A 38.0 kg block at rest on a horizontal frictionless table is connected to the wall via a spring with a spring constant k=17.0 N/m. A 2.20E-2 kg bullet travelling with a speed of 570 m/s embeds itself in the block. What is the amplitude of the resulting simple harmonic motion?
can anyone show me how to answer this question?
can anyone show me how to answer this question?
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Hi, me again
First find the initial velocity of the block and bullet after impact using conservation of momentum
m1v1 +m2v2 = (m1+m2)v3
if m1 is the bullet and m2 is the block then v2 is 0 and v1 is given.
Once you have V2, then we go to the energy equations
a frictionless spring system will continually oscillate between potential energy in the spring and kinetic energy of motion
KE = 1/2 mv^2
KE = 1/2 kx^2
Because there is no friction, these two must be set to equal
At that point the only unknown is x and you can solve for it as it is equal to the amplitude of vibration.
First find the initial velocity of the block and bullet after impact using conservation of momentum
m1v1 +m2v2 = (m1+m2)v3
if m1 is the bullet and m2 is the block then v2 is 0 and v1 is given.
Once you have V2, then we go to the energy equations
a frictionless spring system will continually oscillate between potential energy in the spring and kinetic energy of motion
KE = 1/2 mv^2
KE = 1/2 kx^2
Because there is no friction, these two must be set to equal
At that point the only unknown is x and you can solve for it as it is equal to the amplitude of vibration.
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I get initial block velocity at v = 0.33 m/s
plugging into the energy equations
I get x= 0.493 m
plugging into the energy equations
I get x= 0.493 m
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