Question

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.

Answer 1

Answer:

The magnitude of change in momentum of the ball is $97.5 m$ and impulse is also $97.5 m$

Explanation:

Given:

Velocity of a pitched ball $v _{i} = 47$ $\frac{m}{s}$

Velocity of ball after impact $v_{f} = -50.5$ $\frac{m}{s}$

From the formula of change in momentum,

  $\Delta P = m (v_{f} -v_{i} )$

Here mass is not given in question,

Mass of ball is $m$

Change in momentum is given by,

$\Delta P = m (-50.5 -47)$

$\Delta P = -97.5 m$

Magnitude of change in momentum is

$\Delta P = 97.5 m$

And impulse is given by

 $J = \Delta P$

$J = -97.5 m$

So impulse and

Therefore, the magnitude of change in momentum of the ball is $97.5 m$ and impulse is also $-97.5 m$