# 5.2 Impulse  (Page 2/6)

 Page 2 / 6

Solution for (a)

The first ball bounces directly into the wall and exerts a force on it in the $+x$ direction. Therefore the wall exerts a force on the ball in the $-x$ direction. The second ball continues with the same momentum component in the $y$ direction, but reverses its $x$ -component of momentum, as seen by sketching a diagram of the angles involved and keeping in mind the proportionality between velocity and momentum.

These changes mean the change in momentum for both balls is in the $-x$ direction, so the force of the wall on each ball is along the $-x$ direction.

Strategy for (b)

Calculate the change in momentum for each ball, which is equal to the impulse imparted to the ball.

Solution for (b)

Let $u$ be the speed of each ball before and after collision with the wall, and $m$ the mass of each ball. Choose the $x$ -axis and $y$ -axis as previously described, and consider the change in momentum of the first ball which strikes perpendicular to the wall.

${p}_{\text{xi}}=\text{mu}\text{;}\phantom{\rule{0.25em}{0ex}}{p}_{\text{yi}}=0$
${p}_{\text{xf}}=-\text{mu}\text{;}\phantom{\rule{0.25em}{0ex}}{p}_{\text{yf}}=0$

Impulse is the change in momentum vector. Therefore the $x$ -component of impulse is equal to $-2\text{mu}$ and the $y$ -component of impulse is equal to zero.

Now consider the change in momentum of the second ball.

${p}_{\text{xi}}=\text{mu}\phantom{\rule{0.25em}{0ex}}\text{cos 30º}\text{;}\phantom{\rule{0.25em}{0ex}}{p}_{\text{yi}}=\text{–mu}\phantom{\rule{0.25em}{0ex}}\text{sin 30º}$
${p}_{\text{xf}}=–\text{mu}\phantom{\rule{0.25em}{0ex}}\text{cos 30º}\text{;}\phantom{\rule{0.25em}{0ex}}{p}_{\text{yf}}=-\text{mu}\phantom{\rule{0.25em}{0ex}}\text{sin 30º}$

It should be noted here that while ${p}_{x}$ changes sign after the collision, ${p}_{y}$ does not. Therefore the $x$ -component of impulse is equal to $-2\text{mu}\phantom{\rule{0.25em}{0ex}}\text{cos 30º}$ and the $y$ -component of impulse is equal to zero.

The ratio of the magnitudes of the impulse imparted to the balls is

$\frac{2\text{mu}}{2\text{mu}\phantom{\rule{0.25em}{0ex}}\text{cos 30º}}=\frac{2}{\sqrt{3}}=1\text{.}\text{155}.$

Discussion

The direction of impulse and force is the same as in the case of (a); it is normal to the wall and along the negative $x$ - direction. Making use of Newton’s third law, the force on the wall due to each ball is normal to the wall along the positive $x$ -direction.

Our definition of impulse includes an assumption that the force is constant over the time interval $\Delta t$ . Forces are usually not constant . Forces vary considerably even during the brief time intervals considered. It is, however, possible to find an average effective force ${F}_{\text{eff}}$ that produces the same result as the corresponding time-varying force. [link] shows a graph of what an actual force looks like as a function of time for a ball bouncing off the floor. The area under the curve has units of momentum and is equal to the impulse or change in momentum between times ${t}_{1}$ and ${t}_{2}$ . That area is equal to the area inside the rectangle bounded by ${F}_{\text{eff}}$ , ${t}_{1}$ , and ${t}_{2}$ . Thus the impulses and their effects are the same for both the actual and effective forces.

## Making connections: take-home investigation—hand movement and impulse

Try catching a ball while “giving” with the ball, pulling your hands toward your body. Then, try catching a ball while keeping your hands still. Hit water in a tub with your full palm. After the water has settled, hit the water again by diving your hand with your fingers first into the water. (Your full palm represents a swimmer doing a belly flop and your diving hand represents a swimmer doing a dive.) Explain what happens in each case and why. Which orientations would you advise people to avoid and why?

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