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Problem 3.12 A spring-like device
dp
3
(a) Start from the momentum principle: --------y = F net, y = bs – mg = 0 , since the mass is hanging motionless,
dt
so the momentum is not changing.
( 0.015 kg ) ( 9.8 N/kg )
mg
- = -------------------------------------------------------- = 2300 N/m 3
b = -----3
( 0.04 m ) 3
s
(b) Use a standard scheme for analysis of motion:
1) Choose a system, the mass.
2) Identify what objects exert forces on the system, and draw those forces. The Earth pulls down L0
with a gravitational force of magnitude mg, where as always g = +9.8 N/kg, and the device pulls up3
ward with a force of magnitude bs . Since the stretch at the time of release is greater than the
stretch when the mass was hanging motionless, it must be that the upward force of the device is si
now greater than mg.
vi
3) Use the Momentum Principle to update the momentum.
4) Use the postion update formula to find the new position.
5) Solve for the unknown quantitites.
6) Check.
bs3
m
mg
p f = p i + F net ∆t
〈 0, p new, y, 0〉 = 〈 0, p old, y, 0〉 + 〈 0, bs 3 – mg, 0〉 ∆t
The initial stretch s is (.27-.20) m = 0.07 m.
p new,y = – ( 0.015 kg ) ( 4 m/s ) + [ ( 2300 N/m 3 ) ( 0.07 m ) 3 – ( 0.015 kg ) ( 9.8 N/kg ) ] ( 10 – 3 s )
–4
p new,y = ( 6.4 ×10
kg ⋅ m/s ) – ( 0.06 kg ⋅ m/s ) = – 5.94 ×10
–2
kg ⋅ m/s
–2
p new,y
( – 5.94 ×10 kg ⋅ m/s )- = – 3.96 m/s
- = ------------------------------------------------------v new, y = -----------( 0.015 kg )
m
Speed is v new, y = 3.96 m/s
r = r + v avg ∆t
〈 0, y new, 0〉 = 〈 0, y old, 0〉 + 〈 0, – v y , 0〉 ∆t (downward velocity)
Put the origin at the top of the device, where it is supported, so the initial y is –0.27 m:
y new = ( – 0.27 m ) + ( – 3.96 m/s ) ( 10 –3 s ) = – 0.27396 m
The unstretched length is 0.20 m, so the final stretch is 0.07396 m.
If you use the initial velocity instead of the new velocity, the result is about the same (to 2 significant figures),
because the velocity doesn’t change much in this short time interval of just 1 millisecond:
y new = ( – 0.27 m ) + ( – 4 m/s ) ( 10 – 3 s ) = – 0.27400 m and the final stretch is 0.07400 m.
A particularly accurate approximation would be to average the initial and final speeds and use that average
speed to update the position.
The units check. Moreover, the final speed is smaller than the initial speed, which checks, because the upward
force is greater than mg, so the net force is upward and slows down the downward momentum.