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
• →vB is the velocity with which the bullet is fired
• →vG 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 = →pBx1=mB×→vBx1=mB×0=0
(ii) Final momentum of the bullet in the x direction = →pBx2=mB×→vBx2=mB×→vB
■ So change in momentum in the x direction = Δ→pBx=(→pBx2−→pBx1)=(mB→vB−0)=mB→vB
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: Δ→pBy = 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 →FBG
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
• →vB is the velocity with which the bullet is fired
• →vG 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 = →pBx1=mB×→vBx1=mB×0=0
(ii) Final momentum of the bullet in the x direction = →pBx2=mB×→vBx2=mB×→vB
■ So change in momentum in the x direction = Δ→pBx=(→pBx2−→pBx1)=(mB→vB−0)=mB→vB
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: Δ→pBy = 0
3. Resultant change in momentum Δ→pB is the resultant of Δ→pBx and Δ→pBy
• But Δ→pBy = 0
■ So we get: Δ→pB=Δ→pBx=mB→vB
4. We can write the following four points:
(i) The bullet suffers a change in momentum of mB→vB
(ii) This 'change in momentum' is a vector quantity
(iii) It's magnitude is mB|→vB|
(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 = →pGx1=mG×→vGx1=mG×0=0
(ii) Final momentum of the gun in the x direction = →pGx2=mG×→vGx2=−mG×→vG
■ So change in momentum in the x direction = Δ→pGx=(→pGx2−→pGx1)=(−mG→vG−0)=−mG→vG
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: Δ→pGy = 0
6. Resultant change in momentum Δ→pG is the resultant of Δ→pGx and Δ→pGy
• But Δ→pBy = 0
■ So we get: Δ→pB=Δ→pBx=mB→vB
4. We can write the following four points:
(i) The bullet suffers a change in momentum of mB→vB
(ii) This 'change in momentum' is a vector quantity
(iii) It's magnitude is mB|→vB|
(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 = →pGx1=mG×→vGx1=mG×0=0
(ii) Final momentum of the gun in the x direction = →pGx2=mG×→vGx2=−mG×→vG
■ So change in momentum in the x direction = Δ→pGx=(→pGx2−→pGx1)=(−mG→vG−0)=−mG→vG
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: Δ→pGy = 0
6. Resultant change in momentum Δ→pG is the resultant of Δ→pGx and Δ→pGy
• But Δ→pGy = 0
■ So we get: Δ→pG=Δ→pGx=−mB→vG
7. We can write the following four points:
(i) The gun suffers a change in momentum of −mG→vG
(ii) This 'change in momentum' is a vector quantity
(iii) It's magnitude is mG|→vG|
(iv) It's direction is towards the negative side of the x axis
■ So we get: Δ→pG=Δ→pGx=−mB→vG
7. We can write the following four points:
(i) The gun suffers a change in momentum of −mG→vG
(ii) This 'change in momentum' is a vector quantity
(iii) It's magnitude is mG|→vG|
(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 →FBG
• Let the time duration for which the force acts be Δt
• Consider the product: →FBG×Δt
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 →FBG×Δt=mB→vB
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 →FGB
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 →FBG×Δt=mB→vB
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 →FGB
• The time duration for which the force acts will be the same Δt
• Consider the product: →FGB×Δt
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 →FGB×Δt=−mG→vG
12. So we have the following two information:
(i) The bullet suffers a change in momentum →ΔpB:
→ΔpB=→FBG×Δt=mB→vB
(ii) The gun suffers a change in momentum →ΔpG:
→ΔpG=→FGB×Δt=−mG→vG
13. Take out the first and second part from 12(i) and 12(ii):
(i) →ΔpB=→FBG×Δt
(ii) →ΔpG=→FGB×Δt
14. Applying Newton's third law, we have:
→FBG = -→FGB
15. So we can replace →FBG in 13(i). We get:
→ΔpB=−→FGB×Δt
• But from 13(ii), −→FGB×Δt = −→ΔpG
• So we get: →ΔpB=−→ΔpG
■ 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., →ΔpB=[→pB1−→pB2]
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., →ΔpG=[→pG1−→pG2]
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:
[→pB1−→pB2]=−[→pG1−→pG2]
⇒→pB1−→pB2=−→pG1−→pG2
⇒→pB1+→pG1=→pB2+→pG2
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:
mA=1kg,mB=2kg,→vA2=12ms−1,→vB2=8ms−1,→vC2=4ms−1,mC=?
1. Applying the law of conservation of momentum, we have:
→pA1+→pB1+→pC1=→pA2+→pB2+→pC2
2. But the initial momenta are all zero. So we get:
→pA2+→pB2+→pC2=0
⇒→pA2+→pB2=−→pC2
• That is., the vector →pC2 must be equal and opposite to the resultant of the two vectors →pA2 and →pB2
3. We will denote the resultant as →R
We can write: →R=(→pA2+→pB2)
⇒→R=−→pC2
4. So let us find →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 = |→R|=|(→pA2+→pB2)|=√|→pA2|2+|→pB2|2
• Where:
|→pA2|=mA×|→vA2| = 1 × 12 = 12 kg ms-1.
|→pB2|=mB×|→vB2| = 2 × 8 = 16 kg ms-1.
• Thus we get:
|→R|=√122+162=20kgms−1
6. Now consider the result in (3):
• We have an equality: →R=−→pC2
• This is shown in fig.5.25(d)
• We can write two points:
(i) Magnitude of →R is same as the magnitude of →pC2
(ii) Direction of →R is opposite to the direction of →pC2
7. The first point is important to us. Based on that, we can write:
|→pC2|=20kgms−1
• But |→pC2|=mC×|→vC2|=mC×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:
→p(A+B)=→pA2+→pB2
Where
• →pA2 is the final momentum of the first smaller nucleus A
• →pB2 is the final momentum of the second smaller nucleus B
2. But the original nucleus was at rest. So we have: →p(A+B) = 0
• Substituting this in (1), we get:
→pA2=−→pB2
3. So we can write:
mA×→vA2=−mB×→vB2
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:
mA×|→vA2|=−mB×|→vB2|
7. Considering direction, we can write:
• Direction of mA×→vA2 must be same as the direction of −mB×→vB2
• 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 →vA2 must be same as the diretion of −→vB2
• 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 →FGB×Δt=−mG→vG
12. So we have the following two information:
(i) The bullet suffers a change in momentum →ΔpB:
→ΔpB=→FBG×Δt=mB→vB
(ii) The gun suffers a change in momentum →ΔpG:
→ΔpG=→FGB×Δt=−mG→vG
13. Take out the first and second part from 12(i) and 12(ii):
(i) →ΔpB=→FBG×Δt
(ii) →ΔpG=→FGB×Δt
14. Applying Newton's third law, we have:
→FBG = -→FGB
15. So we can replace →FBG in 13(i). We get:
→ΔpB=−→FGB×Δt
• But from 13(ii), −→FGB×Δt = −→ΔpG
• So we get: →ΔpB=−→ΔpG
■ 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., →ΔpB=[→pB1−→pB2]
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., →ΔpG=[→pG1−→pG2]
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:
[→pB1−→pB2]=−[→pG1−→pG2]
⇒→pB1−→pB2=−→pG1−→pG2
⇒→pB1+→pG1=→pB2+→pG2
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 |
mA=1kg,mB=2kg,→vA2=12ms−1,→vB2=8ms−1,→vC2=4ms−1,mC=?
1. Applying the law of conservation of momentum, we have:
→pA1+→pB1+→pC1=→pA2+→pB2+→pC2
2. But the initial momenta are all zero. So we get:
→pA2+→pB2+→pC2=0
⇒→pA2+→pB2=−→pC2
• That is., the vector →pC2 must be equal and opposite to the resultant of the two vectors →pA2 and →pB2
3. We will denote the resultant as →R
We can write: →R=(→pA2+→pB2)
⇒→R=−→pC2
4. So let us find →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 = |→R|=|(→pA2+→pB2)|=√|→pA2|2+|→pB2|2
• Where:
|→pA2|=mA×|→vA2| = 1 × 12 = 12 kg ms-1.
|→pB2|=mB×|→vB2| = 2 × 8 = 16 kg ms-1.
• Thus we get:
|→R|=√122+162=20kgms−1
6. Now consider the result in (3):
• We have an equality: →R=−→pC2
• This is shown in fig.5.25(d)
• We can write two points:
(i) Magnitude of →R is same as the magnitude of →pC2
(ii) Direction of →R is opposite to the direction of →pC2
7. The first point is important to us. Based on that, we can write:
|→pC2|=20kgms−1
• But |→pC2|=mC×|→vC2|=mC×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:
→p(A+B)=→pA2+→pB2
Where
• →pA2 is the final momentum of the first smaller nucleus A
• →pB2 is the final momentum of the second smaller nucleus B
2. But the original nucleus was at rest. So we have: →p(A+B) = 0
• Substituting this in (1), we get:
→pA2=−→pB2
3. So we can write:
mA×→vA2=−mB×→vB2
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:
mA×|→vA2|=−mB×|→vB2|
7. Considering direction, we can write:
• Direction of mA×→vA2 must be same as the direction of −mB×→vB2
• 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 →vA2 must be same as the diretion of −→vB2
• 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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