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\(t = 12.35\;{\rm{s}}\), the position was\(\left\langle {3.17.2.54, - 9.38} \right\rangle {\rm{m}}\).

At\(t = 12.37\;{\rm{s}}\), the position was\(\left\langle {3.25,2.50, - 9.40} \right\rangle \;{\rm{m}}\).

Late time interval:

At\(t = 14.35\;{\rm{s}}\), the position was\(\left\langle {11.25, - 1.50, - 11.40} \right\rangle \;{\rm{m}}\).

At\(t = 14.37\;{\rm{s}}\), the position was\(\left\langle {11.27, - 1.86, - 11.42} \right\rangle \;{\rm{m}}\).

(a) In the early time interval, from \(t = 12.35\;{\rm{s}}\) to \(t = 12.37\;{\rm{s}}\), what was the average momentum of the ball? The mass of the Ping-Pong ball is \(2.7\) grams \(\left( {2.7 \times {{10}^{ - 3}}\;{\rm{kg}}} \right)\). Express your result as a vector. (b) In the late time interval, from \(t = 14.35\;{\rm{s}}\) to \(t = 14.37\;{\rm{s}}\), what was the average momentum of the ball? Express your result as a vector. (c) In the time interval from \(t = 12.35\;{\rm{s}}\) (the start of the early time interval) to \(t = 14.35\;{\rm{s}}\) (the start of the late time interval), what was the average net force acting on the ball? Express your result as a vector.

Short answer

(a) The average momentum for early time is\(\left\langle {1.08 \times {{10}^{ - 2}}, - 5.4 \times {{10}^{ - 3}}, - 2.7 \times {{10}^{ - 3}}} \right\rangle {\rm{ kg m/s}}\).

(b) The average momentum for late time is \(\left\langle {2.7 \times {{10}^{ - 3}}, - 4.86 \times {{10}^{ - 2}}, - 2.7 \times {{10}^{ - 3}}} \right\rangle \;{\rm{kg}} \cdot {\rm{m/s}}\).

(c) The net force is \(\left\langle {0.75 \times {{10}^{ - 3}},0,0.05 \times {{10}^{ - 3}}} \right\rangle {\rm{ N}}\).

Step-by-step solution

Definition and formulae for average velocity

• The change in position or displacement\(\left( {\Delta x} \right)\)divided by the time intervals (t) in which the displacement happens is the average velocity.
• Depending on the sign of the displacement, the average velocity can be positive or negative.
• Meters per second (m/s or ms-1) is the SI measure for average velocity.
• The average velocity is given by\({v_{avg}} = \frac{{\Delta r}}{{\Delta t}}\)., where\(\Delta r\)is change in position or displacement and\(\Delta t\)is time interval.
• Momentum is given by\(p = mv\), where m is mass and v is velocity

Find the average momentum for early time

(a)

It is given that time interval from\({t_1} = 12.35\;{\rm{ s}}\)to\({t_2} = 12.37{\rm{ }}\;{\rm{s}}\).

Position values are\({r_1} = \left\langle {3.17,2.54, - 9.38} \right\rangle \;{\rm{m}}\)to\({r_2} = \left\langle {3.25,2.50, - 940} \right\rangle \;{\rm{m}}\).

Mass of the ping pong ball is\(2.7\;{\rm{g}}\)or\(2.7 \times {10^{ - 3}}\;\;{\rm{kg}}\).

\(\begin{aligned}{c}{v_{avg}} &= \frac{{\Delta r}}{{\Delta t}}\\ &= \frac{{{r_2} - r{}_1}}{{{t_2} - {t_1}}}\\ &= \frac{{\left\langle {3.25,2.50, - 940} \right\rangle \;{\rm{m}} - \left\langle {3.17,2.54, - 9.38} \right\rangle \;{\rm{m}}}}{{12.37\;{\rm{s}} - 12.35\;{\rm{s}}}}\\ &= \frac{{\left\langle {0.08, - 0.04, - 0.02} \right\rangle \;{\rm{m}}}}{{0.02\;{\rm{s}}}}\\ &= \left\langle {4, - 2, - 1} \right\rangle \;{\rm{m/s}}\end{aligned}\)

Substitute the values in\(p = m{a_{avg}}\).

\(\begin{aligned}{c}p &= m\left( {{v_{{\rm{avg }}}}} \right)\\ &= \left( {2.7 \times {{10}^{ - 3}}\;{\rm{kg}}} \right)(\left\langle {4, - 2, - 1} \right\rangle \;{\rm{m/s}})\\ &= \left\langle {1.08 \times {{10}^{ - 2}}, - 5.4 \times {{10}^{ - 3}}, - 2.7 \times {{10}^{ - 3}}} \right\rangle {\rm{ kg m/s}}\end{aligned}\)

Thus, the average momentum in early time is \(\left\langle {1.08 \times {{10}^{ - 2}}, - 5.4 \times {{10}^{ - 3}}, - 2.7 \times {{10}^{ - 3}}} \right\rangle {\rm{ kg m/s}}\).

Find the late time average velocity

(b)

The average velocity for late time interval is :

\(\begin{aligned}{c}{{\vec v}_{{\rm{avg}}}} &= \frac{{\Delta \vec r}}{{\Delta t}}\\ &= \frac{{\left\langle {11.27, - 1.86, - 11.42} \right\rangle - \left\langle {11.25, - 1.5, - 11.4} \right\rangle }}{{14.37 - 14.35}}\\ &= \;\;\left\langle {1, - 18, - 1} \right\rangle \;{\rm{m/s}}\end{aligned}\)

Substitute the value of \({\vec v_{{\rm{avg}}}}\)in \({\vec p_1} = m{\vec v_{{\rm{avg }}}}\)

\(\begin{aligned}{c}{{\vec p}_1} &= m{{\vec v}_{{\rm{avg }}}}\\ &= \left( {2.7 \times {{10}^{ - 3}}} \right)\left\langle {1, - 18, - 1} \right\rangle \\ &= \left\langle {2.7 \times {{10}^{ - 3}}, - 4.86 \times {{10}^{ - 2}}, - 2.7 \times {{10}^{ - 3}}} \right\rangle \;{\rm{kg}} \cdot {\rm{m/s}}\end{aligned}\)

Thus, the average momentum in late time is\(\left\langle {2.7 \times {{10}^{ - 3}}, - 4.86 \times {{10}^{ - 2}}, - 2.7 \times {{10}^{ - 3}}} \right\rangle \;{\rm{kg}} \cdot {\rm{m/s}}\).

Apply the concept of momentum principle and find average net force.

(c)

According to momentum principle a net force changes the momentum of an object.

So, Net force is given by\({F_{net}} = \frac{{\Delta p}}{{\Delta t}}\).

From above two parts,

\(\begin{aligned}{l}{p_1} &= \left\langle {1.2 \times {{10}^{ - 2}}, - 5.4 \times {{10}^{ - 3}}, - 2.7 \times {{10}^{ - 3}}} \right\rangle \;{\rm{kg}} \cdot {\rm{m/s}}\\{p_2} &= \left\langle {2.7 \times {{10}^{ - 2}}, - 5.4 \times {{10}^{ - 3}}, - 2.6 \times {{10}^{ - 3}}} \right\rangle \;{\rm{kg}} \cdot {\rm{m/s}}\end{aligned}\)

Time given is \({t_1} = 12.35\;{\rm{s to }}{t_2} = 14.35\;{\rm{s}}\).

Substitute the above values in the

\({\vec F_{net}} = \frac{{\Delta \vec p}}{{\Delta t}}\) \(\begin{aligned}{c}{{\vec F}_{net}} &= \frac{{\Delta \vec p}}{{\Delta t}}\\ &= \frac{{\left\langle {2.7 \times {{10}^{ - 2}}, - 5.4 \times {{10}^{ - 3}}, - 2.6 \times {{10}^{ - 3}}} \right\rangle \;{\rm{kg}} \cdot {\rm{m/s}} - \left\langle {1.2 \times {{10}^{ - 2}}, - 5.4 \times {{10}^{ - 3}}, - 2.7 \times {{10}^{ - 3}}} \right\rangle \;{\rm{kg}} \cdot {\rm{m/s}}}}{{14.35 - 12.35}}\\ &= \frac{{\left\langle {1.5 \times {{10}^{ - 2}},0,0.1 \times {{10}^{ - 3}}} \right\rangle }}{2}\\ &= \left\langle {0.75 \times {{10}^{ - 3}},0,0.05 \times {{10}^{ - 3}}} \right\rangle {\rm{ N}}\end{aligned}\)

Thus, the net force in the time interval \({t_1} = 12.35\;{\rm{s }}\)to \({t_2} = 14.35\;{\rm{s}}\) is \(\left\langle {0.75 \times {{10}^{ - 3}},0,0.05 \times {{10}^{ - 3}}} \right\rangle {\rm{ N}}\).