Can someone help me work this problem?

Question

I have worked out this problem and got .77 for part b and that is wrong. I have asked this question already and got some tips, but they aren't working. Can someone work the second part of this problem and tell me how you worked it out?


A 2.50 g bullet, traveling at a speed of 425 m/s, strikes the wooden block of a ballistic pendulum, such as that in Figure 7.14. The block has a mass of 265 g.


http://i21.photobucket.com/albums/b272/Cajunboiler/worstthingever.gif



(a) Find the speed of the bullet/block combination immediately after the collision.


3.9 m/s
(b) How high does the combination rise above its initial position in meters?



Can someone help with part b?

Thanks blue.... I wonder why the website didn't accept .77 That was very close to the answer.

Answer 1

a) (speed is actually closer to 3.97...not a big deal)

b) Conservation of energy:
The sum of potential and kinetic energies when the bullet first hits the block and when it is at the top of the swing is the same:
U(bottom) + K(bottom) = U(top) + K(top)
Now, the speed at the top is zero, so the ewuation becomes:
U(bottom) + K(bottom) = U(top)
Solve for the change in potential energy:
[U(top) - U(bottom)] = K(bottom)
The change in potential energy due to gravity is: (m1+m2)gh
The kinetic energy at the bottom is: (1/2)(m1+m2)v²
So the equation becomes:
(m1+m2)gh = (1/2)(m1+m2)v²
The (m1+m2) terms cancel:
gh = (1/2)v²
Solve for h:
h = v² / (2g) = 0.8 m

Answer 2

Finding the speed of the bullet+block combination:

m1*v1 = (m1 + m2)*Vf, Vf = m1*v1/(m1 + m2),
Vf = 2.5*10^(-3)*425/(2.5*10^(-3) + 265*10^(-3)), Vf = 3.97 m/s (approx)

b. Now about the height, use the conservation of mechanical energy. The kinetic energy at the moment of collision equals the dynamic energy at the highest point:

K = U, (m1+ m2)*Vf^2/2 = (m1+ m2)*g*h, where g = 9.8 m/s^2 the acceleration of gravity and h the maximum height. Solving for h:

h = V^2/2*g = (3.97)^2/2*9.8 = 0.8 m or 80 cm.

Answer 3

Ok here is how i think you solve part (b)

First I calculated the Velocity of the center of mass as follows:

p(i) = p(f) (Momentum conservation)
mv +mv = mv + mv (momentum initial and final mometum of the blockbullet combinaiton)
(.0025)425) + 0 = .2675Vcm
Vcm = 3.972

Now, in this type of collision, Kinetic energy is also conserved. So,

K(i) = U(gf) (Initial Kinetic energy is equal to the final gravitaional potential energy)
(1/2)mv^2 = mgh(f)
h(f)= V^2 / (2g) = .804 m high

I think this is correct, I havent taken a basic physics course for a few years now.

Answer 4

Momentum conservation Law: m1*v1+0*m2 (see fig.a) = (m1+m2)*v2 (see fig.b)
Energy conservation Law: m1*v1^2/2=(m1+m2)*v2^2/2 – when they collided.
And m1*v1^2/2=(m1+m2)*g*h – when in up-most position.
So v2^2=v1^2*m1/(m1+m2) – speed after collision
h=m1*v1^2/(2*(m1+m2)*g) - height