A. An 8 g bullet of 463 m/s speed is shot into a 4.0 kg block, at rest on a fric

Question

A. An 8 g bullet of 463 m/s speed is shot into a 4.0 kg block, at rest on a frictionless horizontal surface. The bullet remains lodged in the block. The block moves into a spring and compresses it until the block stops. The force constant of the spring is 2,500 N/m The collision in this case is:

(A) partial inelastic

(B) perfect inelastic

(C) perfect elastic

(D) none of the above

B. Right after the the collision, ie., the bullet entered the block, the speed of the bloch before it starts to move is:

C.The block moves to the spring and compresses it. The maximum distance of the spring complessed by the block is

D. The potential energy of the spring system after it is fully compressed is

E. The maximum force acting on the block from the spring is

F. How long will the compress take from the beginning till the end?

G. During which part(s) of the process this energy lost?

H. How much mechanical energy is lost during this process from the moving bullet to the full compress of the spring?

Explanation / Answer

part A:

as bullet remains lodged in the block, the collision is perfectly inelastic.

hence option B is correct.

part B:

as external force is not present, total momentum will be conserved.

let speed of the block before it starts to move be v m/s.

then conserving momentum,

momentum before collision=momentum after collision

==>0.008*463=(4+0.008)*v

==>v=0.92415 m/s

part C:

let maximum distance the spring is compressed be x m.

total mechanical energy of the block and bullet system just before it touches the spring is converted to spring potential energy.

==>0.5*(4+0.008)*0.92415^2=0.5*2500*x^2

==>x=sqrt(0.5*(4+0.008)*0.92415^2/(0.5*2500))=0.037 m

so the spring is compressed by 3.7 cm

part D:

potential energy of the spring system=0.5*spring constant*compression^2

=0.5*2500*0.037^2

=1.7112 J

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