The mass of the bullet is 0.01 kg and the mass of the block is 0.99 kg. The bullet is fired into the block andbecomes imbedded in it. The bullet-block combination rises by 0.05 meters. What potential energy is stored inthe pendulum (bullet-block combination)?PE = mgh = a. 50 joules b. 5 joules c. 0.5 joules d. 0.05 joules2. What was the kinetic energy of the pendulum (bullet-block combination) "immediately" after the bullet becameimbedded (i. e. what was the MAXIMUM K. E. of the combination)?KE = PE= (conservation of energy -- no calculation required!) a. 50 joules b. 5 joules c. 0.5 joules d. 0.05 joules3. What was the velocity of the pendulum (bullet-block combination) "immediately" after the bullet becameimbedded (i. e. when its K. E. was maximum)?KE = 1mv2v2 = 2KEv = 2KE = 2 m m a. 0.5 m/sec b. 1.0 m/sec c. 1.5 m/sec d. 2.0 m/s4. What was the momentum of the pendulum (bullet-block combination) "immediately" after the bullet becameimbedded (i. e. when its K. E. was maximum)?p = mv = a. 0 kg m/sec b. 0.25 kg m/sec c. 0.5 kg m/sec d. 1.0 kg m/sec Ballistic Pendulum5/15/082Don RathjenExploratorium Teacher Institute3601 Lyon St., San Francisco, CA [email protected]© 2008 Exploratorium, www. exploratorium5. What was the momentum of the block before the bullet hit it?pblock= mv = (think about this -- mental calculation -- no calculator required!) = a. 0 kg m/sec b. 0.25 kg m/sec c. 0.5 kg m/sec d. 1.0 kg m/sec6. What was the momentum of the bullet before it hit the block?pbullet = (conservation of momentum -- no calculation required!) = a. 0 kg m/sec b. 0.25 kg m/sec c. 0.5 kg m/sec d. 1.0 kg m/sec7. What was the velocity of the bullet before it hit the block (i. e. the "muzzle velocity)?pbullet = mvv = pbullet = m a. 25 m/sec b. 50 m/sec c. 75 m/sec d. 100 m/sec e. 250 m/sec

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