3 parts. A 1 kg part and another 2 kg part moves at right angles to each other. Their velocities are respectively 12 ms-1 and 8 ms-1. If the remaining third part moves with a velocity of 4 ms-1, what is it's mass?
Solution:
• Let the first two parts be named as 'A' and 'B' and the third part be named as 'C'. This is shown in fig.5.25(b) below:
• The given data can be written as:
$\mathbf\small{m_A=1\, \text{kg},\: m_B=2\,\, \text{kg},\: \vec{v_{A2}}=12\, \,ms^{-1},\: \vec{v_{B2}}=8\, \,ms^{-1},\: \vec{v_{C2}}=4\, \,ms^{-1},\: m_C=?}$
1. Applying the law of conservation of momentum, we have:
$\mathbf\small{\vec{p_{A1}}+\vec{p_{B1}}+\vec{p_{C1}}}= {\vec{p_{A2}}+\vec{p_{B2}}+\vec{p_{C2}}}$
2. But the initial momenta are all zero. So we get:
$\mathbf\small{\vec{p_{A2}}+\vec{p_{B2}}+\vec{p_{C2}}=0}$
$\mathbf\small{\Rightarrow \vec{p_{A2}}+\vec{p_{B2}}=-\vec{p_{C2}}}$
• That is., the vector $\mathbf\small{\vec{p_{C2}}}$ must be equal and opposite to the resultant of the two vectors $\mathbf\small{\vec{p_{A2}}}$ and $\mathbf\small{\vec{p_{B2}}}$
3. We will denote the resultant as $\mathbf\small{\vec{R}}$
We can write: $\mathbf\small{\vec{R}=(\vec{p_{A2}}+\vec{p_{B2}})}$
$\mathbf\small{\Rightarrow \vec{R}=-\vec{p_{C2}}}$
4. So let us find $\mathbf\small{\vec{R}}$ first:
• Given that, the two masses A and B move perpendicular to each other
• So their momentum vectors are perpendicular to each other. This is shown in fig.5.25(c) above
5. If two vectors are perpendicular to each other, their resultant can be easily calculated.
• We have:
The magnitude of the resultant = $\mathbf\small{|\vec{R}|=|(\vec{p_{A2}}+\vec{p_{B2}})|=\sqrt{|\vec{p_{A2}}|^2+|\vec{p_{B2}}|^2}}$
• Where:
$\mathbf\small{|\vec{p_{A2}}|=m_A \times |\vec{v_{A2}}|}$ = 1 × 12 = 12 kg ms-1.
$\mathbf\small{|\vec{p_{B2}}|=m_B \times |\vec{v_{B2}}|}$ = 2 × 8 = 16 kg ms-1.
• Thus we get:
$\mathbf\small{|\vec{R}|=\sqrt{12^2+16^2}=20\: kg\: ms^{-1}}$
6. Now consider the result in (3):
• We have an equality: $\mathbf\small{\vec{R}=-\vec{p_{C2}}}$
• This is shown in fig.5.25(d)
• We can write two points:
(i) Magnitude of ${\mathbf\small\vec{R}}$ is same as the magnitude of $\mathbf{\small\vec{p_{C2}}}$
(ii) Direction of $\mathbf{\small\vec{R}}$ is opposite to the direction of $\mathbf\small{\vec{p_{C2}}}$
7. The first point is important to us. Based on that, we can write:
$\mathbf\small{|\vec{p_{C2}}|=20\: kg\: ms^{-1}}$
• But $\mathbf\small{|\vec{p_{C2}}|=m_C \times |\vec{v_{C2}}|=m_C \times 4}$
• So we get: (mc × 4) = 20 ⇒ mc = 5 kg
Solved example 5.12
Two objects, each of mass 5 kg move in a straight line towards each other with the same speed of 3 ms-1. After collision, they stick together. What will be the velocity of the combined mass after collision?
Solution:
• For this problem, there is one-dimensional motion only. See fig.5.26 below:
• So we do not need to use vector notations.
1. Applying the law of conservation of momentum, we have:
pA1 + pB1 = p(A+B)
⇒ mA × vA1 + mB × vB1 = m(A+B) × v(A+B)
2. Substituting the values, we get:
⇒ 5 × 3 + 5 × -3 = 10 × v(A+B)
⇒ 0 = 10 × v(A+B).
⇒ v(A+B) = 0
Solved example 5.13
A nucleus is at rest in the laboratory frame of reference. Show that if it disintegrates into two smaller nuclei the products must move in opposite directions.
Solution:
1. Applying the law of conservation of momentum, we have:
$\mathbf\small{\vec{p_{(A+B)}}=\vec{p_{A2}}+\vec{p_{B2}}}$
Where
• $\mathbf\small{\vec{p_{A2}}}$ is the final momentum of the first smaller nucleus A
• $\mathbf\small{\vec{p_{B2}}}$ is the final momentum of the second smaller nucleus B
2. But the original nucleus was at rest. So we have: $\mathbf\small{\vec{p_{(A+B)}}}$ = 0
• Substituting this in (1), we get:
$\mathbf\small{\vec{p_{A2}}=-\vec{p_{B2}}}$
3. So we can write:
$\mathbf\small{m_A \times\vec{v_{A2}}=-m_B \times\vec{v_{B2}}}$
4. The result in (3) gives the condition.
• This condition must be satisfied when the given stationary nucleus disintegrates into two smaller nuclei
• From this condition, we have to prove that, A and B move in opposite directions.
5. In (3), we have an equality of two vectors. Then we can write the two points:
(i) Magnitude of the two vectors must be equal
(ii) Direction of the two vectors must be the same
6. Considering magnitude, we can write:
$\mathbf\small{m_A \times|\vec{v_{A2}|}=-m_B \times|\vec{v_{B2}}|}$
7. Considering direction, we can write:
• Direction of $\mathbf\small{m_A \times\vec{v_{A2}}}$ must be same as the direction of $\mathbf\small{-m_B \times\vec{v_{B2}}}$
• mA and mB are scalar quantities. They are masses. So they cannot be negative.
8. Thus, the directions of the two vectors are decided entirely by the direction of the velocities
• So we can write:
Direction of $\mathbf\small{\vec{v_{A2}}}$ must be same as the diretion of $\mathbf\small{-\vec{v_{B2}}}$
• That is., A must be travelling in the direction which is the exact opposite to the 'direction in which B is travelling'
• So we can write $\mathbf\small{\vec{F_{GB}}\times\Delta t=-m_G \vec{v_{G}}}$
12. So we have the following two information:
(i) The bullet suffers a change in momentum $\mathbf\small{\vec{\Delta p_B}}$:
$\mathbf\small{\vec{\Delta p_B}=\vec{F_{BG}}\times\Delta t=m_B \vec{v_{B}}}$
(ii) The gun suffers a change in momentum $\mathbf\small{\vec{\Delta p_G}}$:
$\mathbf\small{\vec{\Delta p_G}=\vec{F_{GB}}\times\Delta t=-m_G \vec{v_{G}}}$
13. Take out the first and second part from 12(i) and 12(ii):
(i) $\mathbf\small{\vec{\Delta p_B}=\vec{F_{BG}}\times\Delta t}$
(ii) $\mathbf\small{\vec{\Delta p_G}=\vec{F_{GB}}\times\Delta t}$
14. Applying Newton's third law, we have:
$\mathbf\small{\vec{F_{BG}}}$ = -$\mathbf\small{\vec{F_{GB}}}$
15. So we can replace $\mathbf\small{\vec{F_{BG}}}$ in 13(i). We get:
$\mathbf\small{\vec{\Delta p_B}=-\vec{F_{GB}}\times\Delta t}$
• But from 13(ii), $\mathbf\small{-\vec{F_{GB}}\times\Delta t}$ = $\mathbf\small{-\vec{\Delta p_G}}$
• So we get: $\mathbf\small{\vec{\Delta p_B}=-\vec{\Delta p_G}}$
■ That is., 'change in momenta' are numerically equal, but have opposite signs
16. Now, the change in momentum for the bullet can be written as:
• Change in momentum of bullet = Final momentum of bullet - Initial momentum of bullet
• That is., $\mathbf\small{\vec{\Delta p_B}=[\vec{p_{B1}}-\vec{p_{B2}}]}$
17. Similarly, the change in momentum for the gun can be written as:
• Change in momentum of gun = Final momentum of gun - Initial momentum of gun
• That is., $\mathbf\small{\vec{\Delta p_G}=[\vec{p_{G1}}-\vec{p_{G2}}]}$
18. But according to the result in (15), the two 'change in momenta' are numerically equal, but have opposite signs
• So we can change the sign of (17) and equate it to (16). We get:
$\mathbf\small{[\vec{p_{B1}}-\vec{p_{B2}}]}= {-[\vec{p_{G1}}-\vec{p_{G2}}]}$
$\mathbf\small{\Rightarrow \vec{p_{B1}}-\vec{p_{B2}}}= {-\vec{p_{G1}}-\vec{p_{G2}}}$
$\mathbf\small{\Rightarrow \vec{p_{B1}}+\vec{p_{G1}}}= {\vec{p_{B2}}+\vec{p_{G2}}}$
19. On the left side, we have:
• Sum of momenta before the firing of the gun
On the right side, we have:
• Sum of momenta after the firing of the gun
• So we can write: The total momentum of the (bullet + gun) system is conserved
20. The Law of Conservation of momentum states that:
■ The total momentum of an isolated system of interacting particles is conserved
Note: The 'Recoil of a gun' is essentially a one-dimensional problem. We do not need to use vector notations. But in the above discussion we used them, to get a general understanding of the Law of conservation of momentum
Solved example 5.10
A shell of mass 0.020 kg is fired by a gun of mass 100 kg. If the muzzle speed of the shell is 80 ms-1, what is the recoil speed of the gun?
Solution:
• For this problem, there is one-dimensional motion only. So we do not need to use vector notations.
1. Applying the law of conservation of momentum, we have:
pS1 + pG1 = pS2 + pG2
2. But pS1 = pG1 = 0
• So we get: pS2 = -pG2
⟹ mS × vS2 = mG × vG2
3. Substituting the values, we get:
0.020 × 80 = 100 × vG2
⟹ vG2 = 0.016 ms-1.
Solved example 5.11
In a quarry, a stationary rock explodes into 3 parts. A 1 kg part and another 2 kg part moves at right angles to each other. Their velocities are respectively 12 ms-1 and 8 ms-1. If the remaining third part moves with a velocity of 4 ms-1, what is it's mass?
Solution:
• Let the first two parts be named as 'A' and 'B' and the third part be named as 'C'. This is shown in fig.5.25(b) below:
 |
| Fig.5.25 |
$\mathbf\small{m_A=1\, \text{kg},\: m_B=2\,\, \text{kg},\: \vec{v_{A2}}=12\, \,ms^{-1},\: \vec{v_{B2}}=8\, \,ms^{-1},\: \vec{v_{C2}}=4\, \,ms^{-1},\: m_C=?}$
1. Applying the law of conservation of momentum, we have:
$\mathbf\small{\vec{p_{A1}}+\vec{p_{B1}}+\vec{p_{C1}}}= {\vec{p_{A2}}+\vec{p_{B2}}+\vec{p_{C2}}}$
2. But the initial momenta are all zero. So we get:
$\mathbf\small{\vec{p_{A2}}+\vec{p_{B2}}+\vec{p_{C2}}=0}$
$\mathbf\small{\Rightarrow \vec{p_{A2}}+\vec{p_{B2}}=-\vec{p_{C2}}}$
• That is., the vector $\mathbf\small{\vec{p_{C2}}}$ must be equal and opposite to the resultant of the two vectors $\mathbf\small{\vec{p_{A2}}}$ and $\mathbf\small{\vec{p_{B2}}}$
3. We will denote the resultant as $\mathbf\small{\vec{R}}$
We can write: $\mathbf\small{\vec{R}=(\vec{p_{A2}}+\vec{p_{B2}})}$
$\mathbf\small{\Rightarrow \vec{R}=-\vec{p_{C2}}}$
4. So let us find $\mathbf\small{\vec{R}}$ first:
• Given that, the two masses A and B move perpendicular to each other
• So their momentum vectors are perpendicular to each other. This is shown in fig.5.25(c) above
5. If two vectors are perpendicular to each other, their resultant can be easily calculated.
• We have:
The magnitude of the resultant = $\mathbf\small{|\vec{R}|=|(\vec{p_{A2}}+\vec{p_{B2}})|=\sqrt{|\vec{p_{A2}}|^2+|\vec{p_{B2}}|^2}}$
• Where:
$\mathbf\small{|\vec{p_{A2}}|=m_A \times |\vec{v_{A2}}|}$ = 1 × 12 = 12 kg ms-1.
$\mathbf\small{|\vec{p_{B2}}|=m_B \times |\vec{v_{B2}}|}$ = 2 × 8 = 16 kg ms-1.
• Thus we get:
$\mathbf\small{|\vec{R}|=\sqrt{12^2+16^2}=20\: kg\: ms^{-1}}$
6. Now consider the result in (3):
• We have an equality: $\mathbf\small{\vec{R}=-\vec{p_{C2}}}$
• This is shown in fig.5.25(d)
• We can write two points:
(i) Magnitude of ${\mathbf\small\vec{R}}$ is same as the magnitude of $\mathbf{\small\vec{p_{C2}}}$
(ii) Direction of $\mathbf{\small\vec{R}}$ is opposite to the direction of $\mathbf\small{\vec{p_{C2}}}$
7. The first point is important to us. Based on that, we can write:
$\mathbf\small{|\vec{p_{C2}}|=20\: kg\: ms^{-1}}$
• But $\mathbf\small{|\vec{p_{C2}}|=m_C \times |\vec{v_{C2}}|=m_C \times 4}$
• So we get: (mc × 4) = 20 ⇒ mc = 5 kg
Solved example 5.12
Two objects, each of mass 5 kg move in a straight line towards each other with the same speed of 3 ms-1. After collision, they stick together. What will be the velocity of the combined mass after collision?
Solution:
• For this problem, there is one-dimensional motion only. See fig.5.26 below:
 |
| Fig.5.26 |
1. Applying the law of conservation of momentum, we have:
pA1 + pB1 = p(A+B)
⇒ mA × vA1 + mB × vB1 = m(A+B) × v(A+B)
2. Substituting the values, we get:
⇒ 5 × 3 + 5 × -3 = 10 × v(A+B)
⇒ 0 = 10 × v(A+B).
⇒ v(A+B) = 0
Solved example 5.13
A nucleus is at rest in the laboratory frame of reference. Show that if it disintegrates into two smaller nuclei the products must move in opposite directions.
Solution:
1. Applying the law of conservation of momentum, we have:
$\mathbf\small{\vec{p_{(A+B)}}=\vec{p_{A2}}+\vec{p_{B2}}}$
Where
• $\mathbf\small{\vec{p_{A2}}}$ is the final momentum of the first smaller nucleus A
• $\mathbf\small{\vec{p_{B2}}}$ is the final momentum of the second smaller nucleus B
2. But the original nucleus was at rest. So we have: $\mathbf\small{\vec{p_{(A+B)}}}$ = 0
• Substituting this in (1), we get:
$\mathbf\small{\vec{p_{A2}}=-\vec{p_{B2}}}$
3. So we can write:
$\mathbf\small{m_A \times\vec{v_{A2}}=-m_B \times\vec{v_{B2}}}$
4. The result in (3) gives the condition.
• This condition must be satisfied when the given stationary nucleus disintegrates into two smaller nuclei
• From this condition, we have to prove that, A and B move in opposite directions.
5. In (3), we have an equality of two vectors. Then we can write the two points:
(i) Magnitude of the two vectors must be equal
(ii) Direction of the two vectors must be the same
6. Considering magnitude, we can write:
$\mathbf\small{m_A \times|\vec{v_{A2}|}=-m_B \times|\vec{v_{B2}}|}$
7. Considering direction, we can write:
• Direction of $\mathbf\small{m_A \times\vec{v_{A2}}}$ must be same as the direction of $\mathbf\small{-m_B \times\vec{v_{B2}}}$
• mA and mB are scalar quantities. They are masses. So they cannot be negative.
8. Thus, the directions of the two vectors are decided entirely by the direction of the velocities
• So we can write:
Direction of $\mathbf\small{\vec{v_{A2}}}$ must be same as the diretion of $\mathbf\small{-\vec{v_{B2}}}$
• That is., A must be travelling in the direction which is the exact opposite to the 'direction in which B is travelling'
In the next section we will see Equilibrium of a particle
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