Asked by amanda and leah
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, of force constant 60000 N/m. The cannon fires a 305 kg projectile at a velocity of 174 m/s directed 38.9◦ above the horizontal.If the mass of the cannon and its carriage is 5537.7 kg, find the recoil velocity of the cannon. Answer in units of m/s
PartB
Determine the maximum extension of the
spring. Answer in units of m
PartC
Find the maximum force the spring exerts on the carriage. Answer in units of N
PartB
Determine the maximum extension of the
spring. Answer in units of m
PartC
Find the maximum force the spring exerts on the carriage. Answer in units of N
Answers
Answered by
drwls
A. Apply conservation of momentum in the horizontal direction.
B. Apply conservation of energy during comopression
(1/2)kXmax^2 = (1/2)MV^2
V is the velocity from part a.
Solve for Xmax
c) Fmax = k Xmax
B. Apply conservation of energy during comopression
(1/2)kXmax^2 = (1/2)MV^2
V is the velocity from part a.
Solve for Xmax
c) Fmax = k Xmax
Answered by
amanda and leah
Can you help me set up the conservation of momentum in the X direction please? once I figure that out i'm sure I can get it but I can't quite figure that out
Answered by
drwls
The total horizontal momentum of cannon and projectile remains 0
Mp*Vpcos38.9 + Mc*Vrecoil = 0
Mc = cannon mass
Vrecoil = recoil velocity
Vp = projectile mass
Solve for Vrecoil'
V recoil = -(Mp/Mc)*174*cos38.9 m/s
The minus sign means that it recoils backwards, compared to the direction of firing. .
Mp*Vpcos38.9 + Mc*Vrecoil = 0
Mc = cannon mass
Vrecoil = recoil velocity
Vp = projectile mass
Solve for Vrecoil'
V recoil = -(Mp/Mc)*174*cos38.9 m/s
The minus sign means that it recoils backwards, compared to the direction of firing. .
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