posted by Cass .
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
= 1.82 seconds on table
+ .639 in air
= 2.46 seconds