Question

If the velocity of a pitched ball has a magnitude of 44.5 m/s and the batted ball's velocity is 55.5 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. Find the magnitude of the impulse applied to it by the bat?
If the ball remains in contact with the bat for 2.50 ms, find the magnitude of the average force applied by the bat.

Answer 1

Answer:

ΔP = 14.5 Ns

I = 14.5 Ns

ΔF = 5.8 x 10³ N = 5.8 KN

Explanation:

The mass of the ball is given as 0.145 kg in the complete question. So, the change in momentum will be:

ΔP = mv₂ - mv₁

ΔP = m(v₂ - v₁)

where,

ΔP = Change in Momentum = ?

m = mass of ball = 0.145 kg

v₂ = velocity of batted ball = 55.5 m/s

v₁ = velocity of pitched ball = - 44.5 m/s (due to opposite direction)

Therefore,

ΔP = (0.145 kg)(55.5 m/s + 44.5 m/s)

ΔP = 14.5 Ns

The impulse applied to a body is equal to the change in its momentum. Therefore,

Impulse = I = ΔP

I = 14.5 Ns

the average force can be found as:

I = ΔF*t

ΔF = I/t

where,

ΔF = Average Force = ?

t = time of contact = 2.5 ms = 2.5 x 10⁻³ s

Therefore,

ΔF = 14.5 N.s/(2.5 x 10⁻³ s)

ΔF = 5.8 x 10³ N = 5.8 KN