0.2 6.3 impulse  (Page 2/4)

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.

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.

It should be noted here that while $p_{\mathrm{x}}$ changes sign after the collision, $p_{\mathrm{y}}$ does not. Therefore the $x$ -component of impulse is equal to $-2\text{mu}\text{cos 30º}$ and the $y$ -component of impulse is equal to zero.

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

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 $\mathrm{\Delta }t$ . 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.

Section summary

• Impulse, or change in momentum, equals the average net external force multiplied by the time this force acts:
  $\mathrm{\Delta }\mathbf{p}={\mathbf{F}}_{\text{net}}\mathrm{\Delta }t.$
• Forces are usually not constant over a period of time.

Conceptual questions

Problems&Exercises