9.3 Conservation of linear momentum  (Page 4/10)

A bouncing superball

1. What is the superball’s change of momentum during its bounce on the floor?
2. What was Earth’s change of momentum due to the ball colliding with the floor?
3. What was Earth’s change of velocity as a result of this collision?

(This example shows that you have to be careful about defining your system.)

Strategy

Solution

1. Since this is a one-dimensional problem, we use the scalar form of the equations. Let:
   
   • $p_0=$ the magnitude of the ball’s momentum at time $t_0$ , the moment it was released; since it was dropped from rest, this is zero.
   • $p_1=$ the magnitude of the ball’s momentum at time $t_1$ , the instant just before it hits the floor.
   • $p_2=$ the magnitude of the ball’s momentum at time $t_2$ , just after it loses contact with the floor after the bounce.
   
   The ball’s change of momentum is
   
   $\begin{array}{cc}\hfill \text{Δ}\overrightarrow{p}& ={\overrightarrow{p}}_2-{\overrightarrow{p}}_1\hfill \\ & =p_2\widehat{j}-\left(\text{−}{p}_1\widehat{j}\right)\hfill \\ & =\left(p_2+p_1\right)\widehat{j}.\hfill \end{array}$
   
   Its velocity just before it hits the floor can be determined from either conservation of energy or kinematics. We use kinematics here; you should re-solve it using conservation of energy and confirm you get the same result.
   We want the velocity just before it hits the ground (at time $t_1$ ). We know its initial velocity $v_0=0$ (at time $t_0$ ), the height it falls, and its acceleration; we don’t know the fall time. We could calculate that, but instead we use
   
   ${\overrightarrow{v}}_1=\text{−}\widehat{j}\sqrt{2gy}=\mathrm{-5.4}\text{m/s}\widehat{j}.$
   Thus the ball has a momentum of
   
   $\begin{array}{cc}\hfill {\overrightarrow{p}}_1& =\text{−}\left(0.25\text{kg}\right)\left(\mathrm{-5.4}\text{m/s}\widehat{j}\right)\hfill \\ & =\text{−}\left(1.4\text{kg}·\text{m/s}\right)\widehat{j}.\hfill \end{array}$
   
   We don’t have an easy way to calculate the momentum after the bounce. Instead, we reason from the symmetry of the situation.
   Before the bounce, the ball starts with zero velocity and falls 1.50 m under the influence of gravity, achieving some amount of momentum just before it hits the ground. On the return trip (after the bounce), it starts with some amount of momentum, rises the same 1.50 m it fell, and ends with zero velocity. Thus, the motion after the bounce was the mirror image of the motion before the bounce. From this symmetry, it must be true that the ball’s momentum after the bounce must be equal and opposite to its momentum before the bounce. (This is a subtle but crucial argument; make sure you understand it before you go on.)
   Therefore,
   
   ${\overrightarrow{p}}_2=\text{−}{\overrightarrow{p}}_1=+\left(1.4\text{kg}·\text{m/s}\right)\widehat{j}.$
   Thus, the ball’s change of velocity during the bounce is
   
   $\begin{array}{cc}\hfill \text{Δ}\overrightarrow{p}& ={\overrightarrow{p}}_2-{\overrightarrow{p}}_1\hfill \\ & =\left(1.4\text{kg}·\text{m/s}\right)\widehat{j}-\left(\mathrm{-1.4}\text{kg}·\text{m/s}\right)\widehat{j}\hfill \\ & =+\left(2.8\text{kg}·\text{m/s}\right)\widehat{j}.\hfill \end{array}$
2. What was Earth’s change of momentum due to the ball colliding with the floor?
   Your instinctive response may well have been either “zero; the Earth is just too massive for that tiny ball to have affected it” or possibly, “more than zero, but utterly negligible.” But no—if we re-define our system to be the Superball + Earth, then this system is closed (neglecting the gravitational pulls of the Sun, the Moon, and the other planets in the solar system), and therefore the total change of momentum of this new system must be zero. Therefore, Earth’s change of momentum is exactly the same magnitude:
   
   $\text{Δ}{\overrightarrow{p}}_{\text{Earth}}=\mathrm{-2.8}\text{kg}·\text{m/s}\widehat{j}.$
3. What was Earth’s change of velocity as a result of this collision?
   This is where your instinctive feeling is probably correct:
   
   $\begin{array}{cc}\hfill \text{Δ}{\overrightarrow{v}}_{\text{Earth}}& =\frac{\text{Δ}{\overrightarrow{p}}_{\text{Earth}}}{M_{\text{Earth}}}\hfill \\ & =-\frac{2.8\text{kg}·\text{m/s}}{5.97\times {10}^{24}\text{kg}}\widehat{j}\hfill \\ & =\text{−}\left(4.7\times {10}^{\mathrm{-25}}\text{m/s}\right)\widehat{j}.\hfill \end{array}$
   
   This change of Earth’s velocity is utterly negligible.