Ap Physics C 1991 Question 1?

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Ap Physics C 1991 Question 1?

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Answer 1 · 2018-06-14T02:29:13

a) This is simply conservation of momentum.

$m (v_{0}) + 2 m (0) = (2 m + m) v_{f}$ $m(v_0) + 2m(0) = (2m + m)v_f$

$m v_{0} = 3 m v_{f}$ $mv_0 = 3mv_f$

$v_{f} = \frac{v_{0}}{3}$ $v_f = v_0/3$

b) Now we use conservation of energy (they've given us a real hint when they ask for the kinetic energy of the block and bullet).

Notice that at the bottom, the block has no potential energy and only kinetic energy.

The kinetic energy at the bottom of the loop is $\frac{1}{2} {(\frac{v_{0}}{3})}^{2} (3 m) = \frac{1}{6} v_{0}^{2} m$ $1/2(v_0/3)^2(3m) = 1/6v_0^2 m$

The potential energy at point $P$ $P$ is simply $3 m g r$ $3mgr$ . Our energy equation is therefore

$K E_{initial} = PE + K E_{final}$ $KE_"initial" = "PE" + KE_"final"$

$\frac{1}{6} v_{0}^{2} m = 3 m g r + K E_{final}$ $1/6v_0^2m =3mgr + KE_"final"$

$\frac{1}{6} v_{0}^{2} m - 3 m g r = K E_{final}$ $1/6v_0^2m - 3mgr = KE_"final"$

c) The hardest place for the system to stay on the loop will be at the top, therefore it would make sense to solve for the velocity at the top.

Once again we use conservation of energy:

$K E_{final} + P E_{final} = K E_{top} + P E_{final}$ $KE_"final" + PE_"final" = KE_"top" + PE_"final"$

$\frac{1}{6} v_{min}^{2} m - 3 m g r + 3 m g r = K E_{top} + 6 m g r$ $1/6v_"min"^2m - 3mgr + 3mgr = KE_"top" + 6mgr$

$\frac{1}{6} v_{min}^{2} m - 6 m g r = K E_{top}$ $1/6v_"min"^2m - 6mgr = KE_"top"$

$\frac{1}{6} v_{min}^{2} m - 6 m g r = \frac{1}{2} (3 m) v_{top}^{2}$ $1/6v_"min"^2m - 6mgr = 1/2(3m)v_"top"^2$

$\frac{1}{6} v_{min}^{2} - 6 g r = \frac{3}{2} v_{top}^{2}$ $1/6v_"min"^2 - 6gr = 3/2v_"top"^2$

$\frac{1}{9} v_{min}^{2} - 4 g r = v_{top}^{2}$ $1/9v_"min"^2 - 4gr = v_"top"^2$

Now we need to determine the forces acting on the object at the top. Acting towards the centre of the circle there will be a normal force, force of gravity and centripetal force.

$F_{gravity} + F_{normal} = F_{centripetal}$ $F_"gravity" + F_"normal" = F_"centripetal"$

But at the minimum possible velocity, the normal force will be $0$ $0$ therefore the only forces acting will be gravity and centripetal.

$3 m g = (3 m) \frac{v_{top}^{2}}{r}$ $3mg = (3m)v_"top"^2/r$

$v_{top} = \sqrt{r g}$ $v_"top" =sqrt(rg)$

We know substitute this into the formula derived above.

$\frac{1}{9} v_{min}^{2} - 4 g r = r g$ $1/9v_"min"^2 - 4gr = rg$

$\frac{1}{9} v_{min}^{2} = 5 g r$ $1/9v_"min"^2 = 5gr$

$v_{min} = \sqrt{45 g r} = 3 \sqrt{5 g r}$ $v_"min" = sqrt(45gr) = 3sqrt(5gr)$

Hopefully this helps!

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Ap Physics C 1991 Question 1?

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