Speed Of A Bullet Before It Strikes A Block Of Wood

AI Thread Summary
The discussion revolves around calculating the velocity of a bullet before it strikes a block of wood, given specific parameters like mass, distance, and friction. Two methods were attempted to solve the problem, both leading to the same result of approximately 166.32 m/s for the bullet's velocity. The first method uses energy conservation principles, while the second applies kinematic equations. A suggestion was made to simplify calculations by using g = 10 m/s² for easier computation in introductory physics contexts. The calculations and methods used appear to be correct and on the right track.
amber87
Messages
2
Reaction score
0

Homework Statement



A 10g bullet strikes and becomes embedded in a 0.200kg block of wood placed on a horizontal surface just in front of the gun. If the coefficient of sliding friction between the block and the surface is 0.4 and the impact drives the block a distance of 8m before it comes to rest, what is the velocity of the bullet before it strikes the block of wood?



Homework Equations



E(kinetic)=E(frictional)
1/2(m1+m2)v2^2=μ(m1+m2)gs

or

ma=μmg

a=μg

v^2=u^2+2as

v1(m1)=v2(m1+m2)


The Attempt at a Solution



I've tried 2 different ways to do it which are:
1.)

m1=0.010kg
m2=0.200kg
s=8m
μ=0.4

1. E(kinetic)=E(frictional)
1/2(m1+m2)v^2=μ(m1+m2)gs
1/2(0.210)v^2=0.4(0.210)(9.8)(8)
v2=7.92m/s

m1v1=(m1+m2)v2
0.010v1=0.210*7.92
v1=166.32m/s

2.)

ma=μmg

a=μg
a=0.4*9.8
a=3.92

v^2=0
a=-3.92
s=8m
u^2=?

v^2=u^2+2as
0=u^2+2(-3.92)(8)
u^2=7.92

m1v1=(m1+m2)v2
0.010v1=0.210*7.92
v1=166.32m/s

I'd love to know if I'm on the right track here. Thanks a million in advance.
 
Physics news on Phys.org
amber87 said:

Homework Statement



A 10g bullet strikes and becomes embedded in a 0.200kg block of wood placed on a horizontal surface just in front of the gun. If the coefficient of sliding friction between the block and the surface is 0.4 and the impact drives the block a distance of 8m before it comes to rest, what is the velocity of the bullet before it strikes the block of wood?



Homework Equations



E(kinetic)=E(frictional)
1/2(m1+m2)v2^2=μ(m1+m2)gs

or

ma=μmg

a=μg

v^2=u^2+2as

v1(m1)=v2(m1+m2)


The Attempt at a Solution



I've tried 2 different ways to do it which are:
1.)

m1=0.010kg
m2=0.200kg
s=8m
μ=0.4

1. E(kinetic)=E(frictional)
1/2(m1+m2)v^2=μ(m1+m2)gs
1/2(0.210)v^2=0.4(0.210)(9.8)(8)
v2=7.92m/s

m1v1=(m1+m2)v2
0.010v1=0.210*7.92
v1=166.32m/s

2.)

ma=μmg

a=μg
a=0.4*9.8
a=3.92

v^2=0
a=-3.92
s=8m
u^2=?

v^2=u^2+2as
0=u^2+2(-3.92)(8)
u^2=7.92

m1v1=(m1+m2)v2
0.010v1=0.210*7.92
v1=166.32m/s

I'd love to know if I'm on the right track here. Thanks a million in advance.

I would have used your method 1. and agree with your calculations.
There is a chance, that if this was an introductory physics question, that you were only required to use g = 10. This makes calculations much simpler and able to be done without a calculator. In that case that first velocity is 8, and the initial velocity this 168 m/s,
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top