PHY 121 MIDTERM EXAM 1 - 10/7/98 [Yellow] Solution

1. The relative velocity is v₁-v₂ = 60-(-40) = 100 mi/hr, so t = 50 mi/(100 mi/hr) = 0.5 hr. In that time, train one gets to x = 30 mi, and train two gets to x = 50 - 20 = 30 mi, so they really do meet.

2. $D=vt=\sqrt{v_n^2+v_e^2}t$. v_n= 72/3.6= 20 m/s, so v=23.3 m/s, and D=14.0 m.

3. h=(1/2)(9.8)(0.6)²=1.8 m.

4. The weight, or downward force, is $mg=5.1\times 9.8=50$ N = F_N+T (the upward force), so T=50-40 =10 N.

5. g(r)=g(R_E/r)²=9.8/25=39 cm/s².

6. v²/r=g, so $v=\sqrt{gr}=\sqrt{9.8\times 4.36}=6.5$ m/s.

7. As stated, the position increases from zero to a maximum value, and then comes back down to zero, making a concave-down parabolic shape for x versus t. On the way up, the velocity is positive, then zero at the top, and obviously negative on the way down, decreasing all the time as v=v₀-gt. This means that the acceleration is constant, a=-g=-9.8 m/s².

8. g(calculated)=9.9 m/s². $\delta T/T=0.05$; $\delta \ell/\ell =0.08$,so $\delta g/g=\sqrt{(2\delta T/T)^2+(\delta \ell/\ell)^2} =0.13$.Therefore $g(calc)=9.9\pm 1.3$ m/s², well in agreement with 9.8 m/s².