A 2.25 kg wooden block of mass M rests on a table over a large hole. A 10 gram b

Question

A 2.25 kg wooden block of mass M rests on a table over a large hole. A 10 gram bullet is fired with an initial speed vi is fired upward into the bottom of the block. The bullet passes through the block and emerges with a speed 0.2vi. As a result of impact, the block rises into the air to a maximum height of 18 cm. Assume the bullet makes a perfect hole through the block (i.e., the block does not lose mass) and that “upward” is the +y- direction. a) Find the initial speed of the bullet. b) Find the impulse vector on the bullet. c) The bullet took t = 2 ms to pass through the block, find the average force vector on the block.

Explanation / Answer

velocity of block

1/2*Mv^2 =Mgh

v = 1.878 m/s

linear momentum remain conserve

mbvb = mb*vb + Mb*Vb

10*10^-3 * vi = 10*10^-3 * 0.2 * vi + 2.25 * 1.878

vi = 528.27 m/s

part a )

J = m(vf - vi)

J = 10*10^-3*528.27(1-0.2)

J = 4.226 N-s

part b )

F.dt = m(vf-vi)

vi = 0

vf = 1.878

F = 2.25 * 1.878 /2*10^-3

F = 2112.75 N

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