evan4888
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A 0.010-kg bullet traveling horizontally at 400.0 m/s strikes a 4.0-kg block of wood sitting at the edge of a table. The bullet is lodged into the wood. If the table height is 1.2 m, how far from the table does the block hit the floor?
this is what i did:
lets the mass of bullet = m2
the mass of block of wood = m1
velocity of the block of wood before the collision take place = v1
velocity of the bullet = v2
velocity of the block and bullet after the collision take place = vf
momenta of the block = p1
momenta of the bullet = p2
the momentum of block and bullet after collision equals the sum of momenta of the block and the bullet before the collision:
pf = pi
The initial momentum is the sum of the momenta of the block and bullet
pf = p1 + p2
Since the block is initially at rest, p1 = 0. Then
pf = p2
or
(m1 + m2)(vf) = (m2)(v2)
Solving for vf:
vf = [(0.010)(400.0)]/(4.0+0.010) = 0.9975 m/s
And I am kind of stuck here because I'm not sure how to calculate the distance from the table to the point that the block hits the floor. If it is the projectile motion formula, I guess I'm just not sure what to use and what to plug in.
this is what i did:
lets the mass of bullet = m2
the mass of block of wood = m1
velocity of the block of wood before the collision take place = v1
velocity of the bullet = v2
velocity of the block and bullet after the collision take place = vf
momenta of the block = p1
momenta of the bullet = p2
the momentum of block and bullet after collision equals the sum of momenta of the block and the bullet before the collision:
pf = pi
The initial momentum is the sum of the momenta of the block and bullet
pf = p1 + p2
Since the block is initially at rest, p1 = 0. Then
pf = p2
or
(m1 + m2)(vf) = (m2)(v2)
Solving for vf:
vf = [(0.010)(400.0)]/(4.0+0.010) = 0.9975 m/s
And I am kind of stuck here because I'm not sure how to calculate the distance from the table to the point that the block hits the floor. If it is the projectile motion formula, I guess I'm just not sure what to use and what to plug in.