Caught, Returned or Sent North — Impulse Is a Vector
A ball flies in from the west over a court with a compass rose painted on it, and a bat sends it off again. Four sliders set the ball's mass m, its speed before the hit vi, the exit angle θ measured from east, and the speed after ÷ speed before, vf/vi; they start at 60 g, 20 m/s, 90° and 1.00, and a ratio of 0 means the ball is caught. The bat turns as the sliders move: a smooth bat can push only along the normal at the point of contact, so that normal must point along the impulse. On the rose, the initial momentum pi and the impulse J are drawn from the same point, and the parallelogram they make has the final momentum pf as its diagonal. The flight plays once, 20 times slower, with strobe images of the ball at equal time steps; a timeline lets you stop it at any moment, before or after the hit.
इस सिमुलेशन का उपयोग कैसे करें
- Ball mass m — 0.020 to 0.200 kg, starting at 0.060 kg.
- Speed before vi — 5 to 40 m/s, starting at 20 m/s.
- Exit angle θ — 0° to 180°, measured from east towards the north: 0° keeps the ball heading east, 90° sends it north, 180° straight back west.
- Speed after ÷ before vf/vi — 0 to 1.20, starting at 1.00; 0 means the ball is caught. A change redraws the flight and the vectors without starting it.
- Play/pause, back to the start, a timeline you can drag to any moment, and ½×, ¼×, ⅛× slow motion, on top of the scene's own 20× slow motion.
- Drag anywhere on the scene to look around.
क्या देखें
- At 90° the impulse arrow on the rose points north-west and reads 1.70 N s, while the ball's speed stays 20 m/s.
- Caught, J is 1.20 N s west; sent back at the same speed it is 2.40 N s west — twice as much, because the momentum is reversed, not just removed.
- The dashed copies of pi and J always meet at the tip of pf: the parallelogram closes at every setting.
- The bat turns so that its push points along J: square to the ball's path when the ball is sent straight back, at 45° to it when the ball is sent north.
- At vf/vi = 1 the strobe images are equally spaced before and after the hit; at 0.50 the spacing after the hit halves, and at 0 the ball stays against the bat.
- Doubling the mass or the speed doubles J without turning it; at vf/vi = 1.20 and 180°, J grows to 2.64 N s and the bat has to be swung into the ball.
इसके पीछे की भौतिकी
Take east as positive. Before the hit the ball's momentum is pi = mvi = 6.0 × 10⁻² × 20 = 1.2 kg m/s east, and the impulse is the vector change J = pf − pi. Caught, pf = 0 and J = 0 − 1.2 = −1.2 N s: 1.2 N s westward. Sent back west at the same speed, J = (6.0 × 10⁻²)(−20) − (6.0 × 10⁻²)(20) = −2.4 N s: reversing the momentum doubles the impulse. Sent north at the same speed, pi and pf are at right angles, so |J| = √2 × 1.2 = 1.70 ≈ 1.7 N s, directed north-west. In general the triangle of pi, pf and J gives |J| = m√(vi² + vf² − 2vi·vf cos θ); at equal speeds this is 2mv sin(θ/2), so at 120° it is 2.08 N s, pointing 150° from east. The ball's speed can stay the same while J grows from 0 to 2mv: the impulse is the vector change in momentum, not the difference between the two speeds. The bat stands still whenever that is enough; it has to be swung into the ball when the ball leaves faster than a still bat could send it, and it gives way when the ball keeps moving towards it.