Problems

1. Breakup of projectile in midflight

   Suppose that a projectile breaks up horizontally into two pieces of mass $m_1$ and $m_2$ in midflight. Given $\theta$, $v_0$, and $R_1$, predict $R_2$.

   a) Find $R$. As usual:
   

   $$y = (v_0 \sin \theta) t - \frac{1}{2}gt^2$$ (4.66)

   
   
   
   

   $$t_R(v_0 \sin\theta - \frac{1}{2} g t_R) = 0$$ (4.67)

   
   
   
   

   $$t_R = \frac{2v_0\sin\theta)}{g}$$ (4.68)

   
   
   
   

   $$R = (v_0\cos \theta)t_R = \frac{2v_0^2\sin \theta \cos\theta}{g}.$$ (4.69)

   
   

   b) $R$ is the position of the center of mass. Thus
   

   $$m_1R_1 + m_2R_2 = (m_1+m_2)R$$ (4.70)

   
   
   or
   

   $$R_2 = \frac{(m_1+m_2)R - m_1R_1}{m_2}$$ (4.71)

   
   

2. Ballistic Pendulum (for Lab)

   The usual question gives you the mass of the block ($M$) and the bullet ($m$) and the initial velocity of the bullet $v_0$ and asks for the maximum angle theta through which the pendulum swings after the bullet hits and sticks to the block. In the lab you actually measure the horizontal displacement of the block, but it amounts to the same thing (if you do the trigonometry).

   To do this we use momentum conservation and energy conservation. There is, however, a trick! Momentum is conserved during the collision (before the block has time to swing up against the external force of gravity). Energy is conserved after the collision (when only gravitational and normal forces act). (Kinetic) energy is not conserved during the collision. Momentum is not conserved after the collision (when gravity slows it down).

   The collision is one-dimensional (in the x-direction). Thus (for block $B$ and bullet $b$) we have momentum conservation:
   

   $$p_{b,0} = m v_0 = p_{B+b,f}$$ (4.72)

   
   
   Now if we were foolish we'd evaluate $v_{B+b,f}$ to use in the next step. Being smart, we instead do energy conservation in terms of momentum:
   

   $$E_0 = \frac{p_{B+b,f}^2}{2(M+m)} = \frac{p_{b,0}^2}{2(M+m)}$$

   $$= E_f = (M+m)gH$$

   $$= (M+m)gL(1-\cos\theta)$$ (4.73)

   
   

   Thus:
   

   $$\theta = \cos^{-1}(1 - \frac{(mv_0)^2}{2(M+m)^2gL})$$ (4.74)

   
   
   which only has a solution if $mv_0$ is small enough (which makes sense if you think about it - so think about it).