# Momentum and Projectile Motion

In summary: Thanks for your help!In summary, Homework Statement states that a bullet moving horizontally with speed Vo when it hits and embeds in a block of mass 100m that is at rest on a horizontal frictionless table (in the diagram it isn't placed at the edge of the table, i don't know if this matters) will have the following quantities. a.) the speed of the block as it leaves the tableb.) the change in kinetic energy of the bullet-block system during the impactc.) the distance xIt is assumed that the bullet passes through the block instead of remaining in it.d.) state whether the time it takes the block to reach the floor from the edge of the table would now be

## Homework Statement

Can anyone help me with this problem?

A bullet of mass m is moving horizontally with speed Vo when it hits and embeds in a block of mass 100m that is at rest on a horizontal frictionless table (in the diagram it isn't placed at the edge of the table, i don't know if this matters). The surface of the table is a height h above the floor. After the impact the bullet and the block slide off the table and hit the floor a distance x from the edge of the table.

Derive expressions for the following quantities in terms of m, h, Vo and appropriate constants.
a.) the speed of the block as it leaves the table
b.) the change in kinetic energy of the bullet-block system during the impact
c.) the distance x

Suppose that the bullet passes through the block instead of remaining in it.
d.) state whether the time it takes the block to reach the floor from the edge of the table would now be greater, less, or the same
e.) state whether the distance x would now be greater, less, or the same

## The Attempt at a Solution

i don't think i solved these right at all...

a.) mVo = (m+100m)Vf
Vf = (mVo)/(m+100m)

b.) KEf - KEi = change in KE
1/2(m+100m)(Vf)^2 - 1/2mVo^2 = change in KE

c.) x = Vit + 1/2at^2
x = Vft + 0

i don't know how to figure these out...
d.) would the amount of time increse becasue it has less mass??
e.) would x increase because it has more time to fall??

Thanks for any help!

a.) mVo = (m+100m)Vf
Vf = (mVo)/(m+100m)

b.) KEf - KEi = change in KE
1/2(m+100m)(Vf)^2 - 1/2mVo^2 = change in KE
That's the first step. Keep going. (Don't leave your answer in terms of Vf.)

c.) x = Vit + 1/2at^2
x = Vft + 0
Well? Solve for the horizontal distance. (Please use different letters to represent horizontal and vertical motion.)

You may as well solve these before worrying about d and e. But the hint for those two is: Does the block end up going faster or slower than in the earlier case.

Is this right??
a.) Vf = (mVo)/(m+100m)
(mVo)/m(1+100)
Vo/101

b.) 1/2(m+100m)(Vf)^2 - 1/2mVo^2 = change in KE
1/2(m+100m)(Vo/101)^2 - 1/2mVo^2 = change in KE
1/2m(101)(Vo/101)^2 - 1/2mVo^2 = change in KE
1/2m(Vo)^2 - 1/2mVo^2 = change in KE
0 = change in KE
can that be right?

Is $\frac{101}{(101)^{2}}=1??$

arildno said:
Is $\frac{101}{(101)^{2}}=1??$
no

i'm so confused

So why did you state that it was, when going from the 7th line to your 8th line in your previous post?

Aside from that silly little mistake, your work looks good!

1/2m(101)(Vo/101)^2 - 1/2mVo^2 = change in KE
1/2m(Vo)^2 - 1/2mVo^2 = change in KE

so you can't cancel the 101?

Is this right??
a.) Vf = (mVo)/(m+100m)
(mVo)/m(1+100)
Vo/101
Good.
b.) 1/2(m+100m)(Vf)^2 - 1/2mVo^2 = change in KE
1/2(m+100m)(Vo/101)^2 - 1/2mVo^2 = change in KE
1/2m(101)(Vo/101)^2 - 1/2mVo^2 = change in KE
1/2m(Vo)^2 - 1/2mVo^2 = change in KE
0 = change in KE
can that be right?
You made a mistake simplifying the term that I bolded in red. (As arildno has pointed out.)

so you can't cancel the 101?

You can cancel one of them. (Don't forget the square!)

oh, i see.
so it would be 1/2m(Vo^2/101) - 1/2mVo^2

oh, i see.
so it would be 1/2m(Vo^2/101) - 1/2mVo^2

Good. Now simplify it.

1/2mVo^2(-100/101) = change in KE

for c.) can't i just use Vf=(Vo/101)
x=(Vf)t
x=(Vo/101)t

1/2mVo^2(-100/101) = change in KE
Simplify it further.

for c.) can't i just use Vf=(Vo/101)
x=(Vf)t
x=(Vo/101)t
Nothing wrong with that as a first step, but you can't stop there. (You can't just leave an unknown "t" in your answer.) Now write the equation for the vertical motion and solve both equations simultaneously.

y = Vt + 1/2at^2
h = 0 + 1/2(-10)t^2
t = square root of (h/-5)

x = (Vo/101)(square root of (h/-5))

y = Vt + 1/2at^2
h = 0 + 1/2(-10)t^2
t = square root of (h/-5)

x = (Vo/101)(square root of (h/-5))

You realize that square roots of negative numbers presents a problem, right?

If you take the acceleration as a = -10 m/s^2; then the final height is y = -h (since it falls a distance h).

Thanks for all the help!
I appreciate it!

I have been thinking about parts d and e, and this is what I have decided:

d.) y = 1/2at^2
-h = 1/2(-10)t^2
t = square root of (h/5)
Since the height doesn't change, the time it takes for the block to fall would stay the same.

e.) Since momentum had to be conserved, the velocity of the block would be less if the bullet went through the block.
x = Vt
t = square root of (h/5)
That would mean that the distance would have to be smaller.

Is this correct?

Please ignore sherlock, she is simply a secretary practising her typing and p#ssing everyone off :grumpy:

Thanks, Hootenanny, for pointing that out so pinkpolkadots would not be further confused.

Moonbear said:
Thanks, Hootenanny, for pointing that out so pinkpolkadots would not be further confused.
No problem, she's been doing the same in a few other forums for an hour or so.

pinkpolkadots: I haven't had a detailed look at your problem, but both your answers look good to me (provided x is the distance the block lands from the table).

yes, x was the distance the block falls from the table.

thank you!

thank you!
Twas a pleasure

d.) y = 1/2at^2
-h = 1/2(-10)t^2
t = square root of (h/5)
Since the height doesn't change, the time it takes for the block to fall would stay the same.
Good thinking!

e.) Since momentum had to be conserved, the velocity of the block would be less if the bullet went through the block.
You are correct, but can you explain your reasoning quantitatively? Hint: If the bullet passes through the block, what must be true about the bullet's final speed compared to the block's final speed?

## What is momentum?

Momentum is the measure of an object's mass in motion. It is calculated by multiplying the mass of an object by its velocity.

## What is the law of conservation of momentum?

The law of conservation of momentum states that in a closed system, the total momentum before an event is equal to the total momentum after the event. This means that momentum is conserved and cannot be created or destroyed.

## How is momentum related to projectile motion?

Momentum plays a key role in projectile motion because it determines the direction and speed of an object's motion. As an object moves through the air, its momentum changes due to the forces acting on it, such as gravity and air resistance.

## How does mass affect momentum in projectile motion?

Mass directly affects momentum in projectile motion. A heavier object will have a greater momentum than a lighter object moving at the same velocity. This is because momentum takes into account both an object's mass and its velocity.

## What is the relationship between momentum and kinetic energy?

Momentum and kinetic energy are closely related concepts. While momentum represents the quantity of motion of an object, kinetic energy represents the energy of motion. Both are affected by an object's mass and velocity, and they are both conserved in a closed system.