# Simple question

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?

#### Attachments

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

HallsofIvy