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(PHYS111)[2009](f)midterm~=tkk933^_90150.pdf
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Physics 111 (Fall 2009) Midterm Solution
1) (a) Using Newtons second law at the top of the bridge,
2
vMg . N . M R
2 . 2 .N. Mg . Mv . 1500..9.8 . 20 .. . 8700 N (answer)
R 100
..
(b) If the car loses contact with the bridge at the top, N = 0. Therefore,
2
vMg . M R
. 31.3 ms.1 (answer)

(c) Using Newtons second law along the radial direction,
2
vMg cos.. N . M
R When the car loses contact with the bridge, N = 0. Therefore, v2 302
cos... . 0.9184
gR (9.8)(100) .. 23.3o (answer)
2) (a) Using conservation of energy from A to C,
.K ..U ..Eint . 0
.K . 0 .U ..mg(asin.. bsin . ) Frictional force on slope AB ..kmg cos.
Frictional force on slope BC ..kmg cos .
.E .. mg(acos.. bcos . )
int k
0 . mg(asin.. bsin . ) ..kmg(acos.. bcos . ) . 0
asin.. bsin . ..ka cos...kbcos .
oo
. sin...k cos.. . sin 45 . 0.4cos 45 .
b . a . 10 . 33.26 m . 33.3 m (answer)
.. .... oo ..
. cos .. sin . 0.4cos15 . sin15
. k .. .
(b)
Work done against friction
..Eint . ..U . mg(asin.. bsin . ) . (1.5)(9.8)(10sin 45o . 33.26sin15o)
. 230 J (answer)


(c)
Using the conservation of energy from A to B,


.K ..U ..Eint . 0 .U ..mgasin. .E .. mgacos.
int k
.K . ..U ..E . mgasin. .. mgacos.. mga(sin. .. cos.)
int kk
.(1.5)(9.8)(10)(sin 45o . 0.4cos 45o) . 62.4 J (answer)
1

3) Using conservation of momentum,
x component:
m1u1 .m1v1 cos.. m2v2 cos. (1)

y component: 0 ..m1v1 sin.. m2v2 sin. (2) mv sin.
From (2), v . 11 2 m2 sin.
. sin. cos..
Substituting into (1), m1u1 ..
cos.. .
m1v1
sin.
..
. cos. sin.. sin. cos..

u . v
1 .
.
1
sin.
.. u1 sin.. sin(. ..)v1
o oo
3sin 60 . sin(30 . 60 )v1
o .1 .1
v1 . 3sin 60 . 2.598 ms . 2.60 ms
. m1 sin.. . 0.2sin 30o . o .1
v2 . .. ..v1 . .. o ..3sin 60 . 3 ms (answer)
m sin. 0.1sin 60
. 2 .. .
12
(b) Initial kinetic energy . 2 m1u1
121 2
Final kinetic energy . mv . mv
11 22
22 121 212
Change in kinetic energy . mv . mv . mu
11 2211
22 2
1 21 212

.(0.2)(2.598) . (0.1)(3) . (0.2)(3 ) . 0.225 J (answer)
2 22 Remark: Energy is created during the collision, such as in the case of an explosion.
(c) Change in momentum of the target ball . m2v2
Using the impulse-momentum theorem, average force m2v2 (0.1)(3)
.. . 30.0 N (answer)
t 0.01
4) (a) The rotational inertia of the disk about its center is I . 1 MR2.
2 Using Newtons second law for the rotation of the disk, .. I. , FL .TR . I. (1) Using Newtons second law for the translation of the block, F . ma , T .mg . ma (2)
From (2), T . mg . ma . Substituting into (1),
FL . mgR . maR . I. Since a . R. , Ia
FL . mgR . maR . R
2

FL (33)(0.4)
. g . 9.8
FL . mgR mR (6)(0.2) .2
a... . 0.6 ms (answer)
IM 12
mR . 1. 1.
R 2m (2)(6)
(b)
From (2), T. mg . ma . 6(9.8 . 0.6) . 62.4 N (answer) a 0.6 .

(c)
... . 3 rads 2


R 0.2
1

Since . ..0t . 2 .t2,
2.. 0t . 1 (3)t2
2
2.

t.

. 2.05 s (answer)
3/2
3