table of height 1.00 m. The bullet remains in the block, and after impact the bl
ID: 1584324 • Letter: T
Question
table of height 1.00 m. The bullet remains in the block, and after impact the block lands 2.00 m from the bottom of the table. Determine the initial speed of the bullet . An 7.00 g bullet is fired into a 2.80 kg block initially at rest at the edge of a frictionless 1.00 m 2.00 m 2. A 1.20-kg wooden block rests on a table over a large hole as in the figure below. A 5.40- g bullet with an initial velocity v is fired upward into the bottom of the block and in the block after the collision. remains The block and bullet rise to a maximum height of 21.0 cm. () Find the initia velocity of the bullet using ideas you have leaned in this chapter 8.Explanation / Answer
mass of the bullet m1 = 7 g
= 7*10^ -3 kg
mass of the block m 2 = 2.8 kg
mass of ( bullet + block ) = 2.807 kg
initial speed of the bullet = u 1= ?
initial velocity of the block u2 = 0 m/s
final speed of bullet + block = v ?
applying conservatin of momentum
m1 u1 +m2 u2 = ( m1 + m2 ) v
0.007 u = 2.801 v
initial velocity of the bullet u = 2.807 / 0.007 v
= 401 v .........(1)
final velocity of the system v = distravelled by the block / time of fall
h = 1/2 gt^2
t = ( 2h / g )^( 1/2)
= ( 2* 1m / 9.8 ) ^ ( 1/2)
= 0.451 s
there fore final velocity v = 2 / 0.451 = 4.427 m/s
The initial velocity of the bullet
u = 401 *4.427 = 1775.227m/s
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