Solving for Initial Bullet Speed: Mass, Spring, and Friction

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zyphriss2
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A bullet of mass 1.6×10−3 {\rm kg} embeds itself in a wooden block with mass 0.990 {\rm kg}, which then compresses a spring (k = 160 {\rm N/m}) by a distance 5.0×10−2 {\rm m} before coming to rest. The coefficient of kinetic friction between the block and table is 0.51. What is the initial speed of the bullet? I have been working on this for and hour and a half and have done all of the other 8 problems except this one. Can anyone please help me and explain how it works.
 
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Show what you've done and where you got stuck.

Hint: Work backwards. How fast was the bullet+block moving just after the collision?
 


I have tried working backwards. I found that for the spring to be compressed the stated length that the block+bullet must exert .2 Newtons of force on the spring. Then i found that the summ of the forces in the x direcction is (block+bullet)a-force of friction=.2 and that results in a acceleration of 5.19. Now from here i do not know where to go
 


For the spring compression part of this problem, use energy conservation, not force analysis.
 


I had initially tried that. .5kx^2=.5mv^2 which turned out to make velocity equal to .635 m/s. Now from there I didn't know where to go, the coefficient of friction is messing me up.
 


zyphriss2 said:
I had initially tried that. .5kx^2=.5mv^2 which turned out to make velocity equal to .635 m/s. Now from there I didn't know where to go, the coefficient of friction is messing me up.
You set KE = Spring PE, which isn't true since some of that original energy was lost to friction. You need to take the initial KE and subtract the work done by friction to get the final spring PE.
 


I found that the velocity is = to .9495 m/s, I found this using .5mv^2 - force of friction*distance the spring compressed=.5kx^2. is this correct and from here would I just plug this velocity into mbullet*velocitybullet= m(bullet+block)*.9495?
 


Thank you very much I got it correct.