Determine the velocity of a hardball after it has been thrown

[Murdoc88]

Problem and Given Information

I must determine the velocity of a hardball after it has been thrown into a box of paper and then I measure that the distance the box moved back. The Law of Conservation of Momentum (and other physics principles) will be employed to do this. When the ball is caught in the box, the ball/box system will initially have the same momentum. Then friction (the external force) will eventually reduce the velocity, and hence, the momentum of the box is 0.

mBox = 2.6 Kg
mBall = 0.142 Kg
Fk = 7N
dist. = .57 m

Equations Used

$f_k = \mu_k N$
$Fn = mg$

All kinetic motion equations possibly.

pBall = p'Ball + p'Box

p=mv

Work Attempted

Alright,

What I've done has come to the conclusion I need to find normal force acting on the box which is ~ 25.5 N (25.489 N) and then found the Coefficient of friction. I know that the final velocity of the box is zero, distance is 0.57 m, the Force of kinetic friction on the box is 7 N and that the mass of the box is 2.6 Kg and that the mass of the ball is 0.142 Kg. In the end I need to find the initial velocity in m/s then convert to Mph in order to see how fast I throw.

Any help as usual is greatly appreciated.

 

[nrqed]

Problem and Given Information

I must determine the velocity of a hardball after it has been thrown into a box of paper and then I measure that the distance the box moved back. The Law of Conservation of Momentum (and other physics principles) will be employed to do this. When the ball is caught in the box, the ball/box system will initially have the same momentum. Then friction (the external force) will eventually reduce the velocity, and hence, the momentum of the box is 0.

mBox = 2.6 Kg
mBall = 0.142 Kg
Fk = 7N
dist. = .57 m

Equations Used

$f_k = \mu_k N$
$Fn = mg$

All kinetic motion equations possibly.

pBall = p'Ball + p'Box

p=mv

Work Attempted

Alright,

What I've done has come to the conclusion I need to find normal force acting on the box which is ~ 25.5 N (25.489 N) and then found the Coefficient of friction. I know that the final velocity of the box is zero, distance is 0.57 m, the Force of kinetic friction on the box is 7 N and that the mass of the box is 2.6 Kg and that the mass of the ball is 0.142 Kg. In the end I need to find the initial velocity in m/s then convert to Mph in order to see how fast I throw.

Any help as usual is greatly appreciated.

How did you find the force of kinetic friction??

In any case, if you have the force of kinetic friction, you don't need the coefficient of kinetic friction or the normal force. You can get the acceleration from the force and the mass. Knowing the acceleration (or deceleration if you will since it will be negative) and the distance traveled you can find the initial velocity using a kinematic equation. That's the velocity just after the impact. Using conservation of momentum you can now find the velocity of the ball just before the impact.

 

[Murdoc88]

I found the kinetic force using a Newton scale or whatever it is called.

The 7 N of friction isn't of the box moving, its just the force of the box just slidingind on the table. Therefore I don't know the force of the box moving away when it is hit with the ball. All I know that it has a kinetic force of the box is 7 N. both masses of the ball and box so I have the mass of the system and the distance... that's it

 

[Murdoc88]

any ideas? anyone? I'm really stuck.

 

[denverdoc]

This a lab?

 

[Dick]

nrqed has it exactly right. Conservation of momentum gives you an initial velocity (in terms of the unknown initial ball velocity). You compute the deceleration of the box+ball system using F=ma and your measured frictional force. Now v=v0*-a*t. Find t when v=0 (when the box stops). Put that into x=v0*t-(1/2)*a*t^2. That's the total displacement. What could be simpler? Even if it is a little complicated.

 

[Murdoc88]

It's an older lab, ( we did it in class for practice) its just for review but I really want to do this because we have a lab similar to this coming up on Tuesday.

 

[Dick]

This a lab?

Must be. I saw the Newton meter vs Newton-metre debate (or debat) earlier. I think the OP just measured the frictional force.

 

[Murdoc88]

But thank you for clearing this up Dick. It makes sense now.