Displacement of mass fired by a slingshot problem

[vodkasoup]

Homework Statement

A catapult (slingshot) is used to fire a ball of mass 0.2kg vertically into the air. The elastic is stretched so that there is a tension of 40N in both 'sides' of the slingshot, ie on the elastic on either side of the mass (I am trying to describe a diagram here, so please bear with me...). The angle between the mass and each side of the elastic is 50 degrees. The mass is drawn back a distance of 0.1m.

The question asks me to find how high the ball will travel after being released, assuming the tension in the elastic remains constant.

Homework Equations

F=ma

Equations of motion?

Kinetic/potential energy equations?

The Attempt at a Solution

I have drawn a right-angled triangle with a side of 0.1m opposite to the hypotenuse. From this I have reasoned that the force used to draw back the mass equals the sum of the vertical components of the two tensions, ie 2 times 40cos(50 degrees). This is therefore the force used to fire the mass.

From this I can figure out the vertical acceleration using F=ma. Although I am unsure if I should subtract gravity from F in this calculation. I know that the ball will cease to rise when the vertical acceleration equals the force downwards due to gravity. But I am unsure of how to get this value.

Thanks for any help given.

 

[tiny-tim]

hi vodkasoup!

I know that the ball will cease to rise when the vertical acceleration equals the force downwards due to gravity.

no no no!

acceleration has nothing to do with it

when the ball leaves the catapult, the only force on it is gravity, so the acceleration is a constant, -g

when the ball ceases to rise, its acceleration is still -g

… This is therefore the force used to fire the mass.

From this I can figure out the vertical acceleration using F=ma. Although I am unsure if I should subtract gravity from F in this calculation.

yes, to find the acceleration (using F = ma), you must always use all the forces

however, that only gives you the instantaneous acceleration …

as the ball rises (still on the elastic), the acceleration decreases, doesn't it?

so you'd have to use an integral if you used acceleration

for that reason, use conservation of energy instead ​

 

[vodkasoup]

Hello tiny-tim!

Thanks for your reply, and please excuse my rudeness in taking so long to reply back.

Ah, it seems I have been a bit silly with this one. So I use the energy, or work, done by the catapult from the point when the string is released to the point where the ball leaves the string ( W = F x d = 51.4N x 0.1m = 5.14J , assuming my earlier calculation of force is correct).

The ball will stop rising when all of its kinetic energy has been converted into potential energy, which will equal the energy used to fire the ball (due to the principle of conservation of energy). So at the zenith of its rise, the potential energy of the ball will be 5.14 joules.

So E(p) = mgh ,

Therefore h = E / mg = the answer :)

Gosh, I really need to learn how to use LaTex...

 

[tiny-tim]

hello vodkasoup!

So I use the energy, or work, done by the catapult from the point when the string is released to the point where the ball leaves the string ( W = F x d =

no, same difficulty …

the force in the elastic is not constant, you'd have to find ∫ F.dx if you used the force

sooo … same solution …

use the energy formula for a spring (not the force formula) ​

 

[vodkasoup]

Hi again!

I think, for the introductory course I'm doing, the instructor setting the questions has given the force in the elastic as constant and then neglected to mention it...as the answer I get from using the above is correct, and Hooke's Law etc hasn't been covered.

I shall bear it in mind for the future, though!

Thanks again :)