Post a New Question

physics

posted by 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?

  • physics - ,

    .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

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question