PHY 121 MIDTERM EXAM 1 Practice Fall 1998

1. $v=dx/dt=5\times 4t^3$, t=(80/5)^1/4=16^1/4=2 s, so v=20t³=160 m/s.

2. a=F/m=20 m/s². After 10 s, velocity is at =200 m/s, and position is at ²/2 =1000 m = 1 km. After 60 s, velocity is still 200 m/s, and position is 11 km. After 60.5 s, velocity is reduced: $a\Delta t = -400 \ {\rm m/s^2 \times 0.5 \ s = -200 \ m/s}$, so v=0; position is increased: $\Delta x = v_{av}\Delta t = 100 {\rm \ m/s \times 0.5 \ s =50 \ m}$, so x=11.05 km.

3. $F_{\parallel}= mg{\rm sin}\theta - \mu mg {\rm cos}\theta$. This force acts for a distance $\ell$, where $h=\ell {\rm sin \theta}$, so $\ell=h/{\rm sin \theta}$. Thus, $\Delta (v^2) =2gh(1-\mu / {\tan \theta})$. For $\mu=1/2$ and $\theta = 45^o$, this gives v²=gh = 49(m/s)², and v=7 m/s.

4. ma = -mg + F_N, so F_N=3mg. Therefore the reading is $3\times 60 = 180$ kg.

5. The steamship has to have a southward velocity through the water to cancel the effect of the northward current, so it is not moving either north or south with respect to land. This means it must have a velocity with respect to the water of 20 mi/hr east and 10 mi/hr south, and therefore must be going at an angle ${\rm tan}^{-1}(0.5)= 27^o$ south of east.

6. F/m=a= v²/r=g-T/m=g, so $v=\sqrt{rg}=\sqrt{(9.8)^2}=9.8$ m/s.

7. With motion always in one direction, the position coordinate increases slowly as the car pulls away from the first stop sign, then more rapidly, and finally more slowly again before the next stop. For the same reason, velocity is never negative, but increases from zero, stays steady for a while, and then decreases to zero again. Acceleration is positive at the start, as the velocity increases, then becomes small, and finally becomes negative as the velocity decreases back towards zero: Decreasing velocity is the definition of negative acceleration.

8. The relation between distance and time for the horizontal motion, with no acceleration, is D=v₀ t, or t=D/v₀. For the vertical motion, accelerating from rest, the relation is h=gt²/2=g(D/v₀t)²/2. Substituting numbers gives $h=9.8(0.9\pm 0.1/2.3\pm 0.3)^2/2=0.75$ m. To compute the uncertainty we use the formula sheet with $\delta D/D =0.1/0.9=0.11$, $\delta v/v = 0.3/2.3=0.13$. Now, $\delta/h/h = 2\times ((\delta D/D)^2+ (\delta v/v)^2)^{0.5}=0.34$, where the factor 2 is because the quantities D and v₀ were squared. Finally, we get $h\pm \delta h = 0.75 \times (1\pm 0.34) = 0.75\pm 0.26$ m.