9.4 Types of collisions  (Page 3/10)

Thor vs. iron man

1. Estimate the mass of Thor’s hammer
2. Estimate how much kinetic energy was lost in this collision

Strategy

Solution

1. First, we posit conservation of momentum. For that, we need a closed system. The choice here is the system (hammer + Iron Man), from the time of collision to the moment just before Iron Man and the hammer hit the tree. Let:
   
   • $M_{\text{H}}=$ mass of the hammer
   • $M_{\text{I}}=$ mass of Iron Man
   • $v_{\text{H}}=$ velocity of the hammer before hitting Iron Man
   • v $=$ combined velocity of Iron Man + hammer after the collision
   
   Again, Iron Man’s initial velocity was zero. Conservation of momentum here reads:
   
   $M_{\text{H}}{v}_{\text{H}}=\left(M_{\text{H}}+M_{\text{I}}\right)v.$
   
   We are asked to find the mass of the hammer, so we have
   
   $\begin{array}{ccc}\hfill M_{\text{H}}{v}_{\text{H}}& =\hfill & M_{\text{H}}v+M_{\text{I}}v\hfill \\ \hfill M_{\text{H}}\left(v_{\text{H}}-v\right)& =\hfill & M_{\text{I}}v\hfill \\ \hfill M_{\text{H}}& =\hfill & \frac{M_{\text{I}}v}{v_{\text{H}}-v}\hfill \\ & =\hfill & \frac{\left(200\text{kg}\right)\left(\frac{2\text{m}}{0.75\text{s}}\right)}{10\frac{\text{m}}{\text{s}}-\left(\frac{2\text{m}}{0.75\text{s}}\right)}\hfill \\ & =\hfill & 73\text{kg.}\hfill \end{array}$
   
   Considering the uncertainties in our estimates, this should be expressed with just one significant figure; thus, $M_{\text{H}}=7\times {10}^1\text{kg}$ .
2. The initial kinetic energy of the system, like the initial momentum, is all in the hammer:
   
   $\begin{array}{cc}\hfill K_{\text{i}}& =\frac{1}{2}{M}_{\text{H}}{v}_{\text{H}}^2\hfill \\ & =\frac{1}{2}\left(70\text{kg}\right){(10\text{m/s})}^2\hfill \\ & =3500\text{J.}\hfill \end{array}$
   
   After the collision,
   
   $\begin{array}{cc}\hfill K_{\text{f}}& =\frac{1}{2}\left(M_{\text{H}}+M_{\text{I}}\right)v^2\hfill \\ & =\frac{1}{2}\left(70\text{kg}+200\text{kg}\right){(2.67\text{m/s})}^2\hfill \\ & =960\text{J.}\hfill \end{array}$
   
   Thus, there was a loss of $3500\text{J}-960\text{J}=2540\text{J}$ .

Significance