A point mass m1 = 0.020 kg located at the top corner of a smooth hemispherical bowl m, R. of radius R = 0.450 m is released from rest and hits another point mass m2 = 0.010 kg m2 which is at rest at the bottom of the bowl as shown in the figure. If two masses stick together after the collision, how high above do they go relative to the bottom of the bowl? (Ignore the friction between the %3D masses and the surface of the bowl and take g = 9.8 m/s.)

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Chapter1: Units, Trigonometry. And Vectors
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A point mass m1 =
0.020 kg located at the
top corner of a smooth hemispherical bowl
m,
R.
of radius R = 0.450 m is released from rest
and hits another point mass m2 =
0.010 kg
m2
which is at rest at the bottom of the bowl as
shown in the figure.
If two masses stick together after the collision, how high above do they
go relative to the bottom of the bowl? (Ignore the friction between the
%3D
masses and the surface of the bowl and take g = 9.8 m/s.)
Transcribed Image Text:A point mass m1 = 0.020 kg located at the top corner of a smooth hemispherical bowl m, R. of radius R = 0.450 m is released from rest and hits another point mass m2 = 0.010 kg m2 which is at rest at the bottom of the bowl as shown in the figure. If two masses stick together after the collision, how high above do they go relative to the bottom of the bowl? (Ignore the friction between the %3D masses and the surface of the bowl and take g = 9.8 m/s.)
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