A 12 g bullet passes horizontally through a 15 cm tall, 20 cm long .5 kg block w
ID: 1528183 • Letter: A
Question
A 12 g bullet passes horizontally through a 15 cm tall, 20 cm long .5 kg block which is initially at rest on a 1 meter tall table's edge. The bullet is initially moving 230 m/s, and it passes through the block. After the bullet passes through it, the block hits the ground 134 cm from the edge of the table.
How fast is the block moving after the bullet emerges from it?
How fast is the bullet moving when it merges from the block?
Assuming that the bolck does not move significantly while the bullet is inside it, determine the average force exerted on the block by the bullet while the bullet passes through the block. Use energy!
Explanation / Answer
a) let t is the time taken for the bulet to hit the ground.
use, h = voy*t + (1/2)*g*t^2
h = 0 + (1/2)*g*t^2
==> t = sqrt(2*h/g)
= sqrt(2*1/9.8)
= 0.4517 s
let V is the speed of the block after the bullet emerged from it.
use, x = V*t
==> V = x/t
= 1.3/0.4517
= 2.88 m/s <<<<<<<<<<-------------------------Answer
b) now appy conservation of momentum
m*vi = M*V + m*vf
==> vf = (m*vi - M*V)/m
= vi - M*V/m
= 230 - 0.5*2.88/0.012
= 110 m/s <<<<<<<<<<-------------------------Answer
c) Workdone by the block on bullet = change in kinetic energy
Favg*d*cos(180) = (1/2)*m*(vf^2 - vi^2)
Favg = (1/2)*m*(vi^2 - vf^2)/d
= (1/2)*0.012*(230^2 - 110^2)/0.2
= 1224 N <<<<<<<<<<<-------------------------Answer
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