Momentum, Impulse, and Collision #2

[Tastosis]

Homework Statement

A horizontal force of 0.80 N is required to move a mass M2 = 5kg across the surface with a constant acceleration. With the block initially at rest, a 0.02kg bullet M1 fired horizontally into the block causes the block (with bullet inside) to slide 1.5m before coming to rest again. Find the speed Vo of the bullet.

Homework Equations

M1Vo + M2Vo = (M1 + M2) Vf

The Attempt at a Solution

Should I cancel out M2Vo? Since the block is initially at rest. How do I get Vf? Is it 1.5 m/s?

 

[tiny-tim]

Hi Tastosis!

(try using the X₂ icon just above the Reply box )

With question like this, you need to divide the problem into two parts, the collision and after.

During the collision, use conservation of momentum; after the collision use conservation of energy and the work energy theorem; use v_f from the first part as v_i for the second part.

 

[Tastosis]

Thanks for the tips!

So I use conservation of momentum to solve for V_f which I will use as V_i for the final answer?

 

[tiny-tim]

Yes.

 

[Tastosis]

I'm stuck. How do I get the velocity initial of the bullet? Do I use the 0.80 N to solve for the time?

 

[tiny-tim]

A horizontal force of 0.80 N is required to move a mass M2 = 5kg across the surface with a constant acceleration.

Do I use the 0.80 N to solve for the time?

No, you use the 0.80 N to find the force of friction (to use in the work energy equation)

 

[Tastosis]

Oh yeah.
From where it stands, here is my equation (I'm still in conservation of momentum):
M₁V₁ + 0 (since the block is initially at rest) = (M₁ + M₂) V₂

How do I get V₁ of bullet?

 

[tiny-tim]

M₁V₁ + 0 (since the block is initially at rest) = (M₁ + M₂) V₂

Yup!

Now do the work energy equation, using V₂ as your initial V.

 

[Tastosis]

Conservation of momentum:
V₂ = (M₁ * V₁) / M₁ + M₂

V₂ = (0.02kg * 1 m/s) / 0.02kg + 50kg

V₂ = 0.0004 m/s

Work-energy theorem:
K₁ + W_others = K₂
1/2 MV₁² + W_others = 1/2 MV₂²

V₂² = [(0.02kg * (0.0004 m/s)²) + 2 * 0.80 N] / 0.02kg

V₂² = 80 m/s

V₂ = 8.9 m/s

 

[tiny-tim]

Conservation of momentum:
V₂ = (M₁ * V₁) / M₁ + M₂

V₂ = (0.02kg * 1 m/s) / 0.02kg + 50kg

V₂ = 0.0004 m/s

Where did the 1 m/s come from?

Work-energy theorem:
K₁ + W_others = K₂
1/2 MV₁² + W_others = 1/2 MV₂²

V₂² = [(0.02kg * (0.0004 m/s)²) + 2 * 0.80 N] / 0.02kg

V₂² = 80 m/s

V₂ = 8.9 m/s

Where is your W_others ?

 

[Tastosis]

Where did the 1 m/s come from?


Where is your W_others ?

That's my problem, I don't know how to get the V₁ of the bullet

EDIT: My W_others is the 0.80 N

 

[tiny-tim]

That's my problem, I don't know how to get the V₁ of the bullet

EDIT: My W_others is the 0.80 N

Your V₁ will come from solving both equations simultaneously.

0.80N is a force, isn't it?

W is work.​

 

[Tastosis]

Oh, my bad. So W = Fd

But back to the first equation, I have 2 unknowns, V₂ and V₁?

 

[tiny-tim]

Get V₂ from the second equation, plug it back into the first equation to get V₁

 

[Tastosis]

I finally get it..I think:

From Conservation of momentum:
V₂ = (M₁V₁) / M₁ + M₂

Work-energy:
MV₁² + 2W_other = MV₂²

Substitute:
MV₁² + 2W_other = M [(M₁V₁) / M₁ + M₂]²

 

[tiny-tim]

Work-energy:
MV₁² + 2W_other = MV₂²

You're using the same letters to mean different things.

If they mean the same as in the first equation, then this second equation is wrong.

 

[Tastosis]

I don't get it. I thought I can substitute V₂ from the first equation to the second equation..since it has the same V₂ because the collision was inelastic.

Now I'm confused again lol

 

[tiny-tim]

I don't get it. I thought I can substitute V₂ from the first equation to the second equation

Yes …

your V₂ from the first equation becomes your V₁ in the second equation.

Your V₂ in the second equation is … ?

..since it has the same V₂ because the collision was inelastic.

I've no idea what this means.

 

[Tastosis]

My V₂ from the first equation is...

V₂ = (M₁V₁) / M₁ + M₂, where

M₁ = 0.02kg
M₂ = 50kg
V₁ =