• Support PF! Buy your school textbooks, materials and every day products Here!

Bullet+block+spring physics homework

  • Thread starter Jabababa
  • Start date
  • #1
52
0

Homework Statement



A 15.0 g bullet with a speed of v = 235 m/s is fired into a block of wood that is initially at
rest on a frictionless surface. The block of wood is attached to a horizontal massless
spring with a spring constant k = 195 N/m that is initially uncompressed. The other end
of the spring is attached to a wall as shown in the figure below. The bullet becomes
imbedded in the block of wood which then compresses the spring a distance of 28.0 cm
before momentarily coming to a stop.

a) Determine the mass of the wooden block.
b) What fraction of the kinetic energy is transformed into other forms of energy
during the inelastic collision of the bullet with the wooden block?


Homework Equations





The Attempt at a Solution



for a) is it like this:
kx = 1/2mv^2
k=195
x = 28cm
m = m of bullet + M of block
v = initial speed
and then solve for M
kx = 1/2(m + M)^2.

And 2nd question i dont know how to start.
 

Answers and Replies

  • #2
adjacent
Gold Member
1,549
63
Where is the image,Jabababa?
 
  • #4
1,540
134
for a) is it like this:
kx = 1/2mv^2
k=195
x = 28cm
m = m of bullet + M of block
v = initial speed
and then solve for M
kx = 1/2(m + M)^2.
You are equating kx(force,a vector quantity) with (1\2)mv^2(kinetic energy,a scalar quantity) which is incorrect.

First understand that bullet and block are undergoing inelastic collision.

What happens in an inelastic collision ?
 
Last edited:
  • #5
199
15
for a) is it like this:
kx = 1/2mv^2
k=195
x = 28cm
m = m of bullet + M of block
v = initial speed
and then solve for M
kx = 1/2(m + M)^2.

And 2nd question i dont know how to start.
the equations you wrote isn't correct. Check again.

for 1st question you have to first conserve momentum for bullet-block collision then determine the final velocity of the block+bullet. Now you can apply energy conservation kinetic energy of block+bullet changes to potential energy of spring (1/2 kx^2).

After you solve first question then you would know the kinetic energy after collision, then you can solve for loss of energy by subtracting it from initial energy of bullet. and then calculate the fraction.

Edit: Tanya Sharma got here first, she is also saying the same thing.
 
  • #6
52
0
the equations you wrote isn't correct. Check again.

for 1st question you have to first conserve momentum for bullet-block collision then determine the final velocity of the block+bullet. Now you can apply energy conservation kinetic energy of block+bullet changes to potential energy of spring (1/2 kx^2).

After you solve first question then you would know the kinetic energy after collision, then you can solve for loss of energy by subtracting it from initial energy of bullet. and then calculate the fraction.

Edit: Tanya Sharma got here first, she is also saying the same thing.
oo so i should use this instead:

m(bullet)v + m(Block)v = m(total)v? To find the final velocity? But then on the right hand side i will have 2 unknown variables...


edit: should i be doing this:
m(bullet)v + m(Block)v = m(total)v
1/2mv^2 = 1/2kx^2
then from 1/2mv^2 = 1/2kx^2 i isolate and get m = kx^2/v^2 and sub it into the first momentum equation to find final velocity. Then with final velocity i can find total mass?
 
Last edited:
  • #7
1,540
134
oo so i should use this instead:

m(bullet)v + m(Block)v = m(total)v? To find the final velocity? But then on the right hand side i will have 2 unknown variables...


edit: should i be doing this:
m(bullet)v + m(Block)v = m(total)v
1/2mv^2 = 1/2kx^2
then from 1/2mv^2 = 1/2kx^2 i isolate and get m = kx^2/v^2 and sub it into the first momentum equation to find final velocity. Then with final velocity i can find total mass?
There are two masses,bullet and block,so us two different variables,say m for bullet and M for block.Denote velocity of bullet before collision by 'u' and velocity of (bullet+block) after collision by 'v'.

So your conservation of momentum would look like

mu = (M+m)v .

Here M and v are unknown .

Now what happens after collision ? How would you use the distance travelled by (block+bullet) before stopping ?
 
  • #8
199
15
oo so i should use this instead:

m(bullet)v + m(Block)v = m(total)v? To find the final velocity? But then on the right hand side i will have 2 unknown variables...


edit: should i be doing this:
m(bullet)v + m(Block)v = m(total)v
1/2mv^2 = 1/2kx^2
then from 1/2mv^2 = 1/2kx^2 i isolate and get m = kx^2/v^2 and sub it into the first momentum equation to find final velocity. Then with final velocity i can find total mass?
firstly use different variables as Tanya Sharma already pointed out. You are headed in the right direction. What is the velocity of the block before collision?...............its zero

Yes, then do the substitution.

P.S. : Equations

##m_{bullet}.v_{bullet} + m_{block}.v_{block}= m_{total}.v_{combined}##

##m_{total} = m_{bullet} + m_{block}##

##\frac{1}{2}m_{total}.v_{combined}^2 = \frac{1}{2}k.x^2##
 

Related Threads on Bullet+block+spring physics homework

  • Last Post
Replies
3
Views
5K
  • Last Post
Replies
8
Views
4K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
10
Views
8K
  • Last Post
Replies
3
Views
5K
Replies
13
Views
4K
Replies
2
Views
11K
Replies
3
Views
3K
Replies
3
Views
2K
Replies
13
Views
2K
Top