I have a ball out of plasticine thrown horizontally with speed $v$ on a steel ball that is attached from top to the ground on flexible straps in the middle (see picture). The mass of plasticine is $m$, mass of the steel ball is $M$. The straps have length $l$ in neutral position and affect potential energy while deflection $l^\prime> l$ with $P(l^\prime)=1/2\cdot k\cdot(l^\prime-l)^2$. The hit is completely inelastic.

  1. What would be the total energy and total momentum look like before hit and after it?

  2. What would be the maximum deflection d of the steel ball in dependence of $m$, $M$, $k$, $l$, $v$?

Would be thankful for answers!

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Total energy before collision is purely kinetic that too only of the Plasticine ball and is equal to $\frac{mv^2}{2}$

Total momentum (of system of both balls) before collision = $mv$

Total momentum after collision = $(m+M)u$ as the hit is completely inelastic, the balls stick together.

As the straps are elastic, no impulse would be exerted by it on the system of balls so we could conserve momentum of the the system.

So, $u=\frac{mv}{m+M}$

So the final Kinetic energy of system (after the collision) is $\frac{(mv)^2}{2(m+M)}$

Now the straps will start to stretch, the balls will retard and all Kinetic energy of balls will be converted to Potential energy in straps, So,

Potential energy in straps is $$\frac{k(l'-l)^2}{2}= \frac{(mv)^2}{2(m+M)}$$

From here you can find $l'$ and hence the displacement of the ball.

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