Problem#1
A 75.0-kg firefighter slides down a pole while a constant friction force of 300 N retards her motion. A horizontal 20.0-kg platform is supported by a spring at the bottom of the pole to cushion the fall. The firefighter starts from rest 4.00 m above the platform, and the spring constant is 4 000 N/m. Find (a) the firefighter’s speed just before she collides with the platform and (b) the maximum distance the spring is compressed. (Assume the friction force acts during the entire motion.)
Answer:
Consider the motion of the firefighter during the three intervals:
(1) before, (2) during, and (3) after collision with the platform.
(a) While falling a height of 4.00 m, his speed changes from vi = 0 to v1 as found from
∆E = ∆K + ∆U = K1 – Ki + U1 – Ui
–fh = ½ mv12 – 0 + 0 – mgh
–(300N)(4.00m) = ½ (75.0kg)v12 – (75.0kg)(9.80 m/s2)(4.00m)
3480J = (75.0kg)v12
v1 = 6.81 m/s
(b) During the inelastic collision, momentum is conserved; and if v2 is the speed of the firefighter and platform just after collision, we have
mv1 = (m + M)v2
(75.0kg)(6.81 m/s) = (75.0kg + 20.0kg)v2
v2 = 5.38 m/s
Following the collision and again solving for the work done by non-conservative forces, using the distances as labeled in the figure, we have (with the zero level of gravitational potential at the initial position of the platform):
∆E = ∆K + ∆U = K3 – K2 + Ug3 – Ug2 + Us3 – Us2
–f∆x = 0 – ½ (m + M)v22 + (m + M)g(–∆x) – ½ k∆x2 – 0
–(300N)∆x = – ½ (75.0kg + 20.0kg)(5.38 m/s)2 – (75.0kg + 20.0kg)(9.80 m/s2)∆x – ½ (4000N/m)∆x2
2000∆x2 – 631∆x – 1375 = 0
∆x = {631 ± [6312 – 4(2000)(–1375)]1/2}/4000 = 1.00 m
Problem#2
George of the Jungle, with mass m, swings on a light vine hanging from a stationary tree branch. A second vine of equal length hangs from the same point, and a gorilla of larger mass M swings in the opposite direction on it. Both vines are horizontal when the primates start from rest at the same moment. George and the gorilla meet at the lowest point of their swings. Each is afraid that one vine will break, so they grab each other and hang on. They swing upward together, reaching a point where the vines make an angle of 35.0° with the vertical. (a) Find the value of the ratio m/M. (b) What If? Try this at home. Tie a small magnet and a steel screw to opposite ends of a string. Hold the cen ter of the string fixed to represent the tree branch, and reproduce a model of the motions of George and the gorilla. What changes in your analysis will make it apply to this situation? What If? Assume the magnet is strong, so that it noticeably attracts the screw over a distance of a few centimeters. Then the screw will be moving faster just before it sticks to the magnet. Does this make a difference?
Answer:
(a) George and the gorilla meet at the lowest point of their swings (the speed of both at the lowest point is the same, say u), then
½ mu2 = mgR or ½ Mu2 = MgR
where R is the length of the light vine hanging, then
u = √(2gR)
The collision between George and Gorilla at its lowest point, applies the law of conservation of momentum,
–mu + Mu = (m + M)v
v = (M – m)u/(M +m) (1)
Swinging up: applies the law of conservation of energy,
½ (M + m)v2 = (M + m)gy
where y = R (1 – cos35.00), then
v = [2gR(1 – cos35.00)]1/2 (2)
from (1) and (2) we get
(M – m)u/(M +m) = [2gR(1 – cos35.00)]1/2
(M – m)/(M + m) = 0.425
0.575M = 1.43m
m/M = 0.403
(b) No change is required if the force is different. The nature of the forces within the system of
colliding objects does not affect the total momentum of the system. With strong magnetic
attraction, the heavier object will be moving somewhat faster and the lighter object faster
still. Their extra kinetic energy will all be immediately converted into extra internal energy
when the objects latch together. Momentum conservation guarantees that none of the extra
kinetic energy remains after the objects join to make them swing higher.
Problem#3
A cannon is rigidly attached to a carriage, which can move along horizontal rails but is connected to a post by a large spring, initially unstretched and with force constant k = 2.00 x 104 N/m, as in Figure 2. The cannon fires a 200-kg projectile at a velocity of 125 m/s directed 45.0° above the horizontal. (a) If the mass of the cannon and its carriage is 5 000 kg, find the recoil speed of the cannon. (b) Determine the maximum extension of the spring. (c) Find the maximum force the spring exerts on the carriage. (d) Consider the system consisting of the cannon, carriage, and shell. Is the momentum of this system conserved during the firing? Why or why not?
Answer:
(a) Use conservation of the horizontal component of momentum for the system of the shell, the cannon, and the carriage, from just before to just after the cannon firing.
pxf = pxi
mshellvshell cos45.00 + mcannonvrecoil = 0
(200kg)(125 m/s)cos45.00 + (5000kg)vrecoil = 0
vrecoil = –3.54 m/s
(b) Use conservation of energy for the system of the cannon, the carriage, and the spring from
right after the cannon is fired to the instant when the cannon comes to rest.
∆E = ∆U + ∆K
0 = Ugf – Ugi + Usf – Usi + Kf – Ki
0 = 0 – 0 + ½ kxmax2 – 0 + 0 – ½ mvrecoil2
(2.00 x 104 N/m)xmax2 = (5000kg)(–3.54 m/s)2
xmax = 1.77m
(c) the maximum force the spring exerts on the carriage is
Fs,max = │kxmax│ = (2.00 x 104 N/m)(1.77m) = 3.54 x 104 N
(d) No. The rail exerts a vertical external force (the normal force) on the cannon and prevents it
from recoiling vertically. Momentum is not conserved in the vertical direction. The spring
does not have time to stretch during the cannon firing. Thus, no external horizontal force is
exerted on the system (cannon, carriage, and shell) from just before to just after firing.
Momentum of this system is conserved in the horizontal direction during this interval.
(1) before, (2) during, and (3) after collision with the platform.
(a) While falling a height of 4.00 m, his speed changes from vi = 0 to v1 as found from
∆E = ∆K + ∆U = K1 – Ki + U1 – Ui
–fh = ½ mv12 – 0 + 0 – mgh
–(300N)(4.00m) = ½ (75.0kg)v12 – (75.0kg)(9.80 m/s2)(4.00m)
3480J = (75.0kg)v12
v1 = 6.81 m/s
(b) During the inelastic collision, momentum is conserved; and if v2 is the speed of the firefighter and platform just after collision, we have
mv1 = (m + M)v2
(75.0kg)(6.81 m/s) = (75.0kg + 20.0kg)v2
v2 = 5.38 m/s
Following the collision and again solving for the work done by non-conservative forces, using the distances as labeled in the figure, we have (with the zero level of gravitational potential at the initial position of the platform):
∆E = ∆K + ∆U = K3 – K2 + Ug3 – Ug2 + Us3 – Us2
–f∆x = 0 – ½ (m + M)v22 + (m + M)g(–∆x) – ½ k∆x2 – 0
–(300N)∆x = – ½ (75.0kg + 20.0kg)(5.38 m/s)2 – (75.0kg + 20.0kg)(9.80 m/s2)∆x – ½ (4000N/m)∆x2
2000∆x2 – 631∆x – 1375 = 0
∆x = {631 ± [6312 – 4(2000)(–1375)]1/2}/4000 = 1.00 m
Problem#2
George of the Jungle, with mass m, swings on a light vine hanging from a stationary tree branch. A second vine of equal length hangs from the same point, and a gorilla of larger mass M swings in the opposite direction on it. Both vines are horizontal when the primates start from rest at the same moment. George and the gorilla meet at the lowest point of their swings. Each is afraid that one vine will break, so they grab each other and hang on. They swing upward together, reaching a point where the vines make an angle of 35.0° with the vertical. (a) Find the value of the ratio m/M. (b) What If? Try this at home. Tie a small magnet and a steel screw to opposite ends of a string. Hold the cen ter of the string fixed to represent the tree branch, and reproduce a model of the motions of George and the gorilla. What changes in your analysis will make it apply to this situation? What If? Assume the magnet is strong, so that it noticeably attracts the screw over a distance of a few centimeters. Then the screw will be moving faster just before it sticks to the magnet. Does this make a difference?
Answer:
(a) George and the gorilla meet at the lowest point of their swings (the speed of both at the lowest point is the same, say u), then
½ mu2 = mgR or ½ Mu2 = MgR
where R is the length of the light vine hanging, then
u = √(2gR)
The collision between George and Gorilla at its lowest point, applies the law of conservation of momentum,
–mu + Mu = (m + M)v
v = (M – m)u/(M +m) (1)
Swinging up: applies the law of conservation of energy,
½ (M + m)v2 = (M + m)gy
where y = R (1 – cos35.00), then
v = [2gR(1 – cos35.00)]1/2 (2)
from (1) and (2) we get
(M – m)u/(M +m) = [2gR(1 – cos35.00)]1/2
(M – m)/(M + m) = 0.425
0.575M = 1.43m
m/M = 0.403
(b) No change is required if the force is different. The nature of the forces within the system of
colliding objects does not affect the total momentum of the system. With strong magnetic
attraction, the heavier object will be moving somewhat faster and the lighter object faster
still. Their extra kinetic energy will all be immediately converted into extra internal energy
when the objects latch together. Momentum conservation guarantees that none of the extra
kinetic energy remains after the objects join to make them swing higher.
Problem#3
A cannon is rigidly attached to a carriage, which can move along horizontal rails but is connected to a post by a large spring, initially unstretched and with force constant k = 2.00 x 104 N/m, as in Figure 2. The cannon fires a 200-kg projectile at a velocity of 125 m/s directed 45.0° above the horizontal. (a) If the mass of the cannon and its carriage is 5 000 kg, find the recoil speed of the cannon. (b) Determine the maximum extension of the spring. (c) Find the maximum force the spring exerts on the carriage. (d) Consider the system consisting of the cannon, carriage, and shell. Is the momentum of this system conserved during the firing? Why or why not?
Answer:
(a) Use conservation of the horizontal component of momentum for the system of the shell, the cannon, and the carriage, from just before to just after the cannon firing.
pxf = pxi
mshellvshell cos45.00 + mcannonvrecoil = 0
(200kg)(125 m/s)cos45.00 + (5000kg)vrecoil = 0
vrecoil = –3.54 m/s
(b) Use conservation of energy for the system of the cannon, the carriage, and the spring from
right after the cannon is fired to the instant when the cannon comes to rest.
∆E = ∆U + ∆K
0 = Ugf – Ugi + Usf – Usi + Kf – Ki
0 = 0 – 0 + ½ kxmax2 – 0 + 0 – ½ mvrecoil2
(2.00 x 104 N/m)xmax2 = (5000kg)(–3.54 m/s)2
xmax = 1.77m
(c) the maximum force the spring exerts on the carriage is
Fs,max = │kxmax│ = (2.00 x 104 N/m)(1.77m) = 3.54 x 104 N
(d) No. The rail exerts a vertical external force (the normal force) on the cannon and prevents it
from recoiling vertically. Momentum is not conserved in the vertical direction. The spring
does not have time to stretch during the cannon firing. Thus, no external horizontal force is
exerted on the system (cannon, carriage, and shell) from just before to just after firing.
Momentum of this system is conserved in the horizontal direction during this interval.
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