A 0.00676-kg bullet is fired straight up at a falling wooden block that has a mass of 1.04 kg. The bullet has a speed of 863 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.
2007-06-24 05:28:22 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Physics
The distance of fall = distance of going up.
Therefore the velocity of block before impact = velocity after impact. = U
The velocity of the bullet after impact is also U.
Change in momentum of block = 2 W U-----------W is the mass of wooden block.
Change in momentum of bullet = B (V- U) -----------B is the mass of bullet.
(B + 2W) U = BV
U = V/ (1+ 2W/B)
U = gt.
t = V/ g*(1+ 2W/B)
Substituting the values,
W/B = 1.04 /0.00676= 153.8 and V = 863m/s
t = 0.285 s
2007-06-24 15:50:24 · answer #1 · answered by Pearlsawme 7 · 0⤊ 0⤋
v = at
t = v/a = 883/9.8 = 90.1 seconds
2007-06-24 12:36:44 · answer #2 · answered by jsardi56 7 · 0⤊ 0⤋
jsardi56 is right
2007-06-24 12:58:27 · answer #3 · answered by santosh R 2 · 0⤊ 0⤋