A thin metal rod of mass 1.9 kg and length 0.7 m is at rest in outer space, near

Question

A thin metal rod of mass 1.9 kg and length 0.7 m is at rest in outer space, near a space station (see figure below). A tiny meteorite with mass 0.08 kg traveling at a high speed of 225 m/s strikes the rod a distance 0.2 m from the center and bounces off with speed 60 m/s as shown in the diagram. The magnitudes of the initial and final angles to the x axis of the small mass's velocity are

?i = 26°

and

?f = 82°.

<8.8665,1.65,0>

<0,

  rad/s

(c) What is the increase in internal energy of the objects?
J

Explanation / Answer

length of the rod, L=0.7m

mass the metal rod, m1=1.9 kg

mass of the meteorite, m2=0.08kg

initial speed, vi=225 m/sec

final speed, vf=60 m/sec

theta_i=26 degrees

theta_f=82 degrees

distance, d=0.2 m

a)

initial velocity of m2 in vector form,

Vi=vi*cos(thetai)i + vi*sin(thetai)j

=225*cos(26)i + 225*sin(26) j

=(202.23)i+(98.63)j

and

final velocity of m2 in vector form,

Vf=-vf*cos(thetaf)i + vf*sin(thetaf)j

=-60*cos(82)i + 60*sin(82)j

=(-8.35)i+(59.42)j

by using law of conservation of momentum,

m1*v_rodi+m2*vi=m1*V_rodf+m2*vf

0+0.08*(202.23)i+(98.63)j=1.9*V_rodf+0.08*(-8.35)i+(59.42)j

(1.9/0.08)*V_rodf=(202.23)i+(98.63)j-(-8.35)i+(59.42)j

V_rodf=(0.08/1.9)(210.58)i+(39.21)j

======> Vcm = V_rodf=(8.866)i+(1.65)j

b)

by using conservation of angular momentum,

I*W=(-vi*cos(thetai)-vf*cos(thetaf))*m2*d

I*w=(-vi*cos(thetai)-vf*cos(thetaf))*m2*d

I*W=(-225*cos(26)-60*cos(82))*0.08*0.2

I*W=(-3.37)

1/12*m1*L^2*W = (-3.37)

1/12*1.9*0.7^2*w=(-3.37)

===> w=(-43.44)rad/sec

in the form of vector,

w=(-43.44)k rad/sec

c)

increase in internal energy, U=Ki-Kf

=1/2*m2*vi^2 - (1/2*m2*vf^2 + 1/2*m1*vcm^2 +1/2*I*w^2)

= (1/2*0.08*225^2) - (1/2*0.08*60^2 + 1/2*1.9*(8.866^2 + 1.65^2) + 1/2*(1/12*1.9*0.9^2)*(-43.44)^2)

=1682.73 J

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