Concervation of Momentum: Bullet and Block of Mass

  • #1
A bullet of mass m = 19.7 g traveling horizontally at speed vo = 445 m/s strikes a block of mass M = 854 g sitting on a frictionless, horizontal table. The bullet comes out the other side of the block at speed v = 415 m/s. What is the speed of the block after the bullet exits the block?

Using conservation of momentum, I have the following equation but there are two unknowns (v'1 and v'2) so I need 2 equations but am kind of lost as to what should be my second equation? Can someone please help?

m1v1 + m2v2 = m1v'1 + m2v'2

(.0197kg)(445m/s) + (.854kg)(0) = (.0197kg)v'1 + (.854kg)v'2

8.7665 = .0197v'1 + .854v'2
 

Answers and Replies

  • #2
cepheid
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A bullet of mass m = 19.7 g traveling horizontally at speed vo = 445 m/s strikes a block of mass M = 854 g sitting on a frictionless, horizontal table. The bullet comes out the other side of the block at speed v = 415 m/s. What is the speed of the block after the bullet exits the block?

Using conservation of momentum, I have the following equation but there are two unknowns (v'1 and v'2) so I need 2 equations but am kind of lost as to what should be my second equation? Can someone please help?

m1v1 + m2v2 = m1v'1 + m2v'2

(.0197kg)(445m/s) + (.854kg)(0) = (.0197kg)v'1 + (.854kg)v'2

8.7665 = .0197v'1 + .854v'2
You know v1', it is given in the problem.
 
  • #3
57
0
the velocity on the right side of the equation is 415, it is given in the problem. you can solve for V2 using the same equation
 

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