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ID: 1610089 • Letter: T
Question
To view an interactive solution to a problem that is similar to this one, select Interactive solution 7.24. A 0.0132-kg bullet is fined straight up at a falling wooden block that has a mass of 2.53 kg. The bullet has a speed of 520 m/s when it strikes the block. The block originally was dropped from rest 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 half at the top of the building. Find the time t. Number 0.026 Units sExplanation / Answer
for the block
assuming down direction as negative and up direction as positive
a = acceleration of the block in down direction = - 9.8
Vo = initial velocity of the block when it dropped from top of building = 0 m/s
t = time of travel before collision
Vf = velocity of block just before collision
d = distance travelled before collision
using the equation
d = Vo t + (0.5) a t2
d = 0 (t) + (0.5) (-9.8) t2
d = - 4.9 t2
using the equation
Vf = Vo + at
Vf = - 9.8 t
M = mass of block = 2.53 kg
for bullet :
v = velocity of bullet just before collision = 520 m/s
m = mass of bullet = 0.0132 kg
V = velocity of bullet-block combination after collision
using conservation of momentum
m v + M Vf = (m + M) V
(0.0132) (520) + (2.53) (- 9.8 t) = (0.0132 + 2.53) V
6.864 - 24.8 t = 2.543 V
V = (6.864 - 24.8 t )/2.543
consider the upward motion of bullet-block combination
d = distance travelled = 4.9 t2
Vf' = velocity at the top = 0 m/s
V = initial velocity after collision = (6.864 - 24.8 t )/2.543
a = acceleration = - 9.8
using the equatiion
V'f2 = V2 + 2 a d
02 = ( (6.864 - 24.8 t )/2.543 )2 + 2 (-9.8) (4.9 t2)
t = 0.14 sec
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