What Happens When a Pendulum Hits a Static Object?

[Dracovich]

Ok so i got this assignment to return tomorrow:

"A particle P with mass m is fastened to the end of a string, that has the length L. The particle starts at rest in a position where the string is rigid, and the strings direction makes a 60° with the vertical axis. At it's lowest point of the pendulums swing, the particle P hits an object A, which is at rest on a frictionless horizontal table. The object A has a mass of n*m, where n is a positive number. The collision between P and A is completely elastic. Gravitational acceleration we call g and we ignore wind resistance. So the known variables are: n, m, l, and g.

1. Just before the collision, the particle P has the kinetic energy $$T_0$$, find $$T_0$$.

2. Find the velocity of the object A and the particle P right after the collision, and find out how big a part "q" of the kinetic energy $$T_0$$ has been transferred to A.

3. Investigate q for $$n -> \infty$$. Give a short comment about the observed effect.

4. Find $$cos\theta_0$$, where $$cos\theta_0$$ is the angle between the string and the vertical axis, when the particle P is at it's highest point in the gravitational field.

5. Inveistigate the expresion for $$cos\theta_0$$ for $$n -> \infty$$, give a short comment on the observed effect"

Ok this was translated over to english so i just hope i got it written down right :)

So ok, i think i got 1 and 2, and the most of 3.

First we find the height of P and use conservation of energy to find it's velocity just before it hits A

$$h=L-(cos(\theta)*L)$$
$$mgh=1/2*mv^2 => v=\sqrt{2g(L-(cos60°*L))}$$

And cos60 being0.5 we get $$v=\sqrt{g*L}$$

Then since it's totally elastic, no energy is lost and E before and E after is the same. That coupled with conservation of momentum i have two unknowns with two variables. It's rather tedious handworking but i got to

$$1/2mgl=1/2mu_1^2 + 1/2mnu_2^2$$
$$mgl=mu_1 + mnu_2$$
$$u_1=\sqrt{gl}*((n-1)/(n+1))$$
$$u_2=2*\sqrt{gl}/(n+1)$$

and the to find the fractional quantity of q transferred kinetic energy to A i just took the kinetic energy of the object A right after being hit, and divided it with the kinetic energy of the particle P just before the collision which should give me the % of energy transfered, and i got:

$$\frac{4n}{(n+1)^2}$$

Now for #3, when n goes towards infinity. Obviously (n+1)^2 grows exponantually and much faster then 4n, so when n goes towards infinity, the whole shebang goes towards zero, which i can see mathematicly, but I'm not sure why physicly this is the case. I can also see that this must be correct looking at the formula for kinetic energy, since if V gets very small, then V^2 must be even smaller and the kinetic energy lessens, but i just don't feel like i have an understanding of it :/ Even though i could propably give an answer by simply quoting the formulas that would be satisfactory for this assignment. If i could get some better explenation that would be great too.

Now #4 I'm pretty lost on. First off of course i'll have a new v since it's an unknown variable, so $$v=\sqrt{2g(L-(cos\theta_0*L)}$$. And I've tried playing a bit with that in the formulas, but i can't seem to get anything that makes good sense to me :/

If anyone could give me some tips on #4 and perhaps a better explenation of #3 (and of course if you see something wrong with 1 and 2) that'd be great :)

 

[lippyka]

</code>#4We can use conservation of energy to solve for cosθ₀. Since the particle P is at its highest point in the gravitational field, potential energy = kinetic energy. mgh = 1/2mv²where v is the velocity of the particle at its highest point. Therefore, mgh = 1/2mv² = 1/2m(2gLcosθ₀)²Solving for cosθ₀, cosθ₀ = \sqrt{\frac{h}{L}}#3 As n increases, the fractional quantity of q transferred kinetic energy to A decreases. This is because the momentum of the particle P is conserved, and as the mass of object A increases, the velocity of the object A decreases. Since velocity is squared in the equation for the kinetic energy, a decrease in velocity leads to a decrease in the total kinetic energy.

 

[wais]

1. To find T_0, we can use the formula for kinetic energy: T_0 = 1/2 * m * v^2. Since the particle is at rest, its initial velocity is 0. Therefore, T_0 = 0.

2. To find the velocity of object A and particle P after the collision, we can use conservation of momentum and conservation of energy. Before the collision, the total momentum is 0 since the particle is at rest. After the collision, the total momentum is still 0 since the string is rigid and the object A is at rest. Therefore, the momentum of the particle P must be equal and opposite to the momentum of object A. This means that their velocities must also be equal and opposite.

Using conservation of energy, we can set the initial kinetic energy T_0 equal to the final kinetic energy of the two objects, T_1 and T_2. This gives us the equation: 1/2 * m * v^2 = 1/2 * m * v_1^2 + 1/2 * n * m * v_2^2. We can solve for v_1 and v_2 to get the velocities of the two objects after the collision.

To find the fraction q of kinetic energy transferred to A, we can use the formula: q = T_2 / T_0. Substituting in the values for T_2 and T_0, we get q = 4n / (n+1)^2.

3. As n approaches infinity, the fraction q becomes very small, approaching 0. This means that almost all of the kinetic energy is transferred to the particle A. This makes sense because as n increases, the mass of object A also increases, making it more difficult for the particle P to transfer its kinetic energy to A. Physically, this can be explained by the fact that as n increases, the collision becomes more and more inelastic, meaning that some of the kinetic energy is lost in the form of heat, sound, etc.

4. To find cos\theta_0, we can use the formula for potential energy: mgh = mgL(1-cos\theta_0). Solving for cos\theta_0, we get cos\theta_0 = 1 - h/L. We know that the particle P is at its highest point when its velocity is 0. Therefore