Problem 6 : The ram of a pile driver has a mass m and is released from rest at a height h above the top of the pile which has the mass 3m. To what height does the ram rebound if the coefficient of restitution between the ram and the pile is e and e>(1/3)? If the pile meets a constant resistance F from the material in which it is lodged, calculate how far into the material it penetrates before coming to rest.
This question is as submitted by Baline Patrick through email.
Solution : We need to know the velocities of ram and pile immediately after the inelastic collision. Let the velocity of the ram just before strike be "v", velocity of the ram after the strike be "
v
1
v
1
" velocity of the pile after strike be "
v
2
v
2
".
Let the downward vertical ydirection be positive. We can determine velocity of the ram just before it hits the pile. Applying equation of motion for constant acceleration,
v
2
=
0
+
2
g
h
=
2
g
h
v
2
=
0
+
2
g
h
=
2
g
h
In order to determine velocities immediately after collision, we shall apply conservation of linear momentum in vertical direction (downward direction as positive)
m
1
v
1
i
+
m
2
v
2
i
=
m
1
v
1
f
+
m
2
v
2
f
m
1
v
1
i
+
m
2
v
2
i
=
m
1
v
1
f
+
m
2
v
2
f
Putting values, we have :
⇒
m
v
+
0
x
3
m
=
m
v
1
+
3
m
v
2
⇒
m
v
+
0
x
3
m
=
m
v
1
+
3
m
v
2
⇒
v
=
v
1
+
3
v
2
⇒
v
=
v
1
+
3
v
2
Now, applying equation of restitution, we have :
e
=
v
2
f
−
v
1
f
v
1
i
−
v
2
i
=
v
2
−
v
1
v
−
0
e
=
v
2
f
−
v
1
f
v
1
i
−
v
2
i
=
v
2
−
v
1
v
−
0
⇒
e
v
=
v
2
−
v
1
⇒
e
v
=
v
2
−
v
1
Thus, we have two unknowns “
v
1
v
1
” and “
v
2
v
2
”. Also we have two equations. Adding two equations,
⇒
1
+
e
v
=
4
v
2
⇒
1
+
e
v
=
4
v
2
The velocity of pile, therefore, is :
⇒
v
2
=
1
+
e
v
4
⇒
v
2
=
1
+
e
v
4
Putting this value in the first equation, we get the velocity of the ram after collision,
⇒
v
1
=
v
−
3
v
2
=
1
−
3
e
v
4
⇒
v
1
=
v
−
3
v
2
=
1
−
3
e
v
4
Putting expression of “v”, as obtained earlier, in above two expressions, we have :
⇒
v
2
=
1
+
e
v
4
⇒
v
2
=
1
+
e
v
4
⇒
v
2
=
1
+
e
2
g
h
4
⇒
v
2
=
1
+
e
2
g
h
4
and
⇒
v
1
=
1
−
3
e
2
g
h
4
⇒
v
1
=
1
−
3
e
2
g
h
4
Note that 1  3e < 0 as e > 1/3. It means that the velocity of ram is opposite to the assumed positive reference direction. This further means that it rebounds after the strike in upward direction.
⇒
v
1
=
3
e
−
1
2
g
h
4
⇒
v
1
=
3
e
−
1
2
g
h
4
Let h' be the height attained by the ram after the strike. For the upward motion of ram,
0
=
v
1
2
−
2
g
h
′
0
=
v
1
2
−
2
g
h
′
⇒
h
'
=
v
1
2
2
g
=
3
e

1
2
X
g
h
2
g
X
16
=
3
e

1
2
X
h
16
⇒
h
'
=
v
1
2
2
g
=
3
e

1
2
X
g
h
2
g
X
16
=
3
e

1
2
X
h
16
Now, we consider the motion of pile into the material. There are two forces acting on the pile (i) constant resistance “F” and (ii) force due to gravity “3mg". As given, the net force acts up to decelerate the pile. The deceleration is :
a
=
F
−
3
m
g
3
m
a
=
F
−
3
m
g
3
m
Let "y" be the vertical distance moved by the pile. Applying equation of motion,
0
=
v
2
2
−
2
X
F
−
3
m
g
3
m
X
y
0
=
v
2
2
−
2
X
F
−
3
m
g
3
m
X
y
⇒
y
=
v
2
2
X
3
m
2
F
−
3
m
g
⇒
y
=
v
2
2
X
3
m
2
F
−
3
m
g
⇒
y
=
1
+
e
2
X
6
m
g
h
2
F
−
3
m
g
X
16
=
1
+
e
2
X
3
m
g
h
16
F
−
3
m
g
⇒
y
=
1
+
e
2
X
6
m
g
h
2
F
−
3
m
g
X
16
=
1
+
e
2
X
3
m
g
h
16
F
−
3
m
g