Exercise - 3

How to think about this problem:

1. First, identify what the teacher has given you. Mass m = 3 kg, speed v = 10 m/s at θ = 30° to the normal, contact time Δt = 2 s, and elastic rebound.

2. Next, identify what the teacher is asking you to find. Impulse (J) and Average Force (F_avg) exerted by the wall.

3. Now ask yourself: Which components of velocity change during a wall collision? The parallel component (y-axis) remains unchanged, while the perpendicular component (x-axis) completely reverses direction.

4. Calculate the change in horizontal momentum: Δp_x = 2 m v cos(θ) = 2 × 3 × 10 × cos(30°) = 30√3 N·s. This gives the impulse J.

5. Relate Impulse to Average Force: Impulse = Force × Time (J = F_avg × Δt).

6. Solve for Force: F_avg = J / Δt = (30√3) / 2 = 15√3 N.

The key idea to understand is: impulse is the total change in linear momentum, and average force is simply impulse divided by the contact time duration.

Before moving on, check yourself: Do you see why the y-component of velocity produces zero impulse? If yes, you have understood two-dimensional momentum balance!

How to think about this problem:

1. First, identify what the teacher has given you. Mass m = 3 kg, speed v = 10 m/s at θ = 30° with respect to the vertical (+y-axis), contact time Δt = 2 s, and elastic rebound.

2. Next, identify what the teacher is asking you to find. Impulse (J) and Average Force (F_avg) exerted by the wall.

3. Now ask yourself: How does measuring the angle from the y-axis (wall surface) change the velocity components? The normal component becomes v sin(θ) instead of v cos(θ).

4. Calculate the change in horizontal momentum: Δp_x = 2 m v sin(30°) = 2 × 3 × 10 × (1/2) = 30 N·s. This gives the impulse J.

5. Relate Impulse to Average Force: Impulse = Force × Time (J = F_avg × Δt).

6. Solve for Force: F_avg = J / Δt = 30 / 2 = 15 N.

The key idea to understand is: always check whether the angle is given relative to the normal (x-axis) or the wall (y-axis), as this swaps cos(θ) and sin(θ) in the horizontal momentum calculation.

Before moving on, check yourself: Do you see why sin(30°) is used here instead of cos(30°)? If yes, you have mastered angle resolution in momentum problems!

How to think about this problem:

1. First, identify what the teacher has given you. A bomb of mass 1 kg initially at rest explodes into two equal pieces (0.5 kg each), and one piece moves at +5 m/s.

2. Next, identify what the teacher is asking you to find. The velocity (speed and direction) of the second part.

3. Now ask yourself: Which physics concept applies here? The Law of Conservation of Linear Momentum (Initial Momentum = Final Momentum).

4. Recall the equation: 0 = m₁v₁ + m₂v₂. Since the masses are equal (m₁ = m₂), the equation simplifies directly to v₁ + v₂ = 0.

5. Solve for v₂: v₂ = -v₁ = -5 m/s.

The key idea to understand is: when an object at rest splits into two equal halves, both halves must fly apart with equal speeds in exact opposite directions to keep the total momentum zero.

Before moving on, check yourself: Do you see why equal masses mean equal and opposite velocities? If yes, you have understood the basic idea of two-body momentum balance!

How to think about this problem:

1. First, identify what the teacher has given you. A bomb of mass 1 kg initially at rest explodes into two parts. One part has a mass of 300 g (0.3 kg) and moves at 10 m/s.

2. Next, identify what the teacher is asking you to find. The velocity of the remaining part.

3. Now ask yourself: Which physics concept connects initial rest state to the motion of exploded fragments? The Law of Conservation of Linear Momentum.

4. First calculate the mass of the second fragment: 1 kg - 0.3 kg = 0.7 kg.

5. Set up the conservation equation: 0 = m₁v₁ + m₂v₂ => 0 = (0.3 × 10) + 0.7v₂.

6. Solve for v₂: v₂ = -3/0.7 = -30/7 m/s.

The key idea to understand is: unequal masses moving apart from rest will have inversely proportional speeds, with the lighter mass moving faster and the heavier mass moving slower in the opposite direction.

Before moving on, check yourself: Do you see why converting grams to kilograms is a crucial first step? If yes, you have mastered this problem!

How to think about this problem:

1. First, identify what the teacher has given you. A bomb of mass 1 kg initially at rest explodes into 3 equal pieces, and all pieces move with equal speeds (v = 5 m/s).

2. Next, identify what the teacher is asking you to find. You need to determine the velocity (magnitude and relative angle) of the remaining two parts.

3. Now ask yourself: Which physics concept connects initial rest state and explosion of fragments? This is the Law of Conservation of Linear Momentum.

4. Recall the main vector equation: Initial Momentum = Final Momentum. Since the bomb was at rest, 0 = m₁v₁ + m₂v₂ + m₃v₃.

5. Since the masses are equal, simplify the equation to vector sum: v₁ + v₂ + v₃ = 0, which means the three equal velocity vectors form a closed equilateral triangle.

6. Solve for the relative angle using vector addition formulas: v₂² + v₃² + 2v₂v₃ cos(θ) = v₁², which yields cos(θ) = -1/2, giving θ = 120°.

The key idea to understand is: when a body at rest breaks into 3 equal fragments with equal speeds, they must fly symmetrically in a plane at 120° to each other to maintain zero net momentum.

Before moving on, check yourself: Do you understand why vector directions matter in momentum conservation and how 120° balances the system? If yes, you have understood the basic idea of this problem.

How to think about this problem:

1. First, identify what the teacher has given you. A bomb of mass 1 kg initially at rest explodes into 3 equal parts. The first part moves at 5 m/s, and the other two parts fly off at right angles (90°) to each other with equal speeds.

2. Next, identify what the teacher is asking you to find. You need to calculate the speed of the other two parts.

3. Now ask yourself: Which physics concept connects an explosion at rest to the motion of its fragments? This is the Law of Conservation of Linear Momentum.

4. Recall the main vector equation: Initial Momentum = Final Momentum. Since the bomb was at rest, 0 = m₁v₁ + m₂v₂ + m₃v₃.

5. Since the masses are equal, simplify the vector sum: v₂ + v₃ = -v₁. This means the vector sum of the second and third parts must equal the magnitude of the first part (5 m/s) in opposite direction.

6. Use the Pythagorean theorem for perpendicular vectors: v₂² + v₃² = v₁². Since v₂ = v₃ = v, solve 2v² = 5² to get v = 5/√2 m/s.

The key idea to understand is: when two perpendicular vector components combine, their resultant magnitude is found using v₂² + v₃² = v₁², which must balance out the first part to keep total momentum zero.

Before moving on, check yourself: Do you understand why perpendicular vectors simplify the dot product term to zero? If yes, you have understood the basic idea of this problem.

How to think about this problem:

1. First, identify what the teacher has given you. A bomb at rest explodes into 3 equal pieces. The 1st part goes along +x at 5 m/s, the 2nd part along +y, and the 3rd part along -x.

2. Next, identify what the teacher is asking you to find. You need to determine the magnitudes of velocity (speeds) for the 2nd and 3rd parts.

3. Now ask yourself: Which physics concept applies here? Conservation of Linear Momentum along perpendicular axes (x-axis and y-axis independently).

4. Recall the component form of momentum conservation: Total momentum along x-axis = 0, and total momentum along y-axis = 0.

5. Analyze x-axis: The 1st part (+5) must be balanced entirely by the 3rd part (-x direction), giving v₃ = 5 m/s.

6. Analyze y-axis: Since no part moves in the -y direction to balance a +y motion, the 2nd part cannot move at all, giving v₂ = 0 m/s.

The key idea to understand is: momentum must be conserved along each axis independently. If there is no negative y-momentum, there cannot be any positive y-momentum either.

Before moving on, check yourself: Do you see why component-wise balancing forces the 2nd part's speed to be zero? If yes, you have understood the basic idea of this problem.

How to think about this problem:

1. First, identify what the teacher has given you. A bomb at rest explodes into 3 equal pieces. The 1st part goes along +x at 5 m/s, the 2nd part goes into Quadrant II (135°), and the 3rd part goes into Quadrant III (225°).

2. Next, identify what the teacher is asking you to find. You need to calculate the speeds (magnitudes) of the 2nd and 3rd parts.

3. Now ask yourself: Why must the 2nd and 3rd parts lie on opposite sides of the x-axis? Because to cancel out vertical motion, one part must move upward (+y) and the other downward (-y).

4. Recall momentum conservation in component form: Total x-momentum = 0, and Total y-momentum = 0.

5. Analyze y-axis: Equating +y and -y components reveals that both parts must have equal speeds (v₂ = v₃).

6. Analyze x-axis: The backward x-components of both fragments combined must balance the 5 m/s forward movement of the first fragment.

The key idea to understand is: symmetrical angles above and below the x-axis allow vertical momenta to cancel while their combined horizontal components balance the first part.

Before moving on, check yourself: Do you understand why two angles both above the x-axis (like 40° and 135°) cannot conserve momentum? If yes, you have understood the basic idea of vector balance in explosions.

How to think about this problem:

1. First, identify what the teacher has given you. A bomb at rest explodes into 3 equal pieces, where the first piece moves at 5 m/s.

2. Next, identify what the teacher is asking you to find. You need to find the magnitude of the combined (resultant) velocity of the remaining two parts.

3. Now ask yourself: Which physics concept connects the total momentum before and after an explosion? The Law of Conservation of Linear Momentum.

4. Recall the main vector relation: Initial Momentum = Final Momentum. Since the initial state is at rest, m₁v₁ + m₂v₂ + m₃v₃ = 0.

5. Group the velocity vectors: Combine v₂ and v₃ into a single resultant vector v₂₃ = v₂ + v₃, which simplifies the equation to v₁ + v₂₃ = 0.

6. Solve for the resultant: v₂₃ = -v₁, which directly shows that the resultant vector has the exact same magnitude (5 m/s) but points in the opposite direction.

The key idea to understand is: for any object exploding from rest, the vector sum (resultant) of any subset of fragments must always equal and oppose the combined momentum of the remaining fragments.

Before moving on, check yourself: Do you see why you don't need to know the individual angles or speeds of the second and third parts to find their combined resultant? If yes, you have understood the basic idea of vector balance.

How to think about this problem:

1. First, identify what the teacher has given you. A 1 kg body breaks into ratio 1:1:3 (masses 0.2 kg, 0.2 kg, 0.6 kg). Two lighter pieces move perpendicular to each other at 30 m/s.

2. Next, identify what the teacher is asking you to find. The speed of the heavier 0.6 kg fragment.

3. Now ask yourself: Which physics concept applies here? Law of Conservation of Linear Momentum (Initial Momentum = Final Momentum = 0).

4. Calculate the individual momenta of the lighter pieces: p = m × v = 0.2 kg × 30 m/s = 6 kg·m/s each.

5. Combine perpendicular momenta using the Pythagorean theorem: p₁₂ = √(6² + 6²) = 6√2 kg·m/s.

6. Balance this resultant with the heavier piece's momentum: m₃v₃ = 6√2 kg·m/s. Divide by m₃ (0.6 kg) to isolate v₃ = 10√2 m/s.

The key idea to understand is: the linear momentum vector of the heavier piece must equal and oppose the vector sum of the lighter pieces' momenta to keep total momentum zero.

Before moving on, check yourself: Do you understand how ratio partitioning works to find fragment masses (1/5 and 3/5) and why vector addition is necessary for perpendicular velocities? If yes, you have mastered this problem!