Answer
$18.57m/s$
Work Step by Step
We can find the required velocity as follows:
$m_Av_A+m_Bv_B=(mA+m_B)v$
We plug in the known values to obtain:
$(5000)(20)+2000(15)=(5000+2000)v$
This simplifies to:
$v=18.57m/s$
Engineering Mechanics: Statics & Dynamics (14th Edition)
by Hibbeler, Russell C.
Published by
Pearson
ISBN 10:
0133915425
ISBN 13:
978-0-13391-542-6
Chapter 15 - Kinetics of a Particle: Impulse and Momentum - Section 15.3 - Conservation of Linear Momentum for a System of Particles - Problems - Page 262: 35
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