physics
posted by Cass on .
A 40 gram bullet is fired horizontally from a gun with a momentum of 2.8 (kg*m/s) and embeds itself into a 300 gram block of wood initially at rest on a wooden horizontal surface. After this collision the wood block slides 15 meters before falling off a table with a height of 2m. (neglect air resistance)
Draw a diagram of this scenario and label the parts with the different physics concepts in play.
Determine the horizontal distance travelled by the block after falling off the table.
What is the final velocity of the block.
At what angle does the block hit the ground? (relative to the horizontal)
How much time is elapsed between the bullet hitting the block and the block striking the ground?

.04 kg bullet + 0.3 kg block = .34 kg final
initial momentum = momentum after collision
2.8 = .34 v
v = 8.24 m/s sliding along table
That is the horizontal velocity component until it hits the floor
How long to fall ?
h = (1/2) g t^2
2 = 4.9 t^2
t = .639 seconds falling
how far horizontal in .639 seconds
x = u t = 8.24 * .639 = 5.26 meters
vertical speed component
v = g t = 9.81 (.639) = 6.27 m/s
so velociy is 8.24 i  6.27 j
speed =sqrt(8.24^2 + 6.27*2)
= 10.35 meters/second
tan angle below horizontal = 6.27/8.24
so angle = 37.3 degrees below horizontal
how long on table
15 meters/8.24
= 1.82 seconds on table
+ .639 in air
= 2.46 seconds