Asked by Christina
A 0.5 kg wooden block is dropped from a height of 15 m. At what height must a 5g bullet be traveling upward strike the block in order to momentarily stop its fall? Assume that the time required for the bullet to come to rest in the block in negligibly short, that no internal energy is produced and that the bullet travels with a speed of 320 m/s immediately before striking the block.
Answers
Answered by
Damon
bullet momentum up = .005 * 320 = 1.6 kg m/s
If the momentum of the block down is 1.6 then they both come to a stop
Vdown = g t
momentum down = m g t
so
when m g t = 1.6, they stop
.5*9.81*t = 1.6
t = .326 seconds
how high is the block then?
h = 15 - (1/2)g t^2
h = 15 - 4.9(.326)^2
h = 14.48 meters
If the momentum of the block down is 1.6 then they both come to a stop
Vdown = g t
momentum down = m g t
so
when m g t = 1.6, they stop
.5*9.81*t = 1.6
t = .326 seconds
how high is the block then?
h = 15 - (1/2)g t^2
h = 15 - 4.9(.326)^2
h = 14.48 meters
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