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Applications of Conservation of Momentum
Applications of momentum conservation include recoil, propulsion, collisions, ball-and-bat interactions, vehicle interactions and fragmentation.
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3.3.1 Applications of Conservation of Momentum
16. Recoil
When two parts of a system move apart from rest, their momenta are equal in magnitude and opposite in direction. Initially: p_i = 0 After separation: m₁v₁ + m₂v₂ = 0 Therefore: m₁v₁ = −m₂v₂ A lighter object generally moves faster, while a heavier object moves more slowly in the opposite direction.
17. Recoil Formula
Suppose an object of mass mb moves forward with velocity vb, while a heavier launcher of mass mc recoils with velocity vc. Since the system was initially at rest: 0 = mb vb + mc vc Therefore: mc vc = −mb vb vc = −mb vbmc The negative sign shows that recoil occurs in the opposite direction.
18. Propulsion
Rockets and jet engines operate according to momentum conservation. Hot gases are expelled backward with momentum. The vehicle gains momentum in the opposite direction. For the complete isolated system: Momentum gained by vehicle = −Momentum gained by gases This is an application of Newton’s third law and momentum conservation.
19. Collisions
During a short collision, objects exert internal forces on one another. If the external impulse is negligible: Total momentum before collision = Total momentum after collision Momentum is conserved in: Elastic collisions Inelastic collisions Completely inelastic collisions Kinetic energy, however, is not conserved in every collision.
20. Ball-and-Bat Interaction
When a moving ball interacts with a bat: The ball’s momentum changes. The bat and player-Earth system receive momentum. Equal and opposite interaction forces act. Total momentum is conserved when the complete system is considered and external impulse is negligible. A very short interaction can produce a large average force because: F_avg = ΔpΔt
21. Vehicle Interactions
When two vehicles interact, their individual momenta change. For the selected system: Σ p⃗i = Σ p⃗f Momentum conservation can be applied when external forces during the short interaction are small compared with the internal interaction forces. This analysis does not mean that kinetic energy must also remain constant.
22. Momentum in Fragmentation
If an object initially at rest separates into two parts: p_i = 0 Therefore: m₁v₁ + m₂v₂ = 0 m₁v₁ = −m₂v₂ The momenta are equal in magnitude and opposite in direction. If the masses are equal: m₁ = m₂ then their velocities have equal magnitudes in opposite directions: v₁ = −v₂
23. Solved Example 1: Momentum of an Object
Solved Example 1: Momentum of an Object
Calculate the momentum of a 5 kg object moving at 8 m s⁻¹.
- m = 5 kg
- v = 8 m s⁻¹
- Formula: p = mv
p = (5)(8)
p = 40 kg m s⁻¹
Final answer: p = 40 kg m s⁻¹
24. Book-Based Solved Example 2: Recoil
Book-Based Solved Example 2: Recoil
A 46 g tennis ball is launched from a 1.35 kg launcher. The launcher recoils at 2.1 m s⁻¹. Find the speed of the ball.
- mb = 46 g = 0.046 kg
- mc = 1.35 kg
- Taking the forward direction as positive: v_c = −2.1 m s⁻¹
- Initially, both objects are at rest: p_i = 0
- Conservation: mc vc + mb vb = 0
(1.35)(−2.1) + (0.046)vb = 0
−2.835 + 0.046 vb = 0
0.046 vb = 2.835
vb = 2.8350.046
vb = 61.63 m s⁻¹
Note: Using the rounded intermediate value 2.8, as done in the textbook calculation, gives approximately v_b ≈ 60.87 m s⁻¹
Final answer: vb ≈ 61.6 m s⁻¹
25. Solved Example 3: Two Carts Stick Together
Solved Example 3: Two Carts Stick Together
A 2 kg cart moving at 6 m s⁻¹ strikes a stationary 4 kg cart. The carts move together afterward. Find their common velocity.
- m₁ = 2 kg
- u₁ = 6 m s⁻¹
- m₂ = 4 kg
- u₂ = 0
- Combined mass = 6 kg
- Formula: m₁u₁ + m₂u₂ = (m₁ + m₂)v
(2)(6) + (4)(0) = (2 + 4)v
12 + 0 = 6v
v = 126
v = 2 m s⁻¹
Final answer: v = 2 m s⁻¹
26. Solved Example 4: Opposite Directions
Solved Example 4: Opposite Directions
A 3 kg cart moving right at 4 m s⁻¹ interacts with a 2 kg cart moving left at 5 m s⁻¹. Find their total initial momentum. Take right as positive.
- m₁ = 3 kg
- u₁ = +4 m s⁻¹
- m₂ = 2 kg
- u₂ = −5 m s⁻¹
- Formula: pi = m₁u₁ + m₂u₂
pi = (3)(4) + (2)(−5)
pi = 12 − 10
pi = 2 kg m s⁻¹
Final answer: pi = 2 kg m s⁻¹ rightward
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Q1. Find the momentum of a 6 kg object moving at 7 m s⁻¹.
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Q2. A 2 kg cart moving at 5 m s⁻¹ sticks to a stationary 3 kg cart. Find their common velocity.
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Q3. A 4 kg object moving right at 3 m s⁻¹ and a 2 kg object moving left at 4 m s⁻¹ form a system. Find total momentum.
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Q4. Explain why recoil velocity depends on mass.
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Q5. Explain propulsion using momentum conservation.