imported>Patrick: Created page with ''''(a)''' The kinetic energy is related to the momentum as shown below: $K = \frac{1}{2} m v^2 = \frac{1}{2} \frac{m}{m} m v^2 = \frac{1}{2} \frac{m^2 v^2}{m} = \frac{1}{2} …'$

2011-08-27T16:37:36Z

Created page with ''''(a)''' The kinetic energy is related to the momentum as shown below: <math>K = \frac{1}{2} m v^2 = \frac{1}{2} \frac{m}{m} m v^2 = \frac{1}{2} \frac{m^2 v^2}{m} = \frac{1}{2} …'

New page

'''(a)''' The kinetic energy is related to the momentum as shown below:
<math>K = \frac{1}{2} m v^2 = \frac{1}{2} \frac{m}{m} m v^2 = \frac{1}{2} \frac{m^2 v^2}{m} = \frac{1}{2} \frac{(mv)^2}{m} = \frac{1}{2} \frac{p^2}{m}</math>
 
Given that <math>p_B = p_A</math>, we can compute the ratio of their kinetic energies using the above expression as

<math>\frac{K_B}{K_A} = {\frac{1}{2} \frac{{p_B}^2}{m_B} \over \frac{1}{2} \frac{{p_A}^2}{m_A}} = {\frac{1}{m_B} \over \frac{1}{m_A}} = \frac{m_A}{m_B} = \frac{60}{0.020} = 3000</math>
 
'''(b)''' We can compute the momentum from the kinetic energy

<math>K = \frac{1}{2} \frac{p^2}{m}</math>

<math>2 m K = p^2</math>

<math>p = \sqrt{2 m K}</math>
 
Given that <math>K_B = K_A</math>, we can compute the ratio

<math>\frac{p_B}{p_A} = {\sqrt{2 m_B K_B} \over \sqrt{2 m_A K_A}} = \sqrt{\frac{m_B}{m_A}} = \sqrt{\frac{0.020}{60}} = 0.018</math>

Conservation of Momentum EX 1 - Revision history

imported>Patrick: Created page with ''''(a)''' The kinetic energy is related to the momentum as shown below: K = \frac{1}{2} m v^2 = \frac{1}{2} \frac{m}{m} m v^2 = \frac{1}{2} \frac{m^2 v^2}{m} = \frac{1}{2} …'

imported>Patrick: Created page with ''''(a)''' The kinetic energy is related to the momentum as shown below: $K = \frac{1}{2} m v^2 = \frac{1}{2} \frac{m}{m} m v^2 = \frac{1}{2} \frac{m^2 v^2}{m} = \frac{1}{2} …'$