Bullet in a block hitting a spring

AI Thread Summary
An 8-g bullet is shot into a 4.0-kg block, causing the block to compress a spring by 5.0 cm after the collision. The initial approach incorrectly equated momentum and energy, leading to confusion in calculating the bullet's initial velocity. The correct method involves using conservation of momentum and kinetic energy principles separately. The kinetic energy of the block and bullet system should be expressed as 1/2(m+M)v^2, while the spring's potential energy is 1/2(k)(x)^2. Properly applying these principles will yield the correct initial velocity of the bullet.
preluderacer
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Homework Statement



An 8-g bullet 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 by 5.0 cm. The force constant of the spring is 2200 N/m. In the figure, the initial velocity of the bullet is closest to:


The Attempt at a Solution



what I did was set (mbullet+mblock)vblock=1/2(k)x^2 I solved the the velocity of the block and got 0.686 m/s. Then I set mbullet(v)bullet=(m1+m2)vblock and solved for the vbullet and got 343N. This answer doesn't seem quite right. What am I doing wrong?
 
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preluderacer said:
what I did was set (mbullet+mblock)vblock=1/2(k)x^2 I solved the the velocity of the block and got 0.686 m/s.

The left-hand side of your equation is momentum, the right-hand side is energy. They can not be equal.

ehild
 
i think i got it 1/2(m+M)v^2=1/2(k)(x)^2
 
Last edited:
Write out the kinetic energy of the block with the bullet inside. ehild
 
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