Download PHYS 1901 PHYSICS 1A (ADVANCED) Progressive Test #2

PHYS 1901 PHYSICS 1A (ADVANCED)
Progressive Test #2 - SOLUTION
Question 1
y
TA
37o
TC
29o
37o
29o
TB
x
TC
mg
a)
Block is stationary therefore the sum of forces must be zero.
TC + mg = 0
=> TC - mg = 0
TC = mg = (1.9)(9.8) = 18.62 = 18.6N
(1 mark)
b)
Knot is stationary therefore the sum of forces must be zero.
TA+ TB + TC = 0
x-direction
TA,x = -cos(37) TA
TB,x = cos(29) TB
TC,x = 0
Tx = -cos(37) TA + cos(29) TB = 0
(1 mark)
y-direction
TA,y = sin(37) TA
TB,y = sin(29) TB
TC,y = TC
Ty = sin(37) TA + sin(29) TB - TC = 0
(1 mark)
From Tx:
TA = cos(29) TB / cos(37)
Substituting into Ty and using result from part a)
TB = TC / [ tan(37)cos(29) + sin(29) ]
= 16.3N
(1 mark)
Therefore
TA = cos(29) TB / cos(37)
= 14.79690669…
= 17.8N
(1 mark)
1901_progtest2_SOLN.doc
21/5/07
Question 2
(a) Angular momentum is conserved
(1 mark)
Why?
We can think about it several ways.
(i)
Treat the system as the disk. The vertical forces on the disk due to the chewing gum impose no
torque on the disk.
(ii)
Treat the system as gum + disk. External forces are all vertical, hence no external torque.
(one sensible reason - 1 mark)
(b) Angular velocity of the disk must decrease.
(1 mark)
Why?
We can think about it several ways.
(i)
Treat the system as the disk.
Addition of the gum to the disk then increases the moment of inertia of the disk.
Conservation of AM them means that angular velocity must decrease.
(Initially L = Io o . If we assume a point mass then later L = ( Io + mr2 ) )
(ii)
Treat the system as gum + disk.
Initial AM of gum is zero. Final AM of gum is non-zero.
Conservation of AM them means that disk must loose AM.
(iii)
Treat the system as the disk.
Accelerating the gum in the horizontal direction does impose a torque on the disk so that disk must
loose AM.
(one sensible reason - 1 mark)
(c) I clearly depends on mr2, so clearly the change in angular velocity must also depend on mr2 - i.e.
change in depends on r.
(1 mark)
Io
Io
2
FYI I o o = ( Io + mr ) so = or o = 2 o
2 1 o
Io + mr Io + mr 1901_progtest2_SOLN.doc
21/5/07
Question 3
Conservation of Momentum may be applied since there are no horizontal forces acting on the
system of the two balls.
Momentum before = Momentum after
3 m u – m u = 3m v1 +m v2
2u = 3 v1 + v2
(1)
(2 marks)
Conservation of Kinetic Energy may be applied since the collision is elastic.
3mu 2 + mu 2 = 3mv 12 + mv 22
2
(2)
4u = 3v12 + v 22
(2 marks)
From (1), v2 = 2u 3v1 , so substituting into (2) we get
2
2
4u = 3v12 + (2u 3v1 )
4u 2 = 3v12 + 4u 2 12uv1 + 9v12
0 = 12v12 12uv 1
Therefore
v1 = 0,
or v1 = u .
v1 = u gives v2 = u , reproducing the initial conditions
therefore
v1 = 0
and
(1 mark)
v2 = 2u .
(1 mark)
(b) Before the balls are released, the total Mechanical Energy is given by
M.E. = (m+3m )gh = 4mg x 1.5m
(1 mark)
After the collision, the small ball has reached 6 m and the large ball is stationary ,see above
M.E. = 3m x 0 +mgH = mg x6 m
(2 mark)
Therefore the M.E before the collision = M.E after the collision and ME is conserved
(1 mark)
1901_progtest2_SOLN.doc
21/5/07