Question 1 (10
Points)
1D
Collision and Impulse
A uniform rope of mass m
per unit length
hangs vertically from a hook so that the lower
end of the rope just touches the horizontal table as shown in Fig. 1(a). The rope is then
released from the hook and descends onto the table. The rope is assumed to be
completely flexible and when it reaches the table it stops instantaneously. When a length
y of the rope has fallen (see Fig. 1 (b)) what is the force exerted by the rope on the table
in terms of y, m and g?
Fig. 1
Solution
The rope descends via free fall onto the table. As it starts from zero initial velocity its
velocity v is related to the height y it has fallen by: v(y)=[2gy]
1/2
. The length of the rope
which lands on the table during the time dt at that instant is: v(y)dt and the corresponding
mass is mvdt
Based on the assumption that the rope stops instantaneously after it lands on the table, the
rate at which the rope transfers momentum to the table is:
dp/dt=[mvdt]v/dt=mv
2
=m(2gy). According to Newton’s equation of motion this is the
force exerted on the table by the falling rope. However, there is also rope of length my
already on the table and the force they exert due to gravity on the table is myg.
Thus the total force exerted by the rope on the table is 2mgy+mgy=3mgy.

This
** preview**
has intentionally

**sections.**

*blurred***to view the full version.**

*Sign up*