Question

A 0.00618-kg bullet is fired straight up at a falling wooden block that has a mass of 1.52 kg. The bullet has a speed of 616 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.

Answer 1

here,

m = mass of bullet = 0.00618 kg

v = velocity of bullet before impact = 616 m/s

M = mass of block = 1.52 kg

V = velocity of block (unknown)

U = velocity of bullet/block after impact (unknown)

Momentum of bullet (upwards) = m*v

Momentum of block (downwards) = M*V

Momentum before impact = m*v - M*V .... Using up as positive

Momentum of bullet/block after impact = (m + M)*U

Momentum must be conserved:

m*v - M*V = (m + M)*U

Velocity of block before impact:

V = g*t .... since the block had an initial velocity of 0 and is simply falling

Upward velocity of bullet/block will be the same as the downward velocity of the block at impact  

trajectories under the acceleration of gravity only are completely symmetrical. Meaning that for the bullet/block to rise to the same height as the block started from it must have the same velocity as the block acquired during it's fall.

U = V = g*t

m*v - M*V = m*v - M*U = (m + M)*U

m*v = (m + 2M)*U = (m + 2M)g*t

[m*v/(m + 2M)]/g = t

t = (0.00618 * 616 /( 0.00618 + 2 * 1.52)) / 9.81 s

t = 0.13 s

the time taken t is 0.13 s