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e Employee emal Student eServices t email D2L Brightspace I Need Help? Read n -3 points SerPSET9 9.P.089.MI.SA part of the question, you will not This question has several parts that must be completed sequentially. If you skip a able to come back to the skipped part. Tutorial Exercise A 5.00-g bullet moving with an initial speed of v 445 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 875 N/m. The block moves d 4.70 cm to the right after impact before being brought to rest by the spring- (a) Find the speed at which the bulet emerges from the block. (b) Find the amount of initial kinetic energy of the bullet that is converted into internal energy in bullet-block system during the collision. Part 1 of 5 Conceptualize We expect a speed less than 445 m/s because the bullet gives some momentum to the block. The massive block has a low speed when it starts to compress the spring, so it has relatively little kinetic energy. Most of the bullet's original kinetic energy will become internal energy in the system Part 2 of 5 Categorize We take the process apart into two stages. The first stage is the collision of the bullet and block to set the block in motion, and the second stage is the subsequent compression of the spring. Part 3 of 5-Analyz (a) First we find the speed of the bullet momenturm. We assume for an approximation that the block quickly reaches its maximum velocity V. We represent the mass of the block by M and the mass of the bulet by mThe initial and nnal velocitle of th bullet before and after the collision are represented by vi and v, when it emerges from the block by using the conservation of respectively, as in the figure respectively, as in the figure. By of momentum, we have the following equation. ng for vi from this equation, we have

Explanation / Answer

Mass =5gm

velocity=445 m/s

spring constant,k=875 N/m

d=4.7 cm

So we know that the bullet imparts a velocity V to the block on impact. That kinetic energy is taken up by the spring, hence
0.5*mV2=0.5kx2
(1/2) ( 1 kg ) V2 = (1/2) ( 875 N/m ) ( 0.047 m )2

V = 1.39 m/sec

Then, since momentum is conserved, if S is the exit speed of the bullet we must have

( 0.005 kg ) ( 445 m/sec ) = ( 1 kg )( 1.39 m/sec ) + ( 0.005 kg )S

Hence S = 167 m/sec is the exit velocity.

The various energies are:

Initial bullet: (1/2) ( 0.005 kg ) ( 445 m/sec )2 = 495.06 J
Final bullet: (1/2) ( 0.005 kg ) ( 167 m/sec )2 = 69.72 J
Block/Spring system: (1/2) ( 1 kg ) ( 1.39 m/sec )2 = 0.966 J

The energy lost is therefore:

495- ( 69.72 + 0.966 ) = 424.31 J

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