Mechanics problem (Bullet penetration)

[tanzl]

Homework Statement

A 7.00g bullet, when ired from a gun into a 1.00kg block of wood held in a vise, penetrates the block to a depth of 8.00cm. This block of wood is placed on a frictionless horizontal surface, and a second 7.00g bullet is fired from the gun into the block. To what depth will the bullet penetrate the block in this case?

Homework Equations

The Attempt at a Solution

In the first case, by taking the final velocity as 0 and displacement as 8.00cm.
I can get the initial velocity of bullet in terms of acceleration or time by using kinematics equations.
or I can find the force of the bullet exerts on the block by equating EK = F\Deltar
But they are not seems to be useful in case 2.

In case 2, I am not sure how the mechanics will occur. But, I think the bullet will travel with the same initial velocity, penetrate a shorter distance into the block and bring the block to travel with it with some velocity. I do not know how to put this by using physical principals. I guess I need to use conservation of energy again. By equating
EK of bullet = F\Deltar + EK of (bullet + block). But I do not think I have enough information to do that. Please help me.

 

[Doc Al]

You're on the right track. Hint: What else is conserved during the collision of bullet and block in case 2?

 

[tanzl]

I think momentum is conserved which gives
\frac{1}{2}mv₁ = (M+m)v₂ ...(1)
where m=mass of bullet
M=mass of block
v₁=initial velocity of bullet
v₂=velocity of bullet and block

From the 1st case,
I have \frac{1}{2}mv₁² = F\Deltar₁ ... (2)

From the 2nd case,
\frac{1}{2}mv₁² = F\Deltar₂ + \frac{1}{2}mv₂²... (3)

By equating (3) and (2)
I have F(\Deltar₁-\Deltar₂) = \frac{1}{2}(M+m)v₂²

But I still have one extra unknown (r₂ , F , v₁) for 2 equations.
I have one more question here, if I had assumed that the collision is inelastic, will (3) still correct since the kinetic energy is not conserved anymore.

 

[Doc Al]

I think momentum is conserved which gives
\frac{1}{2}mv₁ = (M+m)v₂ ...(1)
where m=mass of bullet
M=mass of block
v₁=initial velocity of bullet
v₂=velocity of bullet and block

Good. But get rid of that 1/2.

From the 1st case,
I have \frac{1}{2}mv₁² = F\Deltar₁ ... (2)

Good.

From the 2nd case,
\frac{1}{2}mv₁² = F\Deltar₂ + \frac{1}{2}mv₂²... (3)

Good, but that second "m" should be "M + m".

By equating (3) and (2)
I have F(\Deltar₁-\Deltar₂) = \frac{1}{2}(M+m)v₂²

But I still have one extra unknown (r₂ , F , v₁) for 2 equations.

You have 3 equations. In any case, noodle around with the equations, expressing everything in terms of F, r_1, & r_2. See what happens.

I have one more question here, if I had assumed that the collision is inelastic, will (3) still correct since the kinetic energy is not conserved anymore.

The collision is most definitely inelastic. KE is not conserved--total energy is conserved. (Your equation 3 is a total energy equation, not simply a KE equation.)

 

[tanzl]

Oops... I apologise for all the typing errors.

Wow, the unknowns canceled like magic. Thank you so much.
The final answer I got is 7.994cm. Not much different from the first penetration.

 

[Doc Al]

I get a slightly different answer, so you might want to recheck your arithmetic. But there's not much difference.

 

[tanzl]

I recheck my arithmetic. There is nothing wrong.

The final expression I get is
mv₁² = mv₁²*\frac{\Delta r_2}{\Delta r_1} + \frac{m^2}{M+m}*v₁²

Canceled off all the v and m gives
\Deltar₂ = \frac{M}{M+m}*\Deltar₁
= 1/1.007 * 0.08
= 0.07994m

 

[Doc Al]

Canceled off all the v gives
\Deltar₂ = \frac{M}{M+m}*\Deltar₁

Right.

= 1/1.007 * 0.08
= 0.07994m

According to my calculator, 1/1.007 * 0.08 = 0.07944 m.

 

[tanzl]

Yea.. Thank you so much for your help.