A worked-example animation covering two impulse-momentum problems. The first finds impulse, momentum change, and final velocity for a body pushed by a constant force along its direction of motion. The second involves a ball struck by a bat and rebounding at an angle, requiring vector subtraction of momentum to find impulse and average force. Useful for students learning to apply J = F·Δt = Δp in both one-dimensional and two-dimensional collision scenarios.
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Scene 1On-Screen Title: Practice: Impulse and Final Velocity ProblemVisual Prompt: A body of mass $4.0\text{ kg}$ moving at $2.0\text{ m/s}$ acted upon by a constant force of $5.0\text{ N}$ for $2.0\text{ s}$. Question & Answer:Question: A constant force of $5.0\text{ N}$ acts on a body of mass $4.0\text{ kg}$ moving at a velocity of $2.0\text{ m/s}$ for $2.0\text{ s}$ in the direction of its motion. Find: (1) The magnitude of the impulse acting on the body in $\text{N}\cdot\text{s}$, (2) The increase in the body's momentum in $\text{kg}\cdot\text{m/s}$, (3) Its final velocity in $\text{m/s}$. Answer:(1) Impulse: $J = F \times \Delta t = 5.0\text{ N} \times 2.0\text{ s} = \mathbf{10\text{ N}\cdot\text{s}}$ (2) Increase in Momentum: $\Delta p = J = \mathbf{10\text{ kg}\cdot\text{m/s}}$ (3) Final Velocity: $v_f = v_i + \frac{\Delta p}{m} = 2.0 + \frac{10}{4.0} = \mathbf{4.5\text{ m/s}}$ Scene 2On-Screen Title: Practice: Rebound Ball and Average Force ProblemVisual Prompt: A $0.15\text{ kg}$ ball moving horizontally at $30\text{ m/s}$ struck by a bat to rebound at $60^\circ$.Question & Answer:Question: A ball of mass $0.15\text{ kg}$ moves horizontally at $30\text{ m/s}$ and is struck by a bat to rebound at the same speed at an angle of $60^\circ$ above the horizontal. Find: (1) The magnitude of the impulse acting on the ball, (2) If the contact time is $2.0 \times 10^{-2}\text{ s}$, find the magnitude of the average force acting on the ball in Newtons.Answer:(1) Impulse Magnitude: Calculated via momentum change vector components = $\mathbf{4.5\text{ N}\cdot\text{s}}$(2) Average Force: $F_{\text{avg}} = \frac{J}{\Delta t} = \frac{4.5}{2.0 \times 10^{-2}} = \mathbf{225\text{ N}}$Scene 3On-Screen Title: Exercise: Force, Time, and Final VelocityVisual Prompt: A $2.0\text{ kg}$ body moving at $1.0\text{ m/s}$ acted upon by a $2.5\text{ N}$ force for $0.40\text{ s}$.Question & Answer:Question: A force of $2.5\text{ N}$ acts on a body of mass $2.0\text{ kg}$ moving at a speed of $1.0\text{ m/s}$ for $0.40\text{ s}$ in the direction of its motion. Find: (1) The magnitude of the impulse acting on the body, (2) The increase in the body's momentum, (3) Its final velocity.Answer:(1) Impulse: $J = F \cdot \Delta t = 2.5\text{ N} \times 0.40\text{ s} = \mathbf{1.0\text{ N}\cdot\text{s}}$(2) Increase in Momentum: $\Delta p = J = \mathbf{1.0\text{ kg}\cdot\text{m/s}}$(3) Final Velocity: $v_f = v_i + \frac{\Delta p}{m} = 1.0 + \frac{1.0}{2.0} = \mathbf{1.5\text{ m/s}}$ IMPORTANT LANGUAGE RULE: All on-screen text must be in Arabic EXCEPT mathematical equations, formulas, laws, symbols, variables, and numerical expressions. Keep every equation, formula, and law exactly as provided, without translating or changing it. Translate only the explanatory text, headings, labels, and descriptions into Arabic. Arabic text must use proper RTL direction and correct Arabic typography. Do not use English text anywhere else on the screen. Audience: student