Question

3. A pitcher throws a 1 kg baseball toward the batter so that it crosses home plate horizontally and makes contact with the bat. The batter then hits the ball applying an average force of 25 N. The ball and the bat are in contact for 0.5 s. Assume the ball travels along the same line leaving the bat as it followed before contacting the bat. a. Determine the magnitude of the change in momentum of the baseball. b. Determine the magnitude of the impulse delivered by the bat to the baseball. c. The baseball reaches the bat with a velocity of -40 m/s. What is the velocity with which the ball leaves the bat

Answer 1

Answer:

a. 12.5 kgm/s b. 12.5 Ns c. -27.5m/s

Explanation:

a. Determine the magnitude of the change in momentum of the baseball

We know I = Ft = Δp where I = impulse, F = average force = 25 N and t = time which force acts = 0.5 s and Δp = change in momentum of the ball

So, Δp = Ft = 25 N × 0.5 s = 12.5 kgm/s

b. Determine the magnitude of the impulse delivered by the bat to the baseball.

I = Ft where I = impulse, F = average force = 25 N and t = time which force acts = 0.5 s

So, I = Ft = 25 N × 0.5 s = 12.5 Ns

c. The baseball reaches the bat with a velocity of -40 m/s. What is the velocity with which the ball leaves the bat

Since  Δp = m(v - u) where Δp = change in momentum of the ball = 12.5 kgm/s, m =mass of baseball = 1 kg, u = initial velocity of baseball = -40 m/s and v = final velocity of baseball = unknown

So, Δp = m(v - u)

making v subject of the formula, we have

Δp/m = (v - u)

v = Δp/m + u  

Substituting the values of the variables, we have

v = Δp/m + u

v = 12.5 kgm/s/1kg + (-40m/s)

v = 12.5 m/s - 40 m/s

v = -27.5 m/s