A 0.050 kilogram bullet is shot intoand becomes embedded in a 10 kilogram block

Question

A 0.050 kilogram bullet is shot intoand becomes embedded in a 10
kilogram block initially at rest on a horizontal surface.The coefficient of
kinetic friction between the block and the surface is 0.50. The bullet and
block slide and come to rest after traveling 5.0 meters to the right. Find:
[a] the speed of the block just after the bullet enters it
[b] the speed of the bullet just before it entered the block
[c] the energy lost by the bullet-block system during the collision

Write clear documentation supporting computer programs and systems Effectively use Event-Driven Programming Create graphical user interface Write Java program "Accumulate test scores" In this programming project you will design, develop, test and document a Java application. The user enters test scores one at a time and then clicks the Enter Score button. For each entered score, the application adds one to the number of scores, calculates the average score, and determines what the best score is so far. Then, it displays the number of scores, average score, and best score in the three disabled text fields. The user can click the Clear button to reset everything to zero. When the user closes the frame or clicks the Close button, the application exits. The average score is the sum of all scores divided by the number of scores. Assume valid data is entered. If you have trouble getting the labels and text fields to line up properly, try adjusting the frame size. When you use the Flow layout manager, the width of the frame affects how components you add to the frame are lined up. The User's guide should be written using Microsoft Word. The font size should be 12 point. The page margins should be 1 inch. The paragraphs should be with double line spacing. Any figures (screen captures) used should be properly labeled with a title and a figure number.

Explanation / Answer

m1 = 0.05 kg,

m2 = 10 kg

a) let v is the speed of the system after the bbullet embeded


work done by friction = change in kinetic enrgy of the system

mue*(m1+m2)*g*d = 0.5*(m1+m2)*v^2

v = sqrt(2*mue*g*d)

   = sqrt(2*0.5*9.8*5)

= 7 m/s

b) let v1 is the initial speed of the bullet and let v is the speed of the system after the bbullet embeded

m1*v1 = (m1+m2)*v


v1 = (m1+m2)*v/m1

= (0.05+10)*7/0.05

= 1407 m/s

c) ki = 0.5*m1*v1^2 = 49491.225

kf = 0.5*(m1+m2)*v^2 = 246.225

ki - kf = 49245 J

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