Free-Fall Acceleration — Question 9

Question 9

Interval speeds belong to interval midpoint times

A small pointlike bead falls through three noncontact timing levels A, B and C, in that order. Both vertical separations are $L=0.300\ \mathrm{m}$. The A-to-B transit takes $\tau_1=0.200\ \mathrm{s}$ and B-to-C takes $\tau_2=0.100\ \mathrm{s}$. Assume one unknown constant downward acceleration $g>0$, negligible air resistance, and downward motion throughout. Set $t=0$ at A.

1. Find each interval’s average downward speed and the time at which that speed equals the instantaneous speed. Use the correct time separation to infer $g$.

2. Recover the downward speeds at A, B and C. Verify both interval lengths using endpoint speeds.

3. A student divides the difference of interval average speeds by $\tau_2$ to obtain $g$. Explain the timing error and calculate that incorrect result.

4. If the bead was released from rest, how far above A and how long before its A crossing was it released? Explain why that conclusion is conditional on the release assumption.

Question 9 - Solution

Interval speeds belong to interval midpoint times

(a) Place interval means at the midpoints of their own intervals.

$$g=\frac{L/\tau_2-L/\tau_1}{(\tau_1+\tau_2)/2}=\frac{2L(\tau_1-\tau_2)}{\tau_1\tau_2(\tau_1+\tau_2)}.$$

The mean speeds are $L/\tau_1=1.50\ \mathrm{m/s}$ and $L/\tau_2=3.00\ \mathrm{m/s}$. Linear velocity makes these the instantaneous speeds at $t=0.100\ \mathrm{s}$ and $t=0.250\ \mathrm{s}$. Hence $$\boxed{g=\frac{3.00-1.50}{0.250-0.100}=10.0\ \mathrm{m/s^2}}.$$ In general the midpoint separation is $(\tau_1+\tau_2)/2$, not $\tau_2$.

(b) Recover each boundary speed from the same linear history.

$$\boxed{w_A=0.500\ \mathrm{m/s}},\quad \boxed{w_B=2.50\ \mathrm{m/s}},\quad \boxed{w_C=3.50\ \mathrm{m/s}}.$$ Checks: $\tfrac12(0.5+2.5)(0.2)=0.300\ \mathrm{m}$ and $\tfrac12(2.5+3.5)(0.1)=0.300\ \mathrm{m}$. Both speeds and positions are continuous at B.

(c) The averages are not speeds at B and C.

Dividing $1.50\ \mathrm{m/s}$ by $0.100\ \mathrm{s}$ gives the incorrect $15.0\ \mathrm{m/s^2}$. The observations being compared refer to the two interval midpoints, separated by 0.150 s; moving them to endpoints changes the acceleration inference.

(d) Infer a prior release only after specifying its velocity.

For release from rest, the time to A is $w_A/g=\boxed{0.0500\ \mathrm{s}}$ and the release height above A is $w_A^2/(2g)=\boxed{0.0125\ \mathrm{m}}$. Without the initial-rest assumption, a different earlier release height and speed could lead to the same state at A and the same two transits.