Conservation of Momentum?

Question

I would really appreciate help on this problem it's giving me alot of trouble. Thanks. A bullet of mass 6 g strikes a ballistic pendulum of mass 2.0 kg. The center of mass of the pendulum rises a vertical distance of 10 cm. Assuming that the bullet remains embedded in the pendulum, calculate the bullet's initial speed.

Also, please show your work. Thanks

Thanks Edward...but no dice. The answer's wrong :-(

Answer 1

Firstly, note that since the bullet remains embedded in the pendulum, the collision is perfectly inelastic.

Before collision
Initial speed of bullet = u m/s; mass of bullet = 0.006 kg
Initial speed of pendulum = 0; mass of pendulum = 2 kg
Hence,
initial momentum of bullet-pendulum system p(i)
= 0.006*u + 2*0
= (0.006u) kgm/s

Just after collision
Final speed of bullet = final speed of pendulum = v m/s
Total mass of bullet and pendulum = 2 + 0.006 = 2.006 kg
Hence,
final momentum of bullet-pendulum system p(f)
= 2.006*v
= (2.006v) kgm/s

The string attached to the pendulum will not have the time to react and hence there is no external force on the bullet-pendulum system.
Thus, apply conservation of linear momentum.
p(f) = p(i)
=> 2.006v = 0.006u
=> u = 334.33 v ------------- (1)

Now apply principle of conservation of mechanical energy.
KEi + PEi = KEf + PEf
=> (0.5)(2.006)(v^2) + 0 = 0 + (2.006)(9.81)(0.1)
=> v = sqrt(1.962) = 1.40 m/s

Substitute this value into equation (1),
u = 334.33*1.40 = 468.3 m/s
Hence, the bullet's initial speed is 468.3 m/s

Answer 2

A classic conservation problem.

Let m be the mass of the bullet and M be the mass of the pendulum.

Let V0 be the initial velocity of the bullet, and let V be the velocity of the pendulum a split second after the bullet strikes and is embedded in the pendulum.

The initial momentum of the system is m V0. When the bullet strikes the ballistic pendulum, the system momentum becomes (m+M)V. From this point on, mechanical energy is conserved. The kinetic energy of the pendulum (with the bullet still embedded) a split second after the collision is 1/2 (m+M) V^2. Now the center of mass rises a vertical distance h and the pendulum stops for a split second, at which time all of that kinetic energy has been converted into potential energy, (m+M) g h.

So solve 1/2 (m+M) V^2 = (m+M) g h for the only unknown, V0.

Answer 3

This is your conservation of momentum
m1V1=u(m1+m2) not very helpful

However
Ke=Pe
.5m1V^2=(m1+m2)gh

V=sqrt( 2 (m1+m2)g h/m1)
V=sqrt(2(.006 + 2.0) 9.81 x .1/ 0.006)
V=25.6m/s (what a slow bullet)