Calculating Velocity of Bullet After Impact

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  • #1
thefifthlord
25
0

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.
 
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  • #2


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 don't 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


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


thefifthlord said:
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


thefifthlord said:
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.
 

1. How is the velocity of a bullet calculated after impact?

The velocity of a bullet after impact can be calculated using the principles of conservation of momentum and energy. The formula for calculating velocity is: v = √(2gh), where v is the velocity, g is the acceleration due to gravity, and h is the height from which the bullet was dropped.

2. What factors affect the velocity of a bullet after impact?

The velocity of a bullet after impact can be affected by various factors such as the initial velocity of the bullet, the mass of the bullet, the distance traveled, the angle of impact, and the surface on which the bullet impacts.

3. Can the velocity of a bullet after impact be measured?

Yes, the velocity of a bullet after impact can be measured using high-speed cameras, radar technology, or by analyzing the damage caused by the bullet on the target.

4. How does the velocity of a bullet after impact affect its penetration?

The velocity of a bullet after impact plays a crucial role in determining its penetration. Generally, a higher velocity results in deeper penetration as it has more energy to overcome resistance and break through the target.

5. Is there a standard method for calculating the velocity of a bullet after impact?

There is no standard method for calculating the velocity of a bullet after impact as it depends on various factors such as the type of bullet, the target material, and the conditions of the impact. However, the principles of conservation of momentum and energy can be used to derive a formula for calculating the velocity.

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