Conservation of Momentum problem?? im not sure how to approach?

Question

A 5.20 gram bullet moving horizontally at 672 m/s strikes a 700 gram block of wood at rest on a smooth frictionless table. The bullet emerges from the block, traveling in the same direction at 428 m/s. What is the speed of the block after the bullet emerges? How much kinetic energy is converted to other forms of energy during this process?

thanks

Answer 1

way to get your homework done for you :)

Basically the law of conservation of momentum states that in a closed system the total momentum is conserved....i.e. it does not increase or decrease based on an event that occurs.

So we start by finding the event, in this case that is the bullet striking and exiting the block, causing it to move. All you need to do is calculate the total momentum in the system before this event and make it equal to the total momentum in the system after the event.
(hint - always write out your given information and make sure that you have the data in SI units)
Data:
Before event:
mass bullet = 5.2 g = 0.0052kg
velocity bullet = 672 m/s
mass block = 700g = 0.7kg
velocity block = 0m/s

After Event
mass bullet = 5.2 g = 0.0052kg
velocity bullet = 428 m/s
mass block = 700g = 0.7kg
velocity block = ?

Before event:

So.. momentum = mass x velocity (p=m x v)
we are dealing with more than one object so simply calculate the momentum of each object and add it together.

p system = p bullet + p block
p system = (m x v) + (m x v)
p system = (0.0052kg x 672 m/s) + ( 0.7kg x 0m/s)
psystem = 3.4944kgm/s

After:
p system = 3.4944kgm/s
but p system = p bullet + p block
3.4944kgm/s = (0.0052kg x 428m/s) + (0.7kg X v)

now you have to simplify and solve for "v"
you get

v=1.813m/s for the wood block

the energy calculation is simply how much energy required to make the block move = the kinetic energy of the moving block kE=1/2 mass x (velocity squared)
kE =1.15 J

Answer 2

mass of bullet = m= 5.2g
velocity of bullet = v= 672m/s
hence, linear momentum of bullet = p= m*v
= 672*5.2 g m/s
= 3.4944 kg m/s
mass of block =M= 700g
let velocity imparted to block by bullet = V
final velocity of bullet = u = 428 m/s
initial kinetic energy of bullet = k1 = (1/2)mv^2
= (1/2)(.0052)(672)^2
= 1174.1184 J
final kinetic energy of bullet = k2 = (1/2)mu^2
=(1/2)(.0052)(428)^2
=476.2784 J
by conservation of momentum,
mv = MV + mu
=>V= [m(v-u)]/M
=> V = [5.2(672-428)]/700
= 1.812 m/s
hence kinetic energy of block = K
= (1/2)MV^2
=1.149 J
hence energy lost =k1-(k2+K)
=696.69 J

Answer 3

momentum before collision = 0 gm/s
Momentum after collision = p = mv
= (5.20 * 672) + (700 * v)
= 3494.4 + 700v

According to Law of Conservation of Momentum
Momentum before collision = momentum after collision
0 = 3494.4 + 700v
700v + 3494.4 = 0
700v = - 3494.4
v = -3494.4 / 700
= 4.992 gm/s
I have expressed my answer in terms of gm/sec. The S.I unit is kgm/sec. This answer can also be expressed in terms of kgm/sec

Answer 4

I suggest conserving some of your own momentum, and stop needlessly wasting time and energy on pointless hypothetical questions such as this.

Answer 5

m1 u1 +m2 u2 =m1 v1+m2 v2

1/2 m1u1^2+1/2m2u2^2=1/2m1v1^2+1/2m2v2^2