How Does a Cricket Bat Affect the Force on a Ball?

[Littlej4me]

Homework Statement

This is a research problem, these are the figures I have at the moment I don't believe I need anything else to solve this problem (maybe the weight of the bat? it's stationary rather than swinging so I don't think it matters).

A cricket ball weighing 0.159kg is bowled at 161.3km/h. Ignoring air resistance and the energy lost as the ball bounces off the pitch, work out the maximum average force that a cricket bat exerts on the ball if it is in contact with the bat for 1.5ms.

Homework Equations

f=ma

From what I can see I think I need to use the deceleration to 0m/s over 1.5ms in this equation so I'd use:

f=m(change in velocity/time)

The Attempt at a Solution

m = 0.159kg
change in velocity = 44.805m/s
time = 0.0015s

f= 0.159 ( 44.805 / 0.0015 )
f= 4749.33

My problem here is I'm concerned about the ball bouncing off the bat and how I might account for that in the problem (or if I need to?). Thank you ~

 

[Dr. Courtney]

The key is that you have computed the average force while the ball and bat are in contact.

 

[Littlej4me]

The key is that you have computed the average force while the ball and bat are in contact.

I'm sorry, I don't understand what you mean. Does that mean you believe I am correct in my working out?

 

[Dr. Courtney]

I'm sorry, I don't understand what you mean. Does that mean you believe I am correct in my working out?

Rather than tell students they are right or not, I prefer to give them ways to think about it and double check their own work.

You can also double check using the idea of impulse. The impulse on the ball (average force times duration) will be equal to ball's change in momentum.

 

[Littlej4me]

Rather than tell students they are right or not, I prefer to give them ways to think about it and double check their own work.

You can also double check using the idea of impulse. The impulse on the ball (average force times duration) will be equal to ball's change in momentum.

This is the information I have written down in order to explain my reasoning behind why I believe I am correct.

The momentum of a particle is the quantity mv.

Momentum is a vector.

F = ma = d(mv)/dt = dp/dt

The conservation of Momentum

m1u1 + m2u2 = m1v1 + m2v2

Impulse = Ft = Δp
= m.vf - m.vi

F = m(vf - vi)/t

Assuming all of this information is correct I believe that I am correct in my working out. However, as I am not a qualified physicist I prefer to ask the opinion or someone who is able to confirm the knowledge that I believe to be correct. I will not know if I understand the work fully if I do not know whether my working out is correct.

 

[Chestermiller]

I think your initial assessment was correct. You don't know the speed of the ball coming off the bat. The assumption that you made was that the ball comes to a stop once it hits the bat. That's one extreme. The other extreme is where it has the same speed, but in the opposite direction. That would give you twice the force. Since you have no other information available, you can only conclude that it is somewhere in that range.

Chet

 

[insightful]

f=ma

From what I can see I think I need to use the deceleration to 0m/s over 1.5ms in this equation

Is the deceleration to 0m/s the only thing that occurs in that 1.5ms?