Projectiles: Bullet Fired into Block

  • Thread starter kah22
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  • #1
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Homework Statement


A(n) 9 g bullet is fired into a 303 g block that is initially at rest at the edge of a frictionless table of height 1.9 m. The bullet remains in the block, and after impact the block lands
1.5 m from the bottom of the table. The acceleration of gravity is 9.8 m/s/s.
Find the initial velocity (Vi) of the bullet in m/s.

Homework Equations


ΔX = Vi(t) + (1/2)(a)(t^2)

m(Vf-Vi) = -m(Vf-Vi)

The Attempt at a Solution


I separated the equation into vertical and horizontal parts.
Horizontal: ΔX=1.5, a = 0 m/s/s
Vertical: ΔY= -1.9, a= -9.8 m/s/s, Vi = 0 m/s (because it is launched horizontally).
Using the kinematic equation ΔX = Vi(t) + (1/2)(a)(t^2) to solve for the vertical time:
-1.9 = 0(t) + 1/2 (-9.8)t^2
-1.9 = -4.9t^2
t^2 = 0.38775
t = 0.623 seconds

Since the vertical time is equal to the horizontal time, I plugged this time into the same kinematic equation to solve for the horizontal Vi:
1.5 = Vi (0.623) + 1/2(0)(0.623)
1.5 = Vi (o.623)
Vi = 2.41 m/s

This Vi is equal to the velocity at which the block and bullet are launched from the table.
I know the mass of the bullet, the mass of the block and bullet combined (312 g), the Vf (which is the Vi from the equation above), and the Vi of the block and bullet combined (0, since it is at rest). I can use the collision equation to solve for the Vi of the bullet.

m(Vf-Vi) = -m(Vf-Vi)
9(2.41-Vi) = -312(2.41-0)
21.69 -9Vi = -751.92
-9Vi=-773.61
Vi=85.96 m/s

The grading system says this answer isn't correct. I checked my math but I think I am missing something. Can anyone tell me what I am doing wrong?
 

Answers and Replies

  • #2
1,506
18
I agree with your t = 0.623s and your Vi = 2.41m/s
I think your mistake is in the last part. In the collision of the bullet with the block momentum is conserved so if you write:
Momentum of bullet = momentum of (block+bullet) see what you get
I got the velocity of the bullet to be 83.2m/s.....what is the answer, do you know?
 
  • #3
ehild
Homework Helper
15,543
1,914
m(Vf-Vi) = -m(Vf-Vi)
9(2.41-Vi) = -312(2.41-0)
21.69 -9Vi = -751.92
-9Vi=-773.61
Vi=85.96 m/s

According to the red equations m(Vf-Vi) =0, as zero alone is equal to the negative of itself.

Use conservation of momentum in the correct form. You have two objects with different masses and different initial and final velocities. Use proper notations.

ehild
 

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