If the velocity of a pitched ball has a magnitude of 47.0 m/s and the batted ball's velocity is 50.5 m/s in the opposite direction, find the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat.

Respuesta :

Answer:

The magnitude of change in momentum of the ball is [tex]97.5 m[/tex] and impulse is also [tex]97.5 m[/tex]

Explanation:

Given:

Velocity of a pitched ball [tex]v _{i} = 47[/tex] [tex]\frac{m}{s}[/tex]

Velocity of ball after impact [tex]v_{f} = -50.5[/tex] [tex]\frac{m}{s}[/tex]

From the formula of change in momentum,

  [tex]\Delta P = m (v_{f} -v_{i} )[/tex]

Here mass is not given in question,

Mass of ball is [tex]m[/tex]

Change in momentum is given by,

[tex]\Delta P = m (-50.5 -47)[/tex]

[tex]\Delta P = -97.5 m[/tex]

Magnitude of change in momentum is

[tex]\Delta P = 97.5 m[/tex]

And impulse is given by

 [tex]J = \Delta P[/tex]

[tex]J = -97.5 m[/tex]

So impulse and

Therefore, the magnitude of change in momentum of the ball is [tex]97.5 m[/tex] and impulse is also [tex]-97.5 m[/tex]