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Problem 1
A 4kg cat (treat it as a point particle) sits on a horizontal floor eyeing a stationary chair of
mass 10kg which is a horizontal distance of 1.3m away. The seat of the chair is 0.5m
above the floor. The cat jumps up and lands on the seat of the chair just as she reaches the
maximum height of her trajectory. She puts out her claws and hangs on. If the chair sits
on a part of the floor which has just been waxed, is very slippery and therefore
frictionless, what is the momentum of the cat plus chair system just after the cat has
landed?
1) Find horizontal velocity on collision using equations of motion
2)
Use conservation of momentum to find momentum of chair+cat
Define y vertical
x horizontal
At t=0,x=0,y=0,
assume cat leaps with velocity v
0
at angle
θ
.
Horizontal acceleration=0, vertical acceleration = g
Equations of motion
x = v
0
cos
θ
t
v
x
=v
0
cos
θ
y = v
0
sin
θ
t ½ gt
2
v
y
=v
0
sin
θ
gt
At the top of the trajectory
v
y
=0
t=v
0
sin
θ
/g
Substitute for t in the y equation
y = v
0
sin
θ
v
0
sin
θ
/g ½ g v
0
2
sin
2
θ
/g
2
= ½ v
0
2
sin
2
θ
/g
Substitute for t in the x equation
x=v
0
cos
θ
v
0
sin
θ
/g = v
0
2
cos
θ
sin
θ
/g
Divide y by x
y/x= ½ tan
θ
Cat
4kg
v
0
θ
v
x
p
c
Chair
10kg
1.3m
0.5m
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θ
= 2y/x = 0.769
θ
= 37.6
o
v
0
=
√
(2gy)/sin
θ
=
√
(2 9.8 0.5)/sin37.6 = 5.13m/s
Horizontal velocity at the top of
the trajectory
v
x
=v
0
cos
θ
= 5.13cos37.6 = 4.07m/s
By conservation of momentum:
momentum of cat +chair after impact = momentum of
cat before impact = m
c
v
x
= 4x4.07
= 16.3 kgm/s
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Problem 2
The sketch shows a mass, m=3kg, on an inclined plane which is at an angle of 40
o
to the
horizontal. It is attached to a light string which runs over a frictionless, massless pulley
and supports a mass M hanging vertically. The coefficients of friction, static and kinetic,
between the mass m and the plane are
µ
s
=0.4 and
µ
k
=0.3.
(a) If M=5kg, what is the acceleration of the system (magnitude and direction)?
(Caution: make sure your frictional force is pointing in the right direction.)
(b) What is the range of possible values of M such that the system is at rest in
equilibrium?
Free body diagram for the two masses as above
In the first case we assume that the masses are moving and the mass M is accelerating
downwards with an acceleration a
2
nd
law for mass M in the vertical direction
MgT=Ma
For mass m resolve along and perpendicular to the slope
Along the slope
Tmgsin
θ
f=ma
Perpendicular to slope
N = mgcos
θ
Friction force
f=
µ
k
N
Substituting for f and N
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This note was uploaded on 10/07/2009 for the course PHYS 1301W taught by Professor Marshak during the Fall '08 term at Minnesota.
 Fall '08
 Marshak
 Mass

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