Solve Momentum Problem: Speed, Height, KE Before & After

[pb23me]

Homework Statement

A .025kg bullet is fired at 400 m/s into a 5kg wooden blockhanging from a string. the block is at rest before the bullet hits. The bullet becomes embedded in the wooden block.
What is the speed of the combined bullet + block?
How high will the block+bullet rise?
Calculate the total kinetic energy before the collision and after the collision.

Homework Equations

\DeltaKE=Wnet=1/2mv_f²-1/2mv_i²
conservation of momentum= P^net_f=P^net_i
conservation of energy= \DeltaKE=-(\DeltaPE)

The Attempt at a Solution

To find the speed of the block and bullet together, I used the conservation of momentum equation (.025)(400)=V_f(m₁+m₂) and got 2 m/s. I took the initial velocitys of both objects just before impact?? To find how high the block and bullet would rise i used the conservation of energy equation; 1/2(2m/s)²= 9.8(y_f)... and got .204m for y_f.. I set the final velocity equal to zero in this equation. Was that the correct thing to do? Would that really work in reality? Calculating the kinetic energy before the collision you just have the kinetic energy of the bullet so 1/2mv²=2000j Calculating the kinetic energy after the collision; (m¹+m₂)(2m/s)(.5)=5.025j
When they ask for kinetic energy after the collision is this immediately after the collision?
When i used the conservation of momentum equation is the final velocity that i get immediately after impact?
How is it that the bullet is no longer putting a force on the wooden block after impact? This is the logic i used to solve the problem. It's what my proffesor told me.
Wnet after collision... Wnet = \DeltaKE=-Wg since gravity is the only force acting at this point. Then Wnet=-mgy_f... The change in kinetic energy \DeltaKE_f-\DeltaKE_i= 5.025j-2000j=1994.975
I should be able to use this number to calculate y_f...\DeltaKE=Wnet=1994.975=-mgy_f...however this doent work...why is that?

 

[PhanthomJay]

These are very thoughtful questions.

Homework Statement

A .025kg bullet is fired at 400 m/s into a 5kg wooden blockhanging from a string. the block is at rest before the bullet hits. The bullet becomes embedded in the wooden block.
What is the speed of the combined bullet + block?
How high will the block+bullet rise?
Calculate the total kinetic energy before the collision and after the collision.

Homework Equations

\DeltaKE=Wnet=1/2mv_f²-1/2mv_i²
conservation of momentum= P^net_f=P^net_i
conservation of energy= \DeltaKE=-(\DeltaPE)

The Attempt at a Solution

To find the speed of the block and bullet together, I used the conservation of momentum equation (.025)(400)=V_f(m₁+m₂) and got 2 m/s.

good

I took the initial velocitys of both objects just before impact??

you took the final velocity of the bullet/block system after impact and used it as the initial velocity in your conservation of energy equation, which is the correct way of doing it.

To find how high the block and bullet would rise i used the conservation of energy equation; 1/2(2m/s)²= 9.8(y_f)... and got .204m for y_f.. I set the final velocity equal to zero in this equation. Was that the correct thing to do?

again, yes!

Would that really work in reality?

In reality there is always friction, so it doesn't really work that way, but close enough for the first swing anyway, unless there is a lot of friction at the string attachment point that significantly impedes rotation

Calculating the kinetic energy before the collision you just have the kinetic energy of the bullet so 1/2mv²=2000j

yes!

Calculating the kinetic energy after the collision; (m¹+m₂)(2m/s)(.5)=5.025j

math error, you forgot to square the v, it should be 10.05 J

When they ask for kinetic energy after the collision is this immediately after the collision?

yes

When i used the conservation of momentum equation is the final velocity that i get immediately after impact?

yes!

How is it that the bullet is no longer putting a force on the wooden block after impact?

Well immediately after impact, it is, during its deceleration phase, but its force gets reduced rather quickly to 0 in an extremely short time period before the block has had a chance to move much, so you can neglect this, no more acceleration means no more net force

This is the logic i used to solve the problem. It's what my proffesor told me.
Wnet after collision... Wnet = \DeltaKE=-Wg since gravity is the only force acting at this point. Then Wnet=-mgy_f... The change in kinetic energy \DeltaKE_f-\DeltaKE_i= 5.025j-2000j=1994.975
I should be able to use this number to calculate y_f...\DeltaKE=Wnet=1994.975=-mgy_f...however this doent work...why is that?

2 reasons, one is your math error, that initial KE at the start of the swing is 10.5 J (this was just a math error), but the other reason is that you must apply conservation of energy between the start of the swing, where KE_initial = 10.5 J, and the end of the swing, where PE_final = mgy_f, and KE_final = ________??

 

[pb23me]

KE_final=0 so -10.5=-[5.025(9.8)y_f]
y_f=.213m

 

[PhanthomJay]

KE_final=0 so -10.5=-[5.025(9.8)y_f]
y_f=.213m

Decimal slip...it's 10.05 not 10.5... which you should have noted was wrong from your original solution...but otherwise, looks good!