Can someone help me work this problem?

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?Can someone help me work this problem?
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)v虏


So the equation becomes:


(m1+m2)gh = (1/2)(m1+m2)v虏


The (m1+m2) terms cancel:


gh = (1/2)v虏


Solve for h:


h = v虏 / (2g) = 0.8 mCan someone help me work this problem?
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.
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 鈥?when they collided.


And m1*v1^2/2=(m1+m2)*g*h 鈥?when in up-most position.


So v2^2=v1^2*m1/(m1+m2) 鈥?speed after collision


h=m1*v1^2/(2*(m1+m2)*g) - height
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.