Question

a 0.311 kg tennis racket moving 30.3 m/s east makes an elastic collision with a 0.0570 kg ball moving 19.2 m/s west. find the velocity of the tennis ball after the collision.

Answer 1

Answer: 146.12 m/s (to the right)

Step-by-step explanation:

The Conservation of Linear Momentum principle states that the initial momentum $p_{i}$ (before the elastic collision) must be equal to the final momentum $p_{f}$ (after the elastic collision):

$p_{i}=p_{f}$ (1)

Being:

$p_{i}=m_{r}V_{r} + m_{b}V_{b}$

$p_{f}=m_{r}U_{r} + m_{b}U_{b}$

Where:

$m_{r}=0.311 kg$ is the mass of the racket

$V_{r}=30.3 m/s$ is the velocity of the racket before the collision

$m_{b}=0.057 kg$ is the mass of the ball

$V_{b}=-19.2 m/s$ is the velocity of the ball before the collision

$U_{r}=0 m/s$ is the velocity of the racket after the collision  (assuming it is at rest after hitting the ball)

$U_{b}$ is the velocity of the ball the collision

So, we have the following:

$m_{r}V_{r} + m_{b}V_{b}=m_{r}U_{r} + m_{b}U_{b}$  (2)

Finding $U_{b}$:

$U_{b}=\frac{m_{r}V_{r} + m_{b}V_{b}}{m_{b}}$ (3)

$U_{b}=\frac{(0.311 kg)(30.3 m/s) + (0.057 kg)(-19.2 m/s)}{0.057 kg}$ (4)

Finally:

$U_{b}=146.12 m/s$ This is the velocity of the ball after the collision

Answer 2

Answer:

38m/s

Step-by-step explanation: