Question

A 5.00-g bullet is fired into a 900-g block of wood suspended as a ballistic pendulum. The combined mass swings up to a height of 8.00 cm. What was the kinetic energy of the bullet immediately before the collision

Answer 1

Answer:

The value is  $KE_b =0.710 \ J$

Explanation:

From the question we are told that

   The mass  of the bullet is $m_b = 5.00 \ g = 0.005 \ kg$

  The mass  of the wood is  $m_w = 900 \ g = 0.90\ kg$

   The height attained by the combined mass is  $h = 8.0 \ cm = 0.08 \ m$

Generally according to the law of energy conservation    

    $KE _b = PE_c$

Here $KE_b$ is the kinetic energy of the bullet before collision.

and $PE_c$ is the  potential  energy of the combined mass of bullet and wood at the height h which is mathematically represented as

      $PE_m = [m_b + m_w] * g * h$

So

   $KE_b =PE_c = [0.005 + 0.90] * 9.8 *0.08$

=> $KE_b =0.710 \ J$