Question

a volleyball with a mass of 0.267 kg approaches a player horizontally with a speed of 16.0 m/s. the player strikes the ball with her hand, which causes the ball to move in the opposite direction with a speed of 21.2 m/s.
(a) what magnitude of impulse (in kg - m/s) is delivered to the ball by the player?
kg · m/s
(b) what is the direction of the impulse delivered to the ball by the player?
in the same direction as the balls initial velocity
opposite to the balls initial velocity
perpendicular to the balls initial velocity
the magnitude is zero.
(c) if the players hand is in contact with the ball for 0.0600 s, what is the magnitude of the average force (in n) exerted on the players hand by the ball?
n

Explanation:

Step1: Calculate the initial and final momentum

The formula for momentum is \(p = mv\).
Let the initial velocity \(v_{i}=16.0\ m/s\) and final velocity \(v_{f}=- 21.2\ m/s\) (negative because it's in the opposite direction), and mass \(m = 0.267\ kg\).
Initial momentum \(p_{i}=mv_{i}=0.267\times16.0\ kg\cdot m/s = 4.272\ kg\cdot m/s\)
Final momentum \(p_{f}=mv_{f}=0.267\times(-21.2)\ kg\cdot m/s=- 5.6604\ kg\cdot m/s\)

Step2: Calculate the impulse (a)

Impulse \(J=\Delta p=p_{f}-p_{i}\)
\(J=-5.6604 - 4.272=-9.9324\ kg\cdot m/s\)
The magnitude of the impulse \(|J| = 9.93\ kg\cdot m/s\)

Step3: Determine the direction of impulse (b)

Since \(J=-9.9324\ kg\cdot m/s\) (negative sign), it is opposite to the ball's initial velocity.

Step4: Calculate the average force (c)

We know that \(J = F_{avg}\Delta t\), so \(F_{avg}=\frac{J}{\Delta t}\)
\(F_{avg}=\frac{-9.9324}{0.0600}=-165.54\ N\)
The magnitude of the average force \(|F_{avg}| = 166\ N\)

Answer:

(a) \(9.93\ kg\cdot m/s\)
(b) Opposite to the ball's initial velocity
(c) \(166\ N\)