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Group Problem 06
Momentum Energy Work
Name_____________________
1) A 18kg block is at rest on a table
with a coefficient of kinetic friction
k = 0.2. A 2kg ball on a string
swings down from rest and collides
with the block. The ball bounces
back and returns to one half its
original height.
2m
a) How far does the block slide on
the table before coming to a stop?
This problem requires students to construct a solution using energy, momentum, and
work. A common error is to assume the collision is completely elastic and set the energy
lost by the bob equal to that gained by the block during the collision. After students see
they need momentum for the collision they may mistakenly try to find pbob from Ebob
with E 
p2
claiming that p  2mE . Students need (using energy) to find the
2m
velocity with which the bob strikes the block and the velocity with which it recoils to find
pbob . Then the initial velocity of the block can be found, its energy, and distance to stop
from work. I get the following numbers.
pbob =21.4kgm/s Ei ,block =12.7J x =0.36m Please check these!!!
b) Is the collision completely elastic?
Energy lost by the bob is 19.6 and that gained by the block is 12.7J so not completely
elastic. The energy lost by the system is 6.9J.
release position
2) An ingenious friend, with way too much free time,
has designed a spring based launcher to send 2.0 kg
balls of clay up vertically into the sky. The ends of the
spring are designed to pivot around their respective
attachment points so that the spring always behaves in
a linear Hooke's Law fashion. The equilibrium length of
the unstretched spring is just 2.0 m. Its spring constant
is 100. N/m. The launcher starts with the spring 8.0 m
below the release point and then the springs movement
is stopped, as shown, with the spring length, now
horizontal, a distance of 6.0 m. Each clay ball is placed
in a massless cup and there are no frictional forces.
Gravity acts downward with an acceleration of 10. m/s2
Questions a-b pertain to the starting position
a) How much energy is stored in the spring ?
b) What is the acceleration of the ball?
a) E = ½ k ((6x6+8x8)½ m - 2m)2 = ½ 100 N/m (8m)2 = 3200 J
b) Fspring = - k u = - 100 N/m x 8.0 m = -800 N (towards pivot)
Fy = 800 N x 0.80 – 2.0 kg 10 m/s2 = 640 – 20 = 620 N
ay = 620 N / 2.0 kg = 310 m/s2
(c-d) Answer a & b again, but when the cup is at the release position.
c) E = ½ k ( 6m - 2m)2 = ½ 100 N/m (4m)2 = 800 J
d) Fspring = - k u = - 100 N/m x 4.0 m = -400 N (towards pivot)
Fy = – 2.0 kg 10 m/s2 = – 200 = -200 N
ay = -200 N / 2.0 kg = -10 m/s2
e) What is the velocity of the ball at the release position?
½ m v2= 3200 J– 800 J– 2.0 kg 10. m/s2 8 m = 2400 J – 160 J = 2240 J
v = sqrt(2240) m/s
f) What is the power associated with gravity at this position?
P= F v = m g v
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3) The graph, below and at right, shows the potential energy curve of a particle moving
along the x-axis under the influences of a conservative force.
a) In which interval(s) of x are the
forces on the particles to the right?
0< x < 3 m and 6 < x < 9 m
b) At what value(s) of x is the
magnitude of the force a minimum?
At x = 3, 6 & 9 m
c) At what value(s) of x is the
magnitude of the force a maximum
At x = 4 m (slope is biggest)
4) Two masses, 1.0 kg and 2.0 kg, are anchored at positions 3.0 m and 6.0 m,
respectively, from the end of a very, very light 6.0 m long straight rigid rod. The end of
the rod is anchored to a frictionless pivot. Initially the masses are horizontal, but after
releasing them, they swing in a vertical plane.
a) How far from the pivot is the center-of-mass?
c.m. = (1 x 3 + 2 x 6)/(1 + 2) m = 5.0 m
b) If all of the mass is placed at the center-of-mass,
then how much work is done by gravity on this
mass after the rod rotates down 90 degrees?
W = mgh = 3 x 10 x 5 = 150 J
½ I 2 =150 J = 0.5 x 3 x 25 x 2  = 2.0 rad/s
c) How much work was done by gravity on the 1.0
kg mass; on the 2.0 kg? How does this compare to
your answer in part b?
1st mass = 1.0 x 10 x 3.0 = 30 J
2nd mass= 2.0 x 10 x 6.0 = 120 J (total same)
d) What is the angular velocity at the bottom of the
swing?
½ I 2 = ½ (1.0 x 9.0 + 2.0 x 36) 2 = 150 J
 = sqrt(300/81) rad/s = 1.9 rad/s
e) What is the rotational kinetic energy of each mass?
1st mass = ½ 9.0 * 3.7 J = 17 J
2nd mass = ½ 72. * 3.7 J = 133 J
Notice that if we let a 3.0 kg mass swing down from a distance 5.0 m from the pivot
then one obtains a different angular velocity!
½ I 2 = ½ (3.0 x 25) 2 = 150 J or 2 = 4.0 rad2 /s2 so here we get  = 2.0 rad/sec
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