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

To find the impulse delivered to the tennis ball, use the principle of conservation of momentum. The impulse is equal to the change in momentum of the ball. Given the mass and initial and final velocities, calculate the change in velocity and use the conservation of momentum equation to find the impulse.

Step-by-step explanation:

To find the impulse delivered to the tennis ball, we can use the principle of conservation of momentum. The impulse is equal to the change in momentum of the ball. The momentum of an object is the product of its mass and velocity.

Given: m = 0.06 kg, initial velocity = 45 m/s, final velocity = 35 m/s.

The change in velocity is 35 m/s - 45 m/s = -10 m/s. Since the ball is returning horizontally, the change in velocity only affects the speed, not the direction.

Using the principle of conservation of momentum, the initial momentum is equal to the final momentum. The initial momentum is given by m * v_initial and the final momentum is given by m * v_final. Setting them equal:

0.06 kg * 45 m/s = 0.06 kg * 35 m/s + impulse

simplifying the equation, we get:

impulse = 0.06 kg * 45 m/s - 0.06 kg * 35 m/s = 0.6 kg·m/s

Therefore, the impulse delivered to the ball is 0.6 kg·m/s.

Answer 2

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

Step-by-step 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\,(m)/(s) \right)\cdot (\cos 47^(\circ), \sin 47^(\circ))` and
`\vec v_(f) = \left(35\,(m)/(s) \right)\cdot (-1, 0)`, then the impulse delivered to the ball is:

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

`Imp = (0.06\,kg)\cdot (-65.670, -32.911)\,\left[(m)/(s) \right]`

`Imp = \left(-3.941, 1.975\right)\,\left[(kg\cdot m)/(s) \right]`

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