Engineering Mechanics: Statics & Dynamics (14th Edition)

Published by Pearson
ISBN 10: 0133915425
ISBN 13: 978-0-13391-542-6

Chapter 15 - Kinetics of a Particle: Impulse and Momentum - Section 15.2 - Principle of Linear Impulse and Momentum for a System of Particles - Problems - Page 247: 1


$1.75N\cdot s$

Work Step by Step

The vertical and horizontal components of velocity are: $v_x=vcos60^{\circ}$ and $v_y=v sin60^{\circ}$ For vertical distance, we use the equation of motion $y=vsin60^{\circ}\times t-\frac{1}{2}gt^2$ but at the highest point $y=0$ $\implies 0=vsin60^{\circ}\times -\frac{1}{2}gt^2$ $\implies t=\frac{2vsin60^{\circ}}{g}$.....eq(1) Now $v_x=vcos60^{\circ}\times t$ $\implies 12=vcos60^{\circ}\times t$ $\implies t=\frac{12}{vcso60^{\circ}}$......eq(2) Comparing eq(1) and eq(2), we obtain: $\frac{12}{vcos60^{\circ}}=\frac{2vsin60^{\circ}}{g}$ After simplifying the above equation, we have $v=11.66m/s$ Now the impulse can be determined as: $m(v_f-v_i)=I$ We plug in the known values to obtain: $\implies I=0.15\times 11.66$ $I=1.75N\cdot s$
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