=22pt plus 0.2pt minus 0.2pt

PHY 121 MIDTERM EXAM 1 7 October 1998 [Green] Solution

1. The relative velocity is v₁-v₂= 30-(-50) = 80 mi/hr, so t = 80 mi/(80 mi/hr) = 1 hr. In that time, train one gets to x = 30 mi, and train two gets to x = 80 - 50 = 30 mi, so they really do meet.

2. $D=vt=\sqrt{v_n^2+v_e^2}t$. v_n= 36/3.6= 10 m/s, so v=11.2 m/s, and D=7.8 m.

3. h=(1/2)(9.8)(0.7)²=2.4 m.

4. The weight,or downward force, is $mg=10.2\times 9.8=100$ N = F_N+T (the upward force), so T=100-80 =20 N.

5. g(r)=g(R_E/r)²=9.8/100=9.8 cm/s².

6. v²/r=g, so $v=\sqrt{gr}=\sqrt{9.8\times 2.45}=4.9$ 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. $\delta T/T=0.05$; $\delta \ell/\ell =0.02$,so $\delta g/g=\sqrt{(2\delta T/T)^2+(\delta \ell/\ell)^2}\approx 2\delta T/T =0.1$.Therefore $g(calc)=9.9\pm 1.0$ m/s², well in agreement with 9.8 m/s².