A 10 gram bullet is fired horizontally into, and becomes embedded in, a suspende

Question

A 10 gram bullet is fired horizontally into, and becomes embedded in, a suspended block of wood with a mass of 0.890 kg. The length of the cord suspending the block is 1.20 m.

(a) If the block with the embedded bullet swings uward and the cord suspending it ends up at an angle of 48.2 degrees relative to a verticle reference line, what was the initial speed of the bullet?

(b) How does the speed of the block with the embedded bullet immediately after the collision compare with the initial speed of the bullet?

(c) Was the collision elastic? If not, what percentage of the initial kinetic energy was lost?

Explanation / Answer

a) height moved by the block and bullet system = 1.2-1.2cos48.2 = 0.401m

so conserving energy we have

initial energy = 0.5*mass of bullet*velocity of bullet^2

final energy = mass of block and bullet system*acc due to gravity*height = (0.890+0.01)*9.81*0.401 =3.54

so initial energy =final energy

0.5*0.01*v^2=3.54

so v^2 = 3.54/(0.5*0.01) = 708

so v=sqrt(708) =26.6m/s

b)conserving momentum

mv = MV

0.01*26.6 = (0.01+0.890)*V

or V=0.266/.9 =0.29m/s

c)yes the collision is elastic

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.