Conservation of energy and momentum for a bullet fired into a block

[defetey]

Homework Statement

An 8 g bullet is fired into a 2.5 kg block initially at rest at the edge of a frictionless table of height 1 m. The bullet remains in the block and lands 2 m from the bottom of the table after impact. Determine the initial speed
of the bullet.
Answer: 1392 m/s

Homework Equations

\vec{p}_{T}={\vec{p}_{T}}'

E_{T}=E_{T}{}'

E_{K}=\frac{1}{2}mv^{2}

E_{G}=mgh

\Delta \vec{d}=\left ( \frac{\vec v_{1}+\vec v_{2}}{2} \right )\Delta t

\Delta \vec{d}=\vec{v}_{1}\Delta t+\frac{1}{2}\vec{a}\Delta t^{2}

The Attempt at a Solution

Since I know the vertical displacement will be 1 m, the vertical acceleration will be 9.8m/s^{2} and that the initial vertical velocity will be also 0, I can use

\Delta \vec{d}=\vec{v}_{1}\Delta t+\frac{1}{2}\vec{a}\Delta t^{2}

to find the time it takes to hit the ground.

1=(0.5)(9.8)\Delta t^{2}

\Delta t=0.45s

Now for the horizontal, since it is elevated from the ground, it will initially have gravitational potential energy and since it is also going to be hit by the bullet, it will have kinetic energy.

Then, when it hits the ground it will only have kinetic energy. So:

E_{G}+E_{K}=E_{K}{}'

(2.508)(9.8)(1)+(0.5)(2.508)\vec v_{1}^{2}=(0.5)(2.508)\vec v_{2}^{2}

Since there are two variables, we can use

\Delta \bar{d}=\left ( \frac{\vec v_{1}+\vec v_{2}}{2} \right )\Delta t

to get \vec v_{2}.

2=\left ( \frac{\vec v_{1}+\vec v_{2}}{2} \right )(0.45)

\vec v_{2}=8.89-\vec v_{1}

Subbing it in:

(2.508)(9.8)(1)+(0.5)(2.508)\vec v_{1}^{2}=(0.5)(2.508)(8.89-\vec v_{1})^{2}

\vec v_{1}=3.34m/s


Using conservation of momentum:

\vec{p}_{T}={\vec{p}_{T}}'

(0.008)\vec v_{bullet}+0=(2.508)(3.34)

\vec v_{bullet}=1047m/s

Which is different from the answer. I noticed that if I take away the E_{G}, it works. But since it would be elevated, wouldn't it have to have gravitational potential energy, don't I have to include it because energy is not a vector?

EDIT: never mind, think I got it. I forgot to factor in the kinetic energy it would have from going vertically down since I was concentrating so much on the horizontal and thought it would have a vertical \vec v of 0 since the ground would stop it... so it would actually be

(2.508)(9.8)(1)+(0.5)(2.508)\vec v_{1}^{2}=(0.5)(2.508)(8.89-\vec v_{1})^{2}+(0.5)\left [(9.8)(0.45) \right ]^{2}(2.508)

\vec v_{1}=4.44m/s

\vec{p}_{T}={\vec{p}_{T}}'

(0.008)\vec v_{bullet}+0=(2.508)(4.44)

\vec v_{bullet}=1392m/s

 

[kuruman]

I am answering this old thread to have it removed from the unanswered list and to provide a more streamlined solution for future reference.

For the collision part of the motion, use momentum conservation to find the speed of the block + bullet system when it moves off the table. Let $v_0$ be the unknown initial speed of the bullet and $u$ the final speed of the combined system. Let $h$ be the height of the table and $L$ the horizontal landing distance.

Then $$mv_0=(m+M)u\implies u=\frac{m}{m+M}v_0.$$ For the projectile part of the motion, start with the modified projectile trajectory equation¹ $$\frac{\Delta x}{\tan\!\theta-\tan\!\varphi}=\frac{2v_{0x}^2}{g}.$$ Here, $~\Delta x=L~;~~\tan\!\theta=0~;~~\tan\!\varphi=-\dfrac{h}{L}~;~~v_{0x}=u$ so that $$\frac{L^2}{h}=\frac{2u^2}{g}=\frac{2}{g}\left(\frac{m}{m+M}v_0\right)^2\implies v_0=\sqrt{\frac{g}{2h}}\frac{(m+M)}{m}L=1.4\times 10^3~\rm{m/s}.$$

¹ Reference: Equation (4) in https://www.physicsforums.com/insights/how-to-solve-projectile-motion-problems-in-one-or-two-lines/