Question

A 0.00728-kg bullet is fired straight up at a falling wooden block that has a mass of 2.75 kg. The bullet has a speed of 641 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

Answer:

$0.087$ sec

Explanation:

Consider the down direction as negative and upward direction as positive.

$m$ = mass of the bullet  = 0.00728 kg

$M$ = mass of the wooden block = 2.75 kg

$v$ = velocity of the bullet before striking the block = 641 ms⁻¹

$V$ = velocity of the wooden block before collision

$V_{c}$ = velocity of the wooden block + bullet after collision = V

Using conservation of momentum

$m v - MV = (m + M) V$

$(0.00728) (641) - 2.75 V = (0.00728 + 2.75) V$

$(0.00728) (641) - 2.75 V = (0.00728 + 2.75) V$

$V = 0.85$ ms⁻¹

Consider the motion of the wooden block in free fall after being dropped:

$V_{o}$ = Initial velocity of the wooden block = 0 ms⁻¹

$a$ = acceleration of the wooden block = - 9.8 ms⁻²

$t$ = time of travel

$V_{f}$ = Final velocity of the wooden block = - 0.85 ms⁻¹

One the basis of above set of data , we can use the kinematics equation

$V_{f} = V_{o} + a t$

$- 0.85 = 0 + (-9.8) t$

$t = 0.087$ sec