# Simple question

1. Apr 9, 2005

Two simple pendulums of equal length (L=40cm) are suspended from the same point. The pendulum bobs are point-like masses of m1=450g and m2=150g. the more massive bob (m1) is initially drawn back at an angle of $$\theta_0 = 40^o$$ from vertical, as shown(in attached pic). After m1 is released it swings down to $$\theta = 0$$ where it collisdes with m2 and the masses stick together. I have the answers to the following problems, but i would like to know how to do it.

a.) Find the speed of m1 just before the collision. this one was pretty simple, the answer was 1.35m/s and it matches the answer given.

b.) Determind the maximum angle to which the masses swing after the collision. well from problem a.) i found that $$v_i = \sqrt{2gL(1-cos(45))}$$

when the bigger mass hits the smaller mass, $$U_o = mgL(1-cos(\theta))$$ right? and K_o = 0. at the end of the collision, K_f = 1/2mv^2 and U_f = 0. which means $$v_{2f}^2 = 2gL(1-cos(\theta_2))$$ right?

well plugging everything i know to $$V_{2f}^2 = (\frac{(2*m_1)}{m_1+m_2})^2*v_i^2$$ m1 = 450 and L = .4 m right? i get $$V_{2f}^2 = 4.1$$

then setting 4.1 equal to v_2f $$4.1 =2gL(1-cos(\theta_2))$$ and solving for theta, i get 61 degrees. but the real answer is 27, what did i do wrong?

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2. Apr 10, 2005

### einstone

Unlike the total momentum,the total energy is not conserved during this inelastic collision.
Regards,
Einstone.

3. Apr 10, 2005

### HallsofIvy

Staff Emeritus
You say the larger mass is held at 40 degrees in the statement of the problem but use 45 degrees later. Since "Frogpad", giving essentially the same problem in a different post uses 40 degrees, I am going to assume that is correct.

Raising the mass on a 40 cm pendulum arm by 40 degrees means it is raised 40(1- cos(40)) cm vertically and so has potential energy 450g(40(1- cos(40))= 4131186 ergs(g here is the acc due to gravity- 981 cm/s2- rather than "grams") relative to its original position. Just before it strikes the second weight, it is back to its original position and so has 0 potential energy. All the potential energy has been converted to kinetic energy: its kinetic energy is 4131186 ergs. Since kinetic energy is (1/2)mv2= 4131186, we have v2= 8262374/450= 18360 so
v= 135.5 cm/s, just what you got!

Now the two masses stick together so we can think of that as a single mass of 450+350= 800 g. As einstone said, since this is not an elastic collision,energy is not conserved and but momentum is: the initial momentum of the system was the momentum of the first mass (450 g)(135.5 cm/s)= 60976 dynes. The momentum after the collision (800 g)(v cm/s)= 60976 so v= 60976/800= 76.3 cm/s.

Now, calculate the kinetic energy of an 800 g mass moving at 76.3 cm/s. Find the height and then angle that will give potential energy for an 800 g mass equal to that kinetic energy. The difference between that and the original kinetic energy (= original potential energy) is the change in energy due to the collision.