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Oo MIDDLE EAST TECHNICAL UNIVERSITY TIME: TOTAL GRADE
PHYS 105 SECOND MIDTERM EXAM 100 MIN’
1956 Dec. 18, 1999 UTES.
Fill in this part with pen not with pencil :
LAST NAME FIRST NAME STUD. NO. | DEP. SIGNATURE SECT.
GRADES:
Prob. 1 Prob. 2 Prob. 3 Prob. 4
The steps
There are 4 classical type problems. You must solve all of them. Calculator not permitted.
Do not ask proctors any question about the problems; read, understand and solve.
oF the solution of each problem must be shown clearly in the space provided.
Numerical answers must be given with correct SI units and SI unit symbols must be used only.
z ; e iz
=10m/s* 43 1 a.
Oo ‘iat " 1
PROBLEM (1) (25 points)
A 0,5-kg block attached to a spring of relaxed 0.5 kg
length 0.6 m and force constant & = 40 N/m, is k= 40 Nim F=20N
at rest at point A on a frictionless horizontal
surface. A constant horizontal force F = 20 N is wed.
then applied to the block and moves the block A B
to the right along the surface. <—— 0.6m ——>< 0.25 m4
(a) What is the speed of the block when it reaches point B which is 0.25 m to the right of point A?
(2) Woy F)+W Ch ee =
+p
a
ae) (0.25) — £40) (0.25) « ge
Phys 105 Second Midterm
=" Ve jis res
Paget Dec. 18, 1999
(b) When the block reaches point 8, the force F is suddenly removed. In the subsequent motion,
how close will the block get to the wall?
@® (4)
A B
te a= 2.6m —re dain +
cf) Laman
x3
Ey = Ee
Lae z *
av + ska, “si £ kA -x)
ot £220 (0.6-x)° = x= 0-/m
7 PROBLEM (2) (25 points)
A small block of mass m = 0.2 kg is released from rest at point A at a height h = 1 m above the
ground. It slides down the track as shown in the figure. It reaches point B with speed vg = 4 mls.
From point B it slides on a level surface a distance d = 1 m to point C which is at the bottom of a
vertical circular loop of radius R = 0.18 m. This circular section of the track (C + D + E) is
frictionless but the remaining section (A + B + C) is rough.
A m
\\ ve
A
oh
(a) How much work is done by friction as the block slides down from A to B?
@ te by friction Jz E-=,
tng’ moh
a
"
= (0.2 16) -(0.2)¢t2) 4)
= -0-4 79
Phys 105 Second Midterm Page 2 Dec. 18, 1999
(c) Find the x- and y-coordinates of the center of mass of the two-fragment system 3 s after the
explosion.
—
~ — a
@ Kl? = Ra (#20) + ¥ (tot +F at
R sie) = USPS 4 508-1 60)9 F
om o 4
4 6 .
= M04 + 884 yi pa.
PROBLEM (4) (25 points)
A mass m, sliding with a speed of 5 m/s ona horizontal floor collides with another mass mr sliding
with a speed of 3 m/s on the same floor as shown in Fig. I. The speed of mj, after collision is again 5
m/s and itis moving in the direction shown in Fig. Il. Take mi, = 1 kg and mz =4 kg.
Sms
; Vv
2p
Fig. 1 Fig. I
(a) What is the direction of motion of mass mz after the collision, i.c., 6=7
© |
42
= =F sii BF ry #V, sie
5 cos 3F 4 HY ces C
¥
= Vfsin8 =2 mfs a -
2 VytrO = 2 mf Bef 2 mf
a OF 45°
W
Phys 105 Second Midterm Page § Dec. 18, 1999
(b) How much kinetic energy is lost in the collision?
®
lost KE = Ky -kp
2 z
a £2 (Ye -M )
il
az
/ =
- £4(9
z)
(c) Calculate the velocity of the center of mass of the two masses before (or after) the collision.
anh
= (m+) Bea
@ FZ
fofel
beng = FY,
— “A A
PM FG
= — Sia 37 FL 5 cos 57° f =- ae
= ~ 38-49-56 = -5F4F a tn Ms
— = “A
Ap, 2-Dp - £4 +73 pin Mis
Phys 105 Second Midterm Page 6
Dec. 18, 1999
