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

a) block rise to maximum height of 18cm Or 0.18 m.

using energy conservation to find speed just after impact.

m gh = m v^2 /2

v = sqrt(2gh) = sqrt( 2 x 9.8 x 0.18) = 1.88 m/s

Using momentum conservatin for block-bullet system,

0.010 x vi + 2.25 x 0 = 0.010 x 0.2vi + 2.25 x 1.88

vi = 528.75 m/s

b) Impulse = change in momentum

= m (vf - vi) = m (0.2vi - vi) = - 0.8 m vi

= - 0.010 x 0.8 x 528.75 = - 4.23 kg m/s

(negative sign indicates that impulse is vertically downwards)

c) imnpulse = F x delta t

- 4.23 = F x 2 x 10^-3

F = 2115 N

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