Worked examples show how to calculate impulse and average force using the impulse-momentum theorem. A glove catching a ball demonstrates impulse as change in momentum and average force from contact time, while a second scenario compares impulse for catching versus rebounding a ball east, west, and north, highlighting direction and vector components. Useful for students learning momentum concepts and practicing impulse calculations in one and two dimensions.
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Scene 4 On-Screen Title: Exercise: Catching a Ball with a Glove Visual Prompt: A 0.14 kg ball moving at 40 m/s caught by a glove over 2.0×10 −2 s . Question & Answer: Question: A ball of mass 0.14 kg moving at 40 m/s is caught by a glove. Find: (1) The magnitude of the impulse received by the ball, (2) If the contact time is 2.0×10 −2 s , find the magnitude of the average force exerted by the glove on the ball. Answer: (1) Impulse: J=Δp=m(v f −v i )=0.14×(0−40)=−5.6 N⋅s (Magnitude = 5.6 N⋅s ) (2) Average Force: F avg = Δt J = 2.0×10 −2 5.6 =280 N Scene 5 On-Screen Title: Exercise: Eastward Moving Ball Scenarios (Catch, West Rebound, North Rebound) Visual Prompt: A 3.0×10 −2 kg ball moving east at 10 m/s experiencing different collision paths (a, b, c). PNG Question & Answer: Question: A ball of mass 3.0×10 −2 kg moves towards the east at a speed of 10 m/s . Find the magnitude and direction of the impulse acting on the ball for each of the following cases: (i) When the ball is caught, (ii) When it is struck to rebound west at the same speed, (iii) When it is struck to rebound north at the same speed. PNG + 1 Answer: (i) When caught: J=0−(3.0×10 −2 ×10)=−0.3 N⋅s (Magnitude = 0.3 N⋅s , Direction = West) PNG (ii) When rebounding west: J=(3.0×10 −2 ×−10)−(3.0×10 −2 ×10)=−0.6 N⋅s (Magnitude = 0.6 N⋅s , Direction = West) PNG (iii) When rebounding north: Using perpendicular vector components, J= (0.3) 2 +(0.3) 2 ≈0.42 N⋅s (Direction = Northwest) PNG Scene 6 On-Screen Title: Exercise: Angled Rebound ( 120 ∘ ) Visual Prompt: A 0.40 kg ball moving horizontally at 10 m/s rebounding at an angle of 120 ∘ . Question & Answer: Question: A ball of mass 0.40 kg moves horizontally at 10 m/s and is struck to rebound at the same speed in a direction making an angle of 120 ∘ with its initial direction. Find the magnitude of the impulse received by the ball, and the angle between the impulse direction and the initial direction. Answer: Initial and final momentum magnitudes: p i =p f =0.40×10=4.0 kg⋅m/s . Impulse Magnitude ( J ): Using the law of cosines: J= 4 2 +4 2 −2(4)(4)cos(120 ∘ ) = 16+16+16 = 48 ≈6.93 N⋅s Angle with Initial Direction: Due to vector symmetry for equal magnitudes at 120 ∘ , the angle between the impulse and the initial direction is 30 ∘ . Scene 7 On-Screen Title: Exam-Style Question: Ball and Protective Glove Visual Prompt: A 0.50 kg ball moving at 8.0 m/s being brought to rest by a protective glove over 0.040 s . Question & Answer: Question: A ball of mass 0.50 kg moving at a speed of 8.0 m/s is brought to rest by a protective glove in 0.040 s . Find the impulse acting on the ball. Answer: J=Δp=m(v f −v i )=0.50×(0−8.0)=−4.0 N⋅s . Impulse Magnitude: Equal to 4.0 N⋅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