# Concervation of Momentum: Bullet and Block of Mass

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

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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.

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