Ballistic Pendulum


by rbrow039
Tags: ballistic, pendulum
rbrow039
rbrow039 is offline
#1
Oct22-08, 04:06 PM
P: 1
1. The problem statement, all variables and given/known data
A 7.0-g bullet is fired into a 1.5-kg ballistic pendulum. The bullet emerges from the block with a speed of 200 m/s, and the block rises to a maximum height of 12 cm. Find the initial speed of the bullet.

Now I think this is an imperfect inelastic collision because the bullet does not lodge itself in the pendulum. So I assumed since it was inelastic I could ignore conservation of kinetic Energy.
2. Relevant equations
Conservation of Momentum
Conservation of Energy


3. The attempt at a solution
m1v1i +m2v2i=(m1 + m2)vf
(.007kg x v1i)+0=(1.507kg)vf

(PE + KE)collision=(PE + KE)top
0 + (.5 x 1.507 x vf^2)= mgh + .5mv^2
0 + (.5 x 1.507kg x vf^2) = (1.5 kg x 9.81m/s^2 x .12m) + (.5 x .007kg x 200^2)
vf^2=(1.7658J + 140J)/(.7535kg)
vf=13.716m/s

(.007kg x v1i)=1.507kg x 13.716m/s
v1i = 2952 m/s
Now I know this is not the answer because the answer is given as 530m/s but I can't for the life of me figure out what went wrong with the energy calculation. (I'm assuming that's where the big boo boo happened)
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alphysicist
alphysicist is offline
#2
Oct22-08, 11:02 PM
HW Helper
P: 2,250
Hi rbrow039,

Quote Quote by rbrow039 View Post
1. The problem statement, all variables and given/known data
A 7.0-g bullet is fired into a 1.5-kg ballistic pendulum. The bullet emerges from the block with a speed of 200 m/s, and the block rises to a maximum height of 12 cm. Find the initial speed of the bullet.

Now I think this is an imperfect inelastic collision because the bullet does not lodge itself in the pendulum. So I assumed since it was inelastic I could ignore conservation of kinetic Energy.
2. Relevant equations
Conservation of Momentum
Conservation of Energy


3. The attempt at a solution
m1v1i +m2v2i=(m1 + m2)vf
I don't believe this formula is correct; this formula applies to a perfectly inelastic collision (where both objects stick together and move with the same speed vf after the collision). As you remarked, these objects do not stick together after the collision. What would the conservation of momentum equation be for this case? (And notice they give you the speed of the bullet right after the collision.)


(.007kg x v1i)+0=(1.507kg)vf

(PE + KE)collision=(PE + KE)top
0 + (.5 x 1.507 x vf^2)= mgh + .5mv^2
0 + (.5 x 1.507kg x vf^2) = (1.5 kg x 9.81m/s^2 x .12m) + (.5 x .007kg x 200^2)
With the change to the conservation of momentum equation for the collision, do you see how to correct your energy equation?


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