2. (25 pts) A gymnast mounts a balance beam by jumping off of a spring-board. She jumps straight up
before landing on the spring board.
Since there could be some ambiguitites in this problem, I added 5 points to the total score on this
problem.
a) (5 pts) If the gymnast is able to jump to a height of 0.7 m, how fast is she moving when her feet
hit the spring board, which is 20 cm above the ground? (Assume her jumping motion is completely
vertical.)
I suppose you could use kinematics for this, since the person is in free-fall, but we’ve been talking
about energy techniques so I’ll do it that way:
∆K=1
2mv2
−0 = 1
2mv2
∆U=mg(0.2−0.7) = −0.5mg
W= 0
1
2mv2
−(0.5)(9.8)m= 0
v= 3.13 m/s
Note that you don’t actually need the mass of the person to calculate the final speed.
b) (5 pts) The gymnast has a mass of 50 kg and the mass of the spring board is 5 kg. How fast is the
spring board (and the gymnast) moving at the instant after she lands on it?
pi=pf
(50)(3.13) + (5)(0) = (50 + 5)v
v= 2.85 m/s
This is a perfectly inelastic collision since the person lands on the board and remains in contact with
it until it springs them back up.
c) (5 pts) The very stiff spring in the spring board then compresses 15 cm before throwing the gymnast
back up (i.e. the spring board stops 5 cm above the floor). By how much does the kinetic energy of the
gymnast and spring board change as the spring compresses?
Assuming that the person and spring-board come to rest when the spring is fully compressed (or else
it would not be fully compressed!) we have:
∆K=1
2(55)(02)−1
2(55)(2.852) = −222.7 J
d) (5 pts) By how much does the gravitational potential energy of the gymnast and spring board
change?
Remember that after landing on the spring board, it compresses the spring 15 cm. So the vertical
travel while she is on the spring board is 15 cm (down).
∆Ug= (55)(9.8)(−0.15) = −80.85 J
e) (5 pts) How large must the spring constant be in order for the spring board to stop 5 cm above the
floor?
Now we can conserve energy. If we include the spring and the earth in the system, there is no net work
done. So ∆E= 0. The only energy we haven’t already calculated is the potential energy in the spring,
(1/2)kx2. So we have
0 = −222.7−80.85 + 1
2k(0.152)
k= 26982 N/m
Pretty stiff spring, eh?
