Asked by Marc
A bullet of mass m = 5.89 g is fired into a block of mass M = 3.09 kg that is initially at rest on the edge of a smooth table. (The table top is 0.83 m above the floor.) The bullet ends up stuck in the block, and together they fly off the table and fall to the floor.
If the bullet has an initial speed of 870 m/s, what distance D will the block travel horizontally before landing on the floor?
If the bullet has an initial speed of 870 m/s, what distance D will the block travel horizontally before landing on the floor?
Answers
Answered by
Elena
m•v= (m+M) •V,
V = m•v/(m+M) = 5.89•10^-3•870/(5.89•10^-3+3.09) =
=1.294 m/s.
h=g•t^2/2 ,
t = sqrt(2•h/g) = sqrt(2•0.83/9.8) =
= 0.412 s,
D = V•t = 1.294•0.412 = 0.53 m.
V = m•v/(m+M) = 5.89•10^-3•870/(5.89•10^-3+3.09) =
=1.294 m/s.
h=g•t^2/2 ,
t = sqrt(2•h/g) = sqrt(2•0.83/9.8) =
= 0.412 s,
D = V•t = 1.294•0.412 = 0.53 m.
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