A bullet with mass 20 grams travels at 270 m/s is fired vertically ^^^^ into a b
ID: 1514493 • Letter: A
Question
A bullet with mass 20 grams travels at
270 m/s is fired vertically ^^^^ into a block of wood
weighing 13.7 Newtons that is initially at rest.
(A) Calculate how high the bullet and the block will go after the bullet is fully embedded (assume you can ignore gravity during the collision of the block and the bullet)
(B) Calculate the speed block plus the bullet right after the collision ( really important )
(C ) Calculate the work done by the wood on the bullet as it slows the bullet down (again, ignore gravity during the collision) ???
Explanation / Answer
let,
mass of the bullet, m=20 grams
speed of the bullet, ub=270 m/sec
mass of the wood, mw=(13.7)/9.8=1.4 kg
height of the bullet and bock after collision,
A)
by using law of conservation of energy,
1/2*mb*ub^2=(mb+mw)*g*h
1/2*(20*10^-3)*270^2=(20*10^-3+1.4)*9.8*h
==> h=52.38 m
height, h=52.38 m
B)
by using law of conservation momentum,
mb*ub+mw*vw=(mb+mw)*vf
20*10^-3*270+0=((20*10^-3+1.4)*Vf
===> vf=3.802 m/sec
C)
work done=change in K.E
=1/2*(mw+mb)*vf^2-0
=1/2*(1.4+20*10^-3)*3.802^2-0
=10.26 J
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