A ping-pong ball is acted upon by the Earth, air resistance, and a strong wind.

Question

A ping-pong ball is acted upon by the Earth, air resistance, and a strong wind. Here are the positions of the ball at several times. Early time interval: At t = 12.77 s, the position was m At t = 12.82 s, the position was m Late time interval: At t = 15.77 s, the position was m At t = 15.82 s, the position was m (a) In the early time interval, from t = 12.77 s tot = 12.82 s, what was the average momentum of the ball? The mass of the ping-pong ball is 2.7 grams (2.7*10^-3 kg). Express your result as a vector. =.00378,-.00864,-.0459) kg m/s (b) In the late time interval, from t = 15.77 s to t = 15.82 s, what was the average momentum of the ball? Express your result as a vector. = kg m/s

Explanation / Answer

At t=15.77 (11.68,8.53,-61.22)

at t=15.82 (11.81,8.87,-62.07)

Avg momentum= (final position-initial position)*mass/time

= ((11.81,8.87,-62.07)-(11.68,8.53,-61.22))*2.7*10^-3/0.05

(0.00702, 0.01836, -0.0459)

There is slight mistake in your last value

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