A 13 gram bullet is stopped in a 6.38 kg block of wood. The speed of the bullet-plus-wood combination immediately after the collision is 0.593 m/s.
What was the original speed of the bullet in m/s?
2007-12-11 16:39:48 · 4 answers · asked by margarita_9902@sbcglobal.net 2 in Science & Mathematics ➔ Physics
Momentum is always conserved in collisions.
Initial momentum: ( .0013 kg ) ( v )
Final momentum: ( .0013 g + 6.38 kg ) ( .593 m / s )
Set these two expressions equal and solve for the initial speed of the bullet.
2007-12-11 16:43:35 · answer #1 · answered by jgoulden 7 · 0⤊ 0⤋
Momentum
m1u1+m2u2=m1v1+m2v2
(13/1000)u1+(6.38)(0)=((13/1000)+(6.38) )(0.593)
u1=291.6191538 m/s
2007-12-11 16:47:12 · answer #2 · answered by Murtaza 6 · 0⤊ 0⤋
conservation of momentum
(mv)1 = (mv)2
state one - only bulolet moving
state two - bullet and wood
m1 = bullet = 13g = 0.013kg
m2 = wood = 6.38kg
v1 = ???
v2 = initial velocity of wood = 0
v3 = velocity of wood and bullet = 0.593
m1v1 + m2v2 = (m1 + m2)v3
v1 = (m1 + m2)v3/m1
plug n play
2007-12-11 16:45:56 · answer #3 · answered by Kevin 5 · 0⤊ 0⤋
Get a bigger gun!
2016-04-08 22:10:29 · answer #4 · answered by Anonymous · 0⤊ 0⤋