Horizontal Velocity given only vertical displacement and the mass of the two objects.

  • #1

Homework Statement


As part of a forensic experiment, a 50g bullet is fired horizontally into a 2.0kg wooden pendulum. The pendulum with the bullet embedded in it rises 15cm vertically from its initial position and stops.

a) Calculate the velocity of the block and bullet just after the collision.
b) What is the velocity of the bullet just before impact?

Homework Equations


p = m*v
W = F*d = ET
Change in Eg = m*g*change in h
Ek = 1/2*m*v2
Change in p = F*change in t

The Attempt at a Solution


mbullet = 0.05kg
mpendulum = 2kg

In this case i think change in Eg = W in the up direction

m*g*change in h = Wup
(2kg + 0.05kg)(9.8N/kg)(0.15m) = Wup
3.0135N*m = Wup

Wup = Forceup * displacementup
(3.0135N*m) / (0.15m) = Forceup
20.09N = Forceup

Change in p = F*change in t
Change in t = Vav/change in d

Therefore

pup = 20.09N * ((V2 - V1) * 0.5)/0.15m

and here is where i'm lost.. This is part of an 8 question assignment sheet, i've finished the previous 7 but this one really has me stumped.
Would appreciate any help given asap since this is due monday.
 

Answers and Replies

  • #2
kuruman
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Forceup is meaningless. You need to conserve mechanical energy in part (a). The potential energy at maximum height is equal to the kinetic energy at the bottom. Say that with an equation and solve for the bullet plus wood system.
 
  • #3


Tad confused here, as we don't have the height of the pendulum how am i supposed to solve for velocity at the bottom..

Ek = 1/2mv^2
Eg = mgh

We dont have h...

Edit: Do you mean that the change in gravitational potential energy at the new position of 15cm above is equivalent to the kinetic energy?
So like

Ek = Eg
1/2mv^2 = mgh
1/2(2.05kg)v^2 = (2.05kg)(9.8N/Kg)(0.15)
v = 1.714m/s
?
 
Last edited:
  • #4
kuruman
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Yes, change in gravitational potential energy. There is no other form of potential energy.
 
  • #5


Alright, thanks so much for your help!

Would part b be similar aswell?
 
  • #6
kuruman
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Alright, thanks so much for your help!

Would part b be similar aswell?
No. Treat the problem as an inelastic collision and conserve momentum before and after the collision.
 
  • #7


Worked perfectly, again thank you so much for your help. How do i set this question to solved?
 
  • #8
kuruman
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Worked perfectly, again thank you so much for your help. How do i set this question to solved?
I am not sure what you mean.
 

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