Question #605

A baseball being hit is in contact with the bat for 1.30 ms. The ball is moving toward the bat at speed 38.0 m/s just before it is hit, and is moving at 67.5 m/s away from the bat just after it is hit. Assume the ball has a constant acceleration while it is in contact with the bat, and that the ball’s motion is always along a straight line.
(a)Calculate the acceleration of the ball while it is being hit.
(b)What is the ball’s net displacement while it is in contact with the bat?

Expert's answer

a)Constant acceleration a reduces initial velocity and increases inverse velocity for t = 1.3*10¯³ s. General change of velocity is

Δν=(38+67.5) m/s, hence:

Δν=at

a=Δν/t=((38+67.5)m/s)/1.3*10¯³s≈81.154*10³ m/s²

b)Ball’s net displacement is:

Δx=(νο*t)+(at²)/2=38m/s*1.3*10¯³s+(81.154*10³ m/s²*(1.3*10¯³s)²/2)≈0.118m=11.8cm

Answer:a)81.154*10³ m/s²; b)11.8cm

Δν=(38+67.5) m/s, hence:

Δν=at

a=Δν/t=((38+67.5)m/s)/1.3*10¯³s≈81.154*10³ m/s²

b)Ball’s net displacement is:

Δx=(νο*t)+(at²)/2=38m/s*1.3*10¯³s+(81.154*10³ m/s²*(1.3*10¯³s)²/2)≈0.118m=11.8cm

Answer:a)81.154*10³ m/s²; b)11.8cm

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