Ballistic Pendulum Bullet

[rbrow039]

Homework Statement

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.

Homework Equations

Conservation of Momentum
Conservation of Energy

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)

 

[alphysicist]

Hi rbrow039,

Homework Statement

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.

Homework Equations

Conservation of Momentum
Conservation of Energy

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?

 

[baba89]

I would like to point out that the approach taken in this solution is incorrect. The conservation of energy should always be used in any colliding system, regardless of whether it is an elastic or inelastic collision. In this case, since the bullet does not lodge itself in the pendulum, it is an inelastic collision, and the conservation of energy should be used to solve for the initial speed of the bullet. The assumption made in the solution that the conservation of kinetic energy can be ignored is not valid.

Furthermore, the equation used for the conservation of energy in this solution is incorrect. The correct equation for the conservation of energy in this case would be:

(KE)initial + (PE)initial = (KE)final + (PE)final

where (KE)initial and (PE)initial represent the kinetic and potential energy of the bullet before the collision, and (KE)final and (PE)final represent the kinetic and potential energy of the pendulum after the collision.

Using this equation, the correct solution would be:

(0.5 x 0.007kg x v1i^2) + (0) = (0.5 x 1.507kg x 13.716m/s)^2 + (1.5kg x 9.81m/s^2 x 0.12m)

Solving for v1i, we get v1i = 530 m/s, which is the correct answer.

In conclusion, as a scientist, it is important to use the correct equations and assumptions when solving for a problem. In this case, the conservation of energy should always be used in any colliding system, regardless of whether it is an elastic or inelastic collision.