Determine the horizontal component

[Larrytsai]

An 18-g rifle bullet traveling 200 m/s buries itself in a 3.6-kg pendulum hanging on a 2.8-m-long string, which makes the pendulum swing upward in an arc. Determine the horizontal component of the pendulum's displacement.

What I tried was find the momentum of the bullet, then find the velocity of the pendulum when it hits.

Pbefore= (0.018kg)(200)

Pbefore=Pafter
3.6 Kg*m/s = (3.6kg+0.018kg)v`
vf=0.995m/s
vi= 0m/s
d(y)=2.8
d(x)= ?

-y=1/2a(t^2)+vi(t)
-2.8= 1/2(-9.8)(t^2)
t^2=0.571
root "t" = root "0.571"
t= 0.576

x=vt
x=0.995*0.576
x= 0.573m the answer looks wrong... i think.

 

[cryptoguy]

-y=1/2a(t^2)+vi(t)
-2.8= 1/2(-9.8)(t^2)
t^2=0.571
root "t" = root "0.571"
t= 0.576

I believe there's a problem with this part. You're saying that a = -9.8 m/s^2 which means that the object (pendulum bob + bullet) is in free-fall which it's not (the tension force in the string provides acceleration). Try using conservation of energy to find the vertical displacement of the object and then using trib (or Pythagorean theorem) to find horizontal displacement.

 

[elessariitkgp]

Let the initial velocity of bullet be $$v_{0}$$. If the mass of the bullet be denoted by $$m$$, then as per the given data the mass of the pendulum bob is $$\frac{m}{5}$$. Applying linear momentum conservation along the initial direction of motion of the bullet,
$$mv_{0}=\frac{6m}{5} v$$
where $$v$$ is the final velocity of the bullet+bob assembly.
$$v=\frac{5v_{0}}{6}$$
Assuming the initial position of bob to be the zero PE level, we can apply mechanical energy conservation for the system (bullet+bob+Earth)
$$\frac{25}{72}mv^{2}_{0}=mgL(1-cos\theta)$$
where $$\theta$$ is the angular deflection of the string and $$L$$ is the length of the string.
Now, the horizontal displacement of the bob is $$L cos\theta{/tex] and that can be directly found from the equation$$