# Momentum, Impulse, and Collision #2

• Tastosis
In summary: I'm sorry, but your equations are still not clear to me. You have to be much more careful with your notation.You have 2 different V's, but both are called V2.You have 2 different M's, but both are called M.In the first equation, you have 2 terms, M1 V1 and M2 V2. Call them A and B.In the second equation, you have 2 other terms, M V1 and M V2. Call them P and Q.You have to use the values from the first equation to find P and Q in the second equation.But back to the first equation, I have 2 unknowns, V2 and V1?Yes, you
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?

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 vf from the first part as vi for the second part.

Thanks for the tips!

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

Yes.

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?

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

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?

Tastosis said:
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.

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 + Wothers = K2
1/2 MV12 + Wothers = 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

Tastosis said:
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 + Wothers = K2
1/2 MV12 + Wothers = 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 said:
Where did the 1 m/s come from?

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

EDIT: My Wothers is the 0.80 N

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

EDIT: My Wothers 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.​

Oh, my bad. So W = Fd

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

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

I finally get it..I think:

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

Work-energy:
MV12 + 2Wother = MV22

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

Tastosis said:
Work-energy:
MV12 + 2Wother = 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.

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

Tastosis said:
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.

My V2 from the first equation is...

V2 = (M1V1) / M1 + M2, where

M1 = 0.02kg
M2 = 50kg
V1 =

## What is momentum?

Momentum is a physical quantity that measures an object's motion, taking into account its mass and velocity. It is represented by the symbol p and is calculated by multiplying an object's mass (m) by its velocity (v), giving the equation p = m * v.

## What is impulse?

Impulse is the change in an object's momentum over a period of time. It is represented by the symbol J and is calculated by multiplying the average force (F) applied to an object by the time (Δt) over which the force acts, giving the equation J = F * Δt. It is measured in units of Newton-seconds (N*s).

## What is the difference between elastic and inelastic collisions?

Elastic collisions are collisions in which both kinetic energy and momentum are conserved. In other words, the objects bounce off each other with no loss of energy. Inelastic collisions, on the other hand, are collisions in which kinetic energy is not conserved. Some energy is lost during the collision, usually in the form of heat or sound.

## How do you calculate the final velocities of two objects after a collision?

The final velocities of two objects after a collision can be calculated using the conservation of momentum principle, which states that the total momentum before the collision is equal to the total momentum after the collision. This can be represented by the equation m1v1i + m2v2i = m1v1f + m2v2f, where m represents mass and v represents velocity.

## What is the difference between an inelastic collision and a perfectly inelastic collision?

An inelastic collision is one in which some kinetic energy is lost during the collision. A perfectly inelastic collision, on the other hand, is a special case of inelastic collision in which the two objects stick together after the collision and move as one object. In this case, the total kinetic energy after the collision is zero.

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