Bullet through block problem?

Use the law of conservation of momentum, i.e.,

M1V1 + M2V2 = M1V3 + M2V4

where

M1 = mass of the bullet = 6 gm (given) = 0.006 kg.

V1 = initial velocity of bullet = 896 m/sec (given)

M2 = mass of the block = 1.25 kg (given)

V2 = initial velocity of block = 0

V3 = final speed of bullet = 435 m/sec (given)

V4 = final speed of block

Substituting appropriate values,

0.006(896) + 0 = 0.006(435) + 1.25(V4)

and solving for "V4"

V4 = 2.21 m/sec.

Hope this helps.

This problem is related to one of the fundamental principles of physics: the conservation of momentum.

Before the collision, the bullet has a mass, m, of 0.006kg; and a velocity, v, of 896m/s. Its momentum, m*v, is therefore, 5.364kgm/s.

The wooden block (m=1.25kg, v=0) has no momentum: 1.25 * 0 = 0.

So the total momentum before the collision of the bullet and the block is 5.364kgm/s.

After the bullet has passed through the block, the momentum has been redistributed between the block and the bullet.

The bullet is traveling at 435m/s and we will assume it's mass is still 0.006m/s. So its new momentum is 2.61kgm/s.

The momentum in the system is 'conserved'. This means it is the same before and after the collision.

The initial momentum (5.364) minus the bullets new momentum (2.61) leaves 2.754kgm/s for the block. Assuming that the block is still 1.25kg, we can rearrange the equation momentum = mass * velocity so calculate the velocity: v = momentum / mass: 2.2032m/s.

Hope this helped. If you need/want to know more about momentum, wikipedia has a good article on it.

conservation of momentum

Find the momentum of the bullet before it hits the block and subtract the momentum of the bullet after the block and you'll have the momentum of the block after the bullet passes through. Then divide by the block's mass.

Don't listen to that Scott guy, this is obviously not a problem where you're meant to take those things into account because you're not given any of that information.

Use conservation of energy. The difference in kinetic energy of the bullet before and after the event is the amount of kinetic transferred to the block. Relate mass, kinetic energy, and velocity as follows:

Ek = (1/2)mv^2

Doug

Actually, I don't think either of these approaches, momentum or kinetic energy, really work. Here is why - there was a considerable amount of kinetic energy used in breaking through the block. There was a potential energy associated with forcing the molecules in the block of wood apart. This subtracts from the total kinetic energy of the system but we do not know how much.

Impulse = m(vf - vo)

= 0.006(435 - 896) = - 2.766 kg-m/s

v_block = 2.2128 m/s