Answer to Question #204484 in Mechanics | Relativity for Jay

Question

A volleyball is spiked so that its incoming velocity of + 5.0 m/s is changed to an outgoing velocity of -15 m/s. The mass of the volleyball is 0.35 kg. What impulse does the player apply to the ball? Remember an impulse is a change in momentum.

Answer 1

The information given is

• the initial velocity is $V_{i}=+5.0\;\text{m}/\text{s}$
• The final velocity is $V_{f}=-15.0\;\text{m}/\text{s}$
• The mass is $m=0.35\;\text{Kg}$

The impulse is ( the change of momentum )

$I=\Delta P\\ I=P_{f}-P_{i}\\ I=m\;V_{f}-m\;V_{i}\\$

Evaluating numerically.

$I=m\;V_{f}-m\;V_{i}\\ I=0.35\;\text{Kg}\times-15.0\;\text{m}/\text{s}-0.35\;\text{Kg}\times +5.0\;\text{m}/\text{s}\\ I=-5.25\;\text{Kg}\;\text{m}/\text{s}-1.75\;\text{Kg}\;\text{m}/\text{s}\\ I=-7\;\text{Kg}\;\text{m}/\text{s}$

Answer.

The impulse applied by the player is $\displaystyle \color{red}{\boxed{I=-7\;\text{Kg}\;\text{m}/\text{s}}}$