Problema Solution
A 500kg gun fires a 1.5 kg shell with a muzzle velocity of 650 m/s. a. Find the recoil velocity of the gun. b. If the recoil is opposite a resistive force of 1,600 N, find the time in which the gun is brought to rest.
Answer provided by our tutors
applying conservation of momentum
m1v1=m2v2
m1=500 kg
v1=?
m2= 1.5 kg
v2= 650 m/sec
so substituting these values in the conservation of momentum equation we get
500 v1= 1.5*650
so v1= 1.95 m/sec
final velocity of the gun= 0
initial velocity of the gun = 1.95 m/sec
we know
force= mass * (change in velocity)/time
so time= mass*change in velocity/force
so time = 500* 1.95/1600
so time = 0.6094 sec
so the gun will be brought to rest in 0.6094 seconds