Practice question

Given: A 0.25 kg ball moving at 18 m/s at 53° to a wall rebounds at the same speed and angle. What is the impulse perpendicular to the wall? (Take sin 53° = 0.8, cos 53° = 0.6 ) These values define the system as per NCERT data. Formula: Initial: p_{x1 = m v cos 53° = 0.25 × 18 × 0.6 = 2.7 kg m/s. This is standard NCERT relation. Substitution & Calculation: Perpendicular component (x-axis) reverses upon rebound. . Final: p_{x2 = -m v cos 53° = -0.25 × 18 × 0.6 = -2.7 kg m/s . Impulse: Δ p_x = -2.7 - 2.7 = -5.4 N s . Magnitude: |Δ p_x| = 5.4 N s . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.