Impulse

A(n) 8.9g bullet is fired into a(n) 132g block
of wood at rest on a horizontal surface and
stays inside. After impact, the block slides
13m before coming to rest.

The acceleration of gravity is 9.8 m/s^2/

If the coeffcient of friction between the
surface and the block is 0.5, find the speed of
the bullet before impact. Answer in units of
m/s.


For this problem, the only direction I know where to go in is to solve for the force of friction. Because we have the coefficient of friction and the normal force (9.8 * .132kg) the friction force = .6468N. So now where do I go with this?

Quote:

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Originally Posted by vesperka

A(n) 8.9g bullet is fired into a(n) 132g block
of wood at rest on a horizontal surface and
stays inside. After impact, the block slides
13m before coming to rest.

The acceleration of gravity is 9.8 m/s^2/

If the coeffcient of friction between the
surface and the block is 0.5, find the speed of
the bullet before impact. Answer in units of
m/s.


For this problem, the only direction I know where to go in is to solve for the force of friction. Because we have the coefficient of friction and the normal force (9.8 * .132kg) the friction force = .6468N. So now where do I go with this?

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Just apply it to $\displaystyle F\Delta{t}=M\Delta{v}$

Hmm, I'm definitely plugging it in wrong because i'm getting 4.91m/s as my answer.

.6486N = (.132kg)V

I don't have a value for time so I'm guessing we don't need it, but what else can I do here? Sorry if I seem a little dense on the subject but I have a make-up test for this subject tomorrow so I don't understand it all that well :\

Quote:

---

Originally Posted by vesperka

A(n) 8.9g bullet is fired into a(n) 132g block
of wood at rest on a horizontal surface and
stays inside. After impact, the block slides
13m before coming to rest.

The acceleration of gravity is 9.8 m/s^2/

If the coeffcient of friction between the
surface and the block is 0.5, find the speed of
the bullet before impact. Answer in units of
m/s.


For this problem, the only direction I know where to go in is to solve for the force of friction. Because we have the coefficient of friction and the normal force (9.8 * .132kg) the friction force = .6468N. So now where do I go with this?

---

But are you sure that this is an impulse problem...it sounds more like a perfectly inelastic collison to me?

As Mathstud28 noted, this is a momentum problem.

First, let us consider the block, with the bullet embedded, coasting to a stop following the impact. Then we have that the normal force has magnitude N=Mg, where g=9.8 m/s^2 is the accelleration of gravity and M is the mass of the block + bullet (132 g + 8.9 g= 141 g (in significant figures)).
Now, the friction force is thus given by $\displaystyle F=\mu{N}=\mu{M}g$, and thus the accelleration experienced by the block during this time is $\displaystyle a=-\frac{F}{M}=-\mu{g}=-4.9$ m/s^2, where the minus sign is because we are decellerating. If $\displaystyle v_i$ is the initial velocity (just after impact) of the block+bullet, then from the basic kinematic equation $\displaystyle v_f^2-v_i^2=2ad$, we have
$\displaystyle 0^2-v_i^2=2\cdot(13)\cdot(-4.9)$
$\displaystyle v_i^2=127.4$ m^2/s^2
$\displaystyle v_i=11.3$ m/s
This means the momentum of the block+bullet immediately following their (inelastic) collision is (0.141 kg)*(11.3 m/s)=1.59 kg*m/s
Since momentum is conserved in the collision, and the block is initially at rest, this is thus the momentum of the bullet before the collision. Dividing the above momentum by the mass of the bullet
v=(1.59 kg*m/s)/(0.0089 kg)=179 m/s

--Kevin C.

Yeah I think you're right, so maybe it isn't impulse. Still, how do I go about solving this? The perfectly inelastic collision equation is

http://upload.wikimedia.org/math/2/7...0be90a32b4.png

but I have no velocities.

Quote:

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Originally Posted by vesperka

Yeah I think you're right, so maybe it isn't impulse. Still, how do I go about solving this? The perfectly inelastic collision equation is

http://upload.wikimedia.org/math/2/7...0be90a32b4.png

but I have no velocities.

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the $\displaystyle V_i$ for the block is 0 and you must calculate the other using a force equation I think

You're answer is correct, but is there a way to solve it without using any kinematics? I'm not positive about this, but I'm pretty sure our teacher only wants us to use energy, momentum, and impulse equations on the test :x

Nonetheless, you're reasoning is great so your help is very much appreciated :D

Quote:

---

Originally Posted by vesperka

You're answer is correct, but is there a way to solve it without using any kinematics? I'm not positive about this, but I'm pretty sure our teacher only wants us to use energy, momentum, and impulse equations on the test :x

Nonetheless, you're reasoning is great so your help is very much appreciated :D

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Well, you could do this slightly differently.

You know how much work is done to stop the bullet-block (bb) combination. It is simply the friction force applied over 13 m. (By the way, make sure to add the mass of the bullet to the normal force, though it is likely to be too small to influence the final answer.) Since work done is equal to the change in kinetic energy, we can find the speed of the bb just after the collision.

Then you can apply your momentum equation in reverse to find the initial speed of the bullet.

-Dan