An 8.0 g bullet is fired horizontally into a 9 kg block of wood and sticks in it.The block,which is free to move,has a velocity of 40cm/s after impact.Find the initial velocity of a bullet.
2006-10-09 23:35:31 · 15 answers · asked by gen 2 in Science & Mathematics ➔ Physics
450 metres/second
2006-10-09 23:40:29 · answer #1 · answered by jquittenton2 3 · 0⤊ 0⤋
With the Principle of Conservation of Momentum, the equation will be, m1v1 + m2v2 = MV. Here we assume
m1 = mass of bullet = 8g
v1 = inivitial velocity of bullet
m2 = mass of wooden block = 9 kg = 9000gms
v2 = initial velocity of wooden block = 0 cm/s
M = Mass of wooden block + bullet = 9008 gms
V = Final velocity of the combined mass = 40cms/s
Hence 8*v1 + 0 = 9008*40
Therefore v1 = (9008*40)/8 = 45,040cm/s i.e.450.4 Meters/sec
So initial velocity of horizontally fired bullet = 450.4 m/s
2006-10-10 06:48:51 · answer #2 · answered by Manindomb 2 · 0⤊ 0⤋
can not be answered, some energy is used to make a tunnel in the wood for the bullet. The question should be that after collision the bullet stops and the block moves at 40cm/s
2006-10-10 07:54:38 · answer #3 · answered by gjmb1960 7 · 0⤊ 1⤋
Holden is close but is an order of magnitude out.
The final mass is 9.008 kg, the final velocity 0.4 m/s which gives:
9.008 x 0.4 = 3.6032 kg m /s of momentum
Assuming a completely elastic process, i.e. no energy lost in the collision, all of this momentum will be conserved. Thus to get the initial velocity you divide by the initial mass (0.008 kg)
3.6032 x 0.008 = 450.4 m/s
2006-10-10 06:49:24 · answer #4 · answered by Vanguard 3 · 0⤊ 0⤋
8.0g=0.008kg,40cm/s=0.4m/s,
let initial velocity of the bullet be v
let the mass of the bullet be m
let the mass of the block be M
let the velocity when they join be V
law of conservation of linear momentum sates that
momentum before impact=momentum after impact
mv=[M+m]V
0.008*v=[9+0.008]*0.4
0.008*v=9.008*0.4
v=9.008*0.4/0.008
v=450.04m/s
the initial velocity of the bullet is 450.04m/s or 4.5cm/s.
2006-10-10 06:57:12 · answer #5 · answered by flyguy 2 · 0⤊ 0⤋
This will look real bad if I get this wrong :)
Momentum of block + bullet after imapct = (9 + 0.008) x 0.4 = 3.6032 kgm/s
Momentum of block + bullet = momentum of bullet before impact
Momentum of bullet = .008 x u = 3.6032
3.6032 / .008 = 450.4 m/s
2006-10-10 06:45:45 · answer #6 · answered by tekn33k 3 · 0⤊ 0⤋
8*v+9000*0 = 9008*40 by conservation of momentum so calculate v.
2006-10-10 08:26:28 · answer #7 · answered by Anonymous · 0⤊ 0⤋
The momentum of the Block/bullet = 9008 gram * 40 cm/s
9.008 * .0040 m/s = .36032 kgm/s = .008 * V
V = 450.4 m/s
2006-10-10 06:40:47 · answer #8 · answered by Holden 5 · 1⤊ 0⤋
45,000 cm/s or about 1500 feet per second, which is reasonable for a handgun.
Holden is wrong to add the mass of the bullet, because it isn't being accelerated, it is being decelerated
on second thought I think he may be right.
2006-10-10 06:38:43 · answer #9 · answered by Anonymous · 0⤊ 0⤋
0 because the final velocity is stopped
2006-10-10 06:38:24 · answer #10 · answered by Anonymous · 0⤊ 1⤋