Question 9.124: A 0.550 kg ball falls directly down onto concrete, hitting i......

Fundamentals of Physics Extended – Instructor’s Solutions Manual [1360823]

A 0.550 kg ball falls directly down onto concrete, hitting it with a speed of 12.0 m/s and rebounding directly upward with a speed of 3.00 m/s. Extend a y axis upward. In unit-vector notation, what are (a) the change in the ball’s momentum, (b) the impulse on the ball, and (c) the impulse on the concrete?

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(a) The change in momentum (taking upwards to be the positive direction) is

$\Delta \vec{p}=(0.550 \,kg )[(3 \,m / s ) \hat{j}-(-12 \,m / s ) \hat{j}]=(+8.25 \,kg\ m / s ) \hat{j}$ .

(b) By the impulse-momentum theorem (Eq. 9-31) $\vec{J}=\Delta \vec{p}=(+8.25 N s ) \hat{j}$

$\Delta \vec{p}=\vec{J}$ (linear momentum–impulse theorem). (9-31)

(c) By Newton’s third law, $\overrightarrow{J_c}=-\overrightarrow{J_b}=(-8.25\, N s ) \hat{j}$ .

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