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(PHYS111)[2008](f)midterm~=tkk933^_55243.pdf
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Physics 111 (Fall 2008) Midterm Solution
1. (a) 12
yv gt
. 0 sin.t .
2
1000 . 12

1000 . 300* sin30 . *9.8 t
3600 2
10.7s (ans) or t .. s rej )

.t . 19.2 (
(b) 1000 .
vy . v0 sin .. gt . 300* sin30 . 9.8*10.7 .146m / s
3600
1000 .

vx . v0 cos .. 300* cos30 . 72.2m / s
3600
22
22
v .
vx . vy . 146 . 72.2 .163m / s
v 146
tan .. y .. 2.03 vx 72.2
.. . 63.7.
(c) 1000 .

d . y cot .. v0 cos .t .1000*
3 . 300* cos30 *11.7 . 963m (ans)
3600
2. (a) FN cos.. mg
2
mv
FC .. FN sin.. mg tan .
R
. 18.6m /s (ans)
(b)
fs ..sN
N cos.. fs sin.. mg
2
N sin.. f cos.. mv sR

Dividing, v2 N sin.. fs cos. N sin. ..sN cos. sin. ..s cos.
.. .

gR N cos.. fs sin. N cos. ..sN sin. cos. ..s sin.
oo
. sin. .. cos..
. sin10 . 0.5cos10 . -1
v .
gR s (9.8)(200) 38.1ms
.. ... .. oo .. .
cos. .. sin. cos10 . 0.5sin10
. s .. .
3. (a) 121 2
PE . kd . *80*0.1 . 0.4 (
J ans )
22
(b) 12 12

2 . kd . fd .1.2*9.8*0.1 . *80*0.1 . 2*0.1 . 0.576J (ans)
KE . m gd
22

1

(c)

Wf ..fd ..2*0.1 ..0.2 J

So the work done against friction is 0.2J (ans)

(d)


1 2
2KE
2*0.576
KE . (m1 .m2)v ..
v
.

.0.76m / s
1 . 2 1.2
2(mm ) 0.8 .
Pmgv .1.2*9.8*0.76 .8.93J / s
. 2
(e) 12
m2 gh . kh . fh
2
(1.2)(9.8)h .1 (80)h2 .2h
2 h(40h .9.76) .0
9.76
.h ..0.244 m (answer) or h .0 (reject)
40

4. (a) Using the parallel axis theorem, the rotational inertia of the rod about the pivot point is
12 2 .. 1 . 22 2
Irod. ML .Mh ..(0.2)(0.8) .(0.2)(0.3) .0.02867 .0.0287 kgm (ans)
12 .12 .
(b) Initial angular momentum of the mud ball:

o2 .1
Li.mvr sin..(0.05)(6)(0.5)sin 70 .0.1410 .0.141kgm s (ans)
(c) Let k be the distance between the mud ball and the pivot point. Final rotational inertia of the system:
2 22
I .I .mk .0.02867 .(0.05)(0.5) .0.04117 kgm
tot rod
Using the conservation of angular momentum,
Li .Itot.
L 0.1410
.. i ..3.4248 .3.42 rads.1 (ans)Itot 0.04117
(d) Initial kinetic energy:

1 21 . 2 m
Ki . mv ... .(0.05)(6) .0.9 J
2 .2 . Kinetic energy after the impact: 12 ..1 . 2
Kf . Itot. ..(0.04117)(3.4248) .0.2414 J
2 .2 .

Energy loss:
Ki .Kf .0.9 .0.2414 .0.6586 .0.659 J (ans)

(e) Let . be the angular displacement after the impact.
Using the conservation of energy,
12

Itot..Mgh .mgk ..Mghcos..mgk cos.
2
2
1 Itot. ..Mh .mk.g(1.cos.)
2
2


0.2414 ..(0.2)(0.3) . (0.05)(0.5).(9.8)(1. cos. ) . 0.833(1. cos. ) 0.2414
cos.. 1.. 0.7102 . .. 44.7o (ans)
0.833
3