Answer to Question #245081 in Molecular Physics | Thermodynamics for Kelani

Question

A 0.279-kg volleyball approaches a player horizontally with a speed of 12.1 m/s. The player strikes the ball with her fist and causes the ball to move in the opposite direction with a speed of 21.5 m/s.

(a) What impulse is delivered to the ball by the player? (Take the direction of final velocity to be the positive direction. Indicate the direction with the sign of your answer.)

___ kg · m/s

(b) If the player's fist is in contact with the ball for 0.0600 s, find the magnitude of the average force exerted on the player's fist.

___ N

Answer 1

(a) What impulse is delivered to the ball by the player? (Take the direction of final velocity to be the positive direction. Indicate the direction with the sign of your answer.)

To solve this we have to consider the velocities defined as $v_{1}=-[12.1\,m/s]\widehat{i}$ and $v_{2}=+[21.5\,m/s]\widehat{i}$. Then we find the impulse as:

$\vec{J}=\vec{p_2}-\vec{p_1}=m(\vec{v_2}-\vec{v_1}) \\ \vec{J}=(0.279\,kg)(+21.5\widehat{i}-(-12.1\widehat{i}) )m/s \\ \vec{J}=(0.279)\big(21.5+12.1 \big)[kg\cdot m/s]\widehat{i} \\ \vec{J}=(0.279)(33.6)[kg\cdot m/s]\widehat{i} \\ \vec{J}=[+9.3744\,kg\cdot m/s]\widehat{i}$

The positive sign for $\vec{J}$ means that goes into the same direction as the final velocity (the one determined as the positive direction) and since there is only a coordinate in x for the displacement we also have $\mid \vec{J} \mid=9.3744\,{kg\cdot m/s}$.

In conclusion, the impulse delivered to the ball by the player is +9.37 kg · m/s.

(b) If the player's fist is in contact with the ball for 0.0600 s, find the magnitude of the average force exerted on the player's fist:

We use the impulse-force theorem and find the average force as

$\vec{J}=\vec{p_2}-\vec{p_1}=\sum{\vec{F}}\cdot \Delta t \\ \implies \sum{{\vec{\mid F \mid}}}=\cfrac{{\mid \vec{J} \mid}}{\Delta t}$

Then we substitute and find

$\sum{{\vec{\mid F \mid}}}=\cfrac{{\mid [9.3744\,{kg\cdot m/s}] \widehat{i} \mid}}{0.0600\,s} \\ \text{ } \\ \implies \sum{{\vec{\mid F \mid}}}=\frac{9.3744}{0.0600}\,N\approxeq 156.24\,N$

In conclusion, the average force exerted on the player's fist is 156.24 N.

Reference:

• Sears, F. W., & Zemansky, M. W. (1973). University physics.