Question

A 15 g bullet is fired at 610 m/s into a 4.0 kg block that sits at the edge of a 75-cm-

high table. The bullet embeds itself in the block and carries it off the table. How far from the

point directly below the table's edge does the block land?

Answer 1

First we need to determine the initial velocity of the bullet-block combination using the conservation of momentum equation. Momentum is conserved in this collision because no external forces are acting in the horizontal direction. p represents velocity, m represents mass and v represents velocity.

Σp0=Σpfmbullet Vbullet=Σmbullet+Mblockvf

(0.015)(610)=(0.015+4)vf

vf=2.278ms

Now we need to determine the time required to fall from the table to the floor using a kinematics equation. d represents the vertical displacement, v0 is the initial velocity in the vertical direction, t represents time and a represents the acceleration due to gravity.

d = v0t+1/2at^2

0.75 = 0+4.9t^2

t=0.3912s

The horizontal displacement of the block is simply the horizontal velocity we calculated earlier times the time required for the block to fall. This is because there is no acceleration in the horizontal direction.

d=vt

d=(2.78)(0.3912)

d= 1.0875m