Impulse, Momentum, and Baseball

[Soaring Crane]

A baseball has mass 0.145 kg.

A. If the velocity of a pitched ball has a magnitude of 45.0 m/s and the battered ball's velocity is 55.0 m/s in the opposite direction, find the magnitude of the change in momentum of the ball.

delta p = m*delta v = m*(v2 -v1) = .145 kg (-55.0 m/s - 45.0 m/s) = -14.5 kg*m/s

B. Find magnitude of impulse applied to it by bat.

J = F*delta t = delta p = -14.5 kg*m/s ?? Is it the same as the above?

C. If the ball remains in contact with the bat for 2.00 ms, find the magnitude of the average force applied by bat.

J = Avg. F*delta t

delta t = 2.00 ms(1s/1000 ms) = .002 s

Avg. F = J/delta t = (-14.5 kg*m/s)/.002 s = -7250 N ?

Thanks.

 

[Doc Al]

All good. (Since they ask for magnitudes only, your answers should be positive.) And yes, since impulse equals change in momentum, the answers to A and B are the same.

 

[gamarrapaul]

why should the answer be positive ? iam kind of confused I thought it depended on what velocity is positive and negative

 

[dst]

It comes from the formula for magnitude - 2d = sqrt(x^2 + y^2). Even if x or y were negative, -x * -x = +x^2. There are no square roots of negative numbers (apart from complex numbers).

 

[pinkyjoshi65]

The answers should be positive, since they are asking for the magnitudes. The negative sign gives you the direction.