A 0.0147-kg bullet is fired straight up at a falling wooden block that has a mas

Question

A 0.0147-kg bullet is fired straight up at a falling wooden block that has a mass of 2.94 kg. The bullet has a speed of 760 m/s when it strikes the block. The block originally was dropped from rest from the top of a building and had been falling for a time t when the collision with the bullet occurs. As a result of the collision, the block (with the bullet in it) reverses direction, rises, and comes to a momentary halt at the top of the building. Find the time t.

Please give answer to 4 significant figures.

Explanation / Answer

m1v1 + m2v2 = (m1+m2) vf

but you also know that since the block started from rest and accelerated to v2, then returned to it's original position, the v2 must = vf but in the opposite direction. right? { because an object falling from rest has Vf^2 = Vo^2 + 2ax. and an object rising and slowing down also goes by Vf^2 = Vo^2 + 2ax. the falling case, Vo = 0 and a is positive so ...Vf^2 = 0 + 2ax . ie Vf^2 = 2ax. for an object rising to a point and slowing to a stop, Vf = 0. so 0 = Vo^2 + 2ax. but in that case "a" rising has an opposite sign to "a" falling. since it is slowing down vs speeding up. so ... Vo^2 -2ax = 0. Vo^2 = 2ax. Since Vf^2 = 2ax = Vo^2, therefore Vf^2 falling = Vo^2 rising...... square rooting both sides Vf = +/- Vo. but we know they are in opposite directions so Vf = -Vo}

now let's choose a direction for (+). since acceleration is towards the earth let's say down is (+)

in which case

(0.0147 kg) x (-760 m/s) + (2.94kg) x v2 = (0.0147 kg + 2.94kg) x (- v2)

5.8947 kg v2 = 11.172 kgm/s

v2 = 1.98526 m/s

that is the velocity of the block when it is struck by the bullet...


next.... from equations of motion...

Vf = Vo + at

t = (Vf - Vo)/a = (1.89526 m/s - 0) / (9.8 m/s^2) = 0.1934 s

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