Calculating Bullet Speed from Impact with Lumber

AI Thread Summary
A 35.0-g bullet embeds itself in a 4.7-kg piece of lumber, which then moves at 8.0 m/s, prompting a discussion on calculating the bullet's original speed using conservation of linear momentum. The principle states that the total momentum before the collision equals the total momentum after the collision. Participants emphasize the importance of applying this principle correctly, particularly in the context of the IB program. Additionally, there is a suggestion to start new threads for different problems to maintain clarity. The conversation highlights the relevance of understanding momentum conservation in solving physics problems.
shann0nsHERE
Messages
6
Reaction score
0
A 35.0-g bullet strikes a 4.7-kg stationary piece of lumber and embeds itself in the wood. The piece of lumber and bullet fly off together at 8.0 m/s. What was the original speed of the bullet?
I don't know where to begin...
 
Physics news on Phys.org
This one too pleeasee

A thread holds a 1.5-kg and a 4.50-kg cart together. After the thread is burned, a compressed spring pushes the carts apart, giving the 1.5 kg cart a speed of 27 cm/s to the left. What is the velocity of the 4.5-kg cart?



I would be so grateful if someone could help me..
 
Just go through the conservation of linear momentum.
 
Hi there. Is conservation of linear momentum something you have studied in class? :smile:

Also: I would start a new thread for each problem, else things get confusing/messy.
 
Yes, I am in the ib program but i was absent for that lesson..
ha thanks for the heads up, i just started this
Andd thank you both for the help
 
shann0nsHERE said:
Yes, I am in the ib program but i was absent for that lesson..
ha thanks for the heads up, i just started this
Andd thank you both for the help
Okay, well paraphrasing Conservation of momentum says that if no external forces act on a system (which is true in this case), then the total linear momentum is conserved.

That is, \sum mv_{initial}=\sum mv_{final}

In this case,

M_{bullet}V_{bullet}+M_{block}V_{block}=M_{(block+bullet)}V'_{(block+bullet)}

where I used the ' symbol to indicate velocity after the collision

And Yes, they are both conservation of momentum.
 
Last edited:
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
View full post »

Similar threads

How Does a Bullet Impact Affect the Angular Velocity of a Door?
Replies
3
Views
3K
Calculating Force and Impulse of a Bullet Hitting a Man
Replies
1
Views
1K
Linear Momentum to Angular Momentum
Replies
6
Views
2K
Thermodynamics, bullet melting ice problem
Replies
4
Views
2K
What was the initial temperature of the lead bullet before it melted completely?
Replies
3
Views
1K
Is My Physics Calculation on Bullet and Block Interaction Correct?
Replies
14
Views
4K
Inelastic collision problem: Bullet striking a wood block
Replies
2
Views
3K
Just to double check, Linear Momentum
Replies
1
Views
1K
Momentum problem -- Two bullets fired into a block
Replies
7
Views
2K
Velocity of bullet before hitting a pendulum
Replies
11
Views
3K

Hot Threads

Recent Insights

Back
Top