Exercise - 4

How to think about this problem:

1. First, identify what the teacher has given you: Initial horizontal velocity u_x = 10 m/s, initial vertical velocity u_y = 0, and vertical acceleration due to gravity g downwards.

2. Next, identify what the teacher is asking you to find: Velocity vector v⃗(t) and speed (magnitude) at time t.

3. Now ask yourself: How do horizontal and vertical motions behave independently in 2D kinematics? Horizontal velocity remains constant because a_x = 0, while vertical velocity increases linearly downwards due to gravity (a_y = -g).

4. Calculate component velocities: v_x = 10 m/s and v_y = -gt m/s.

5. Combine components into vector form: v⃗(t) = 10î - gt ĵ.

6. Compute total magnitude using Pythagorean theorem: v = √(v_x² + v_y²) = √(100 + g²t²).

The key idea to understand is: projectile motion is a combination of uniform horizontal motion and uniformly accelerated vertical motion.

Before moving on, check yourself: Do you see why displacement equations like (1/2)gt² cannot be directly substituted for velocity? If yes, you have mastered velocity vectors in kinematics!

How to think about this problem:

1. First, identify what the teacher has given you: Initial horizontal speed u_x = 10 m/s, initial vertical speed u_y = 0, time t = 2 s, and g = 9.8 m/s².

2. Next, identify what the teacher is asking you to find: Velocity vector v⃗ and total speed at t = 2 s.

3. Now ask yourself: How do we compute component velocities at a specific timestamp? Horizontal velocity remains fixed at 10 m/s, while vertical velocity gains 9.8 m/s every second downwards (v_y = -9.8 × 2 = -19.6 m/s).

4. Form the vector: Combine horizontal (10î) and downward vertical (-19.6ĵ) components into 10î - 19.6ĵ m/s.

5. Compute total speed: Apply Pythagorean theorem √(10² + 19.6²) = √(100 + 384.16) = √484.16 ≈ 22.00 m/s.

The key idea to understand is: to find exact numerical velocity at any time t, substitute t directly into the general parametric equations v_x = u_x and v_y = -gt.

Before moving on, check yourself: Can you calculate the angle θ below the horizontal using tan(θ) = v_y / v_x? If yes, you have fully mastered 2D projectile kinematics!

How to understand and solve this problem step-by-step:

1. How to understand the problem:
- The ball is inside an accelerating car up to time t_release = 4 s. Due to inertia, the ball shares the exact horizontal velocity (v_x) of the car at the instant it is dropped.
- Once dropped into the air, the car's engine can no longer push the ball. Hence, horizontal acceleration drops to zero (a_x = 0 m/s²), and the ball enters horizontal projectile motion under gravity alone (a_y = g = 10 m/s²).

2. Step 1: Find the initial horizontal velocity (v_x) at the moment of release (t_release = 4 s).
- Apply 1D kinematics equation along the horizontal axis: v_x = u_x + (a_car × t_release).
- v_x = 0 + (5 m/s² × 4 s) = 20 m/s. This value stays constant throughout the rest of the flight because a_x = 0 m/s².

3. Step 2: Determine the acceleration of the ball at t = 6 s.
- Since only gravity acts on the free-falling ball, acceleration along the y-axis is a_y = 10 m/s², while a_x = 0 m/s².
- The net acceleration vector is a_net = 10 m/s² downwards.

4. Step 3: Find the vertical velocity component (v_y) gained after release.
- Calculate time elapsed in free fall: Δt = t - t_release = 6 s - 4 s = 2 s.
- Apply 1D kinematics along vertical axis: v_y = u_y + (g × Δt) = 0 + (10 m/s² × 2 s) = 20 m/s.

5. Step 4: Combine orthogonal velocity components to find total magnitude (v_net).
- Use Pythagorean addition: v_net = √(v_x² + v_y²).
- v_net = √((20 m/s)² + (20 m/s)²) = √(400 + 400) = √800 = 20√2 m/s.

Key Takeaway: Vehicle acceleration affects an object only while inside/connected. Once released, inertia retains the current horizontal velocity v_x, while gravity g becomes the sole acceleration factor!

How to understand and solve this problem step-by-step:

1. How to understand the problem:
- In any collision where no net external force acts on the system, total linear momentum is conserved.
- The 1 kg moving ball transfers its entire momentum to the heavier 2 kg target ball.

2. Step 1: Calculate total initial momentum (p_initial) before impact.
- p_initial = m₁u₁ + m₂u₂ = (1 kg × 100 m/s) + (2 kg × 0 m/s) = 100 kg·m/s.

3. Step 2: Calculate momentum carried away by the second ball (p₂_final).
- p₂_final = m₂v₂ = (2 kg × 50 m/s) = 100 kg·m/s.

4. Step 3: Find remaining momentum left for the 1 kg ball (p₁_final).
- p₁_final = p_initial - p₂_final = 100 kg·m/s - 100 kg·m/s = 0 kg·m/s.
- Therefore, v₁ = p₁_final / m₁ = 0 / 1 = 0 m/s.

Key Takeaway: Since the 2 kg ball absorbs all 100 kg·m/s of the initial momentum, the 1 kg ball is left with zero momentum and comes to a complete halt immediately after the collision!

How to understand and solve this problem step-by-step:

1. How to understand the problem:
- This problem combines two fundamental concepts: Conservation of Mechanical Energy during the block's vertical swing, and Conservation of Linear Momentum during the collision.
- The length of the string (5 m) is extra information and is not needed to calculate the height rise.

2. Step 1: Find the speed of the block (V_block) right after impact using energy conservation.
- The kinetic energy gained by the block immediately after collision converts into gravitational potential energy at its highest point:
(1/2) M (V_block)² = M g h ⇒ V_block = √(2 g h).
- Substituting values: V_block = √(2 × 9.8 × 0.1) = √1.96 = 1.4 ms⁻¹.

3. Step 2: Set up the momentum conservation equation.
- Momentum before collision: p_initial = m_bullet × u_bullet + M_block × 0 = (0.01 kg × 400 ms⁻¹) = 4 kg·ms⁻¹.
- Momentum after collision: p_final = m_bullet × v_bullet + M_block × V_block = (0.01 × v_bullet) + (2 × 1.4).

4. Step 3: Equate initial and final momentum to find v_bullet.
- 4 = 0.01 v_bullet + 2.8
- 0.01 v_bullet = 1.2 ⇒ v_bullet = 120 ms⁻¹.

Key Takeaway: Whenever an object swings upward to height h after a collision, calculate its post-collision speed using V = √(2gh) first, then apply momentum conservation to find the remaining unknowns!