Question

Lonnie pitches a baseball of mass 0.500 kg. The ball arrives at home plate with a speed of 35.0 m/s and is batted straight back to Lonnie with a return speed of 50.0 m/s. If the bat is in contact with the ball for 0.055 s, what is the impulse experienced by the ball

Answer 1

Answer:

Explanation:

The impulse equation is

Δp = FΔt, where Δp = final momentum - initial momentum, F is the Force exerted on an object, and Δt is the change in time. In this equation,the entire right side defines the impulse. In other words, FΔt is the impulse; thus the change in momentum an object experiences is due to its change in impulse and is directly proportional to it.

Therefore, once we find the change in momentum, that is the impulse the object experiences. Δp = final momentum - initial momentum, where

p = mv and p is momentum.

$p_f=(.500)(50.0)$ so

$p_f=25.0$ and

$p_i=(.500)(35.0)$ so

$p_i=17.5$; therefore,

Δp = 25.0 - 17.5 = 7.5$\frac{kg*m}{s}$ which is the unit for momentum