{"id":1143,"date":"2021-06-27T19:08:16","date_gmt":"2021-06-27T19:08:16","guid":{"rendered":"https:\/\/books.compclassnotes.com\/rothphys110-2e\/?p=1143"},"modified":"2022-01-22T23:42:41","modified_gmt":"2022-01-22T23:42:41","slug":"section-7-1-v2","status":"publish","type":"post","link":"https:\/\/books.compclassnotes.com\/rothphys110-2e\/2021\/06\/27\/section-7-1-v2\/","title":{"rendered":"Chapter 7: Linear momentum"},"content":{"rendered":"\n

7.1 Linear momentum<\/h2>\n\n\n\n

When Isaac Newton was formulating his famous laws of motion, he was not thinking in terms of position, velocity, and acceleration. Instead, he worked with what he called the \u201cquantity of motion.\u201d This was a particular quantity that by itself described the motion of an object. Today we call it linear momentum,<\/em> and usually just refer to it as momentum. An object’s momentum is its mass times its velocity:<\/p>\n\n\n

\\[
\n\\vec{p} = m\\vec{v} \\tag{7.1}
\n\\]<\/p>\n\n\n\n

Notice the arrows on top of the symbols for momentum and velocity: these are vector quantities. When working with one-dimensional motion (only moving forward or backwards along a single axis), a positive or negative sign is all we need to tell us direction, and we drop the vector notation. When we work with two-dimensional motion, formal vector notation becomes important (see chapter 2<\/a> to review vectors).<\/p>\n\n\n\n

The SI units for momentum are kilograms times meters per second (kg\u00b7m\/s). There is no special name for this unit.<\/p>\n\n\n\n

Example 7.1<\/h4>\n\n\n\n

In a world-record-breaking sprint, Jamaican athlete Usain Bolt ran 100 m in 9.58 s. If his mass was 80 kg at the time, what was his momentum? For simplicity, let’s neglect any fluctuations in his speed; in effect we’re finding his average momentum over this time.<\/p>\n\n\n

\\[
\np = mv = \\left(80\\ \\textrm{kg}\\right)\\left(\\frac{100\\ \\textrm{m}}{9.85\\ \\textrm{s}}\\right) = 835\\ \\textrm{kg}\\cdot\\textrm{m\/s}
\n\\]<\/p>\n\n\n\n

Example 7.2<\/h4>\n\n\n\n

A falling ball is traveling at 8.25 m\/s when it hits the ground. It bounces, and leaves the ground with the same speed, but in the exactly opposite direction. The ball is in contact with the floor for 7.3 ms. What was the change in the ball’s momentum, if the mass is 0.75 kg?<\/p>\n\n\n\n

Say the ball is initially going in the negative direction, and returns in the positive direction.<\/p>\n\n\n

\\[
\n \\begin{align*}
\n \\Delta p &= p_f – p_i \\\\
\n &= mv_f – mv_i \\\\
\n &= m\\underbrace{(v_f – v_i)}_{\\Delta v} \\\\
\n &= (0.75\\ \\textrm{kg})\\left((8.25\\ \\textrm{m\/s}) – (-8.25\\ \\textrm{m\/s})\\right) \\\\
\n &= 12.4\\ \\textrm{kg}\\cdot\\textrm{m\/s}
\n \\end{align*}
\n\\]<\/p>\n\n\n\n

I want to point out two details here:<\/p>\n\n\n\n

  • When the mass does not change, \\(\\Delta p = m\\Delta v\\)<\/li>
  • When the ball was traveling in the negative direction, \\(v = -8.25\\ \\textrm{m\/s}\\). Including the minus sign when you substitute in the value for \\(v_i\\) to do the final calculation is essential<\/em>. <\/li><\/ul>\n\n\n\n

    Practice 7.1<\/h4>\n\n\n\n\n