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how do i find the exit speed of the bullet after it leaves the block?

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the question is A 5.00-g bullet moving with an initial speed of vi = 440 m/s is fired into and passes through a 1.00-kg block as shown in the figure below. The block, initially at rest on a frictionless, horizontal surface, is connected to a spring with force constant 850 N/m. The block moves d = 4.20 cm to the right after impact before being brought to rest by the spring.

 

I tried using 1/2mv^2=1/2kx^2 and m1v1+m2v2=m1v1+m2v2 but i am getting the wrong answer i dont know what to do now?

asked Oct 15, 2013 in Intro College Physics by anonymous
recategorized Oct 15, 2013 by LeiBao

1 Answer

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The bullet and the block is in a semi-inelastic collision so kinetic energy is lost but momentum would conserve. The question gives that the block is stopped after compressing the spring for a distance of d. With this you can find the velocity of the block after the collision. So you will have two equations to solve the problem:

$$m_1V_{1i}=m_1V_{1f}+m_2V_{2f}$$
$$\frac{k}{2}d^2=\frac{m_2}{2}V_{2f}^2$$

The second question asks for the change of mechanical energy of the bullet-block system.
answered Oct 15, 2013 by LeiBao (360 points)
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