   Chapter 6, Problem 43P

Chapter
Section
Textbook Problem

A 12.0-g bullet is fired horizontally into a 100-g wooden block that is initially at rest on a frictionless horizontal surface and connected to a spring having spring constant 150 N/m. The bullet becomes embedded in the block. If the bullet-block system compresses the spring by a maximum of 80.0 cm, what was the speed of the bullet at impact with the block?

To determine
Initial speed of the bullet

Explanation

Given Info: Mass of the bullet is m=12.0×103kg , Mass of the block is M=100×103kg , initial velocity of the block is vB=0 , spring constant is k=150 N/m and compression of spring is x=80×102m ., mass of the bullet is m=12.0×103kg , Mass of the block is M=100×103kg , and initial velocity of the block is vB =0 .

Explanation:

Apply conservation of mechanical energy to block-embedded bullet.

(KE+PE)f=(KE+PE)i

• KE is the kinetic energy.
• PE is the potential energy.

Use 1/2(m+M)V2 for (KE)f , 0 for (PE)f , 0 for (KE)i , and 1/2kx2 for (PE)i .

12(m+M)V2+0=0+12kx212(m+M)V2=12kx2

• m is the mass of the bullet.
• M is the mass of the block.
• k is the spring constant.
• x is the compression of spring.
• V is the speed of the block-bullet combination after impact.

Rewrite the above equation to solve for V .

V=kx2m+M

Substitute 12

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