Question

A tennis ball of mass of 0.06 kg is initially traveling at an angle of 47o to the horizontal at a speed of 45 m/s. It then was shot by the tennis player and return horizontally at a speed of 35 m/s. Find the impulse delivered to the ball.

Answer 1

Answer:

The impulse delivered to the ball is $Imp = \left(-3.941, 1.975\right)\,\left[\frac{kg\cdot m}{s} \right]$.

Explanation:

By Impulse Theorem, the motion of the tennis ball is modelled after the following expression:

$Imp = m\cdot (\vec v_{f} - \vec v_{o})$ (1)

Where:

$m$ - Mass of the ball, in kilograms.

$\vec v_{o}$ - Vector of the initial velocity, in meters per second.

$\vec v_{f}$ - Vector of the final velocity, in meters per second.

$Imp$ - Impulse, in meters per second.

If we know that $m = 0.06\,kg$, $\vec v_{o} = \left(45\,\frac{m}{s} \right)\cdot (\cos 47^{\circ}, \sin 47^{\circ})$ and $\vec v_{f} = \left(35\,\frac{m}{s} \right)\cdot (-1, 0)$, then the impulse delivered to the ball is:

$Imp = (0.06\,kg)\cdot \left[\left(35\,\frac{m}{s} \right)\cdot (-1,0) -\left(45\,\frac{m}{s} \right)\cdot (\cos 47^{\circ}, \sin 47^{\circ})\right]$

$Imp = (0.06\,kg)\cdot (-65.670, -32.911)\,\left[\frac{m}{s} \right]$

$Imp = \left(-3.941, 1.975\right)\,\left[\frac{kg\cdot m}{s} \right]$

The impulse delivered to the ball is $Imp = \left(-3.941, 1.975\right)\,\left[\frac{kg\cdot m}{s} \right]$.