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- Homework Statement
- A catapult consists of a massless cup attached to a massless spring of length

l and of spring constant k.

i) If a ball of mass m is loaded into the cup and the

catapult pulled back to extend the spring to a total length x, what velocity

does the ball reach when launched horizontally?

ii) The catapult is then used to launch the ball vertically. If the spring is

extended to the same total length of x before release, to what velocity does

the catapult now accelerate the ball?

Q: Why does h = x - l

- Relevant Equations
- 1/2kx^2

1/2 mv^2

mgh

My solutions: When ball is launched horizontally, assuming its velocity is entirely in the horizontal dimension, there is no interaction of the ball with the gravitational field, thus no change in GPE, so all of the EPE (elastic potential energy ) of the spring is transferred to KE of the ball.

(1/2) mv^2= 1/2 (k(x-l)^2)

i) v = √ (k/m) * (x-l)ii) When launched vertiically, now the mass is interacting with gravtiational field, as well as possessing maxmium kinetic energy (thus speed) at the moment of release.

so, total energy of system (ball) = KE +GPE = (1/2) mv^2 + mgh

1/2 (k(x-l)^2) = (1/2) mv^2 + mgh

Q1) But the solution mentions that h = x-l , why is this? :

With the diagram I have drawn ( attached), I cannot understand why h= x-l

I'm assuming the height from which it is released intially is zero. But does the ball have to gain a potential energy of the height it is displaced by (x-l) ? I don't quite understand why this.

1/2 (k(x-l)^2) = (1/2) mv^2 + mg(x-l)

when gives v velocity as the maximum velocity reached. It is assuming the greatest velocity reached by the ball is the instant it leaves the catapult, then it must travel a vertical distance of x-l to do so, and thus, gains Mg(x-l) as well as KE. But I don't understand why the maximum velocity is when the ball leaves the catapult.

(1/2) mv^2= 1/2 (k(x-l)^2)

i) v = √ (k/m) * (x-l)ii) When launched vertiically, now the mass is interacting with gravtiational field, as well as possessing maxmium kinetic energy (thus speed) at the moment of release.

so, total energy of system (ball) = KE +GPE = (1/2) mv^2 + mgh

1/2 (k(x-l)^2) = (1/2) mv^2 + mgh

Q1) But the solution mentions that h = x-l , why is this? :

With the diagram I have drawn ( attached), I cannot understand why h= x-l

I'm assuming the height from which it is released intially is zero. But does the ball have to gain a potential energy of the height it is displaced by (x-l) ? I don't quite understand why this.

1/2 (k(x-l)^2) = (1/2) mv^2 + mg(x-l)

when gives v velocity as the maximum velocity reached. It is assuming the greatest velocity reached by the ball is the instant it leaves the catapult, then it must travel a vertical distance of x-l to do so, and thus, gains Mg(x-l) as well as KE. But I don't understand why the maximum velocity is when the ball leaves the catapult.