Bullet through block problem?

Question

A bullet with a mass of 6.00 g is fired through a 1.25 kg block of wood on a frictionless surface. The initial speed of the bullet is 896 m/s, and the speed of the bullet after it exits the block is 435 m/s. At what speed does the block move after the bullet passes through it?How would you go about solving this...it's been a while since I learned this. Thanks :]

Mr. Toni Zulauf Sr. · Accepted Answer

Use the law of conservation of momentum, i.e.,M1V1 + M2V2 = M1V3 + M2V4whereM1 = 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 = 0V3 = final speed of bullet = 435 m/sec (given)V4 = final speed of blockSubstituting appropriate values,0.006(896) + 0 = 0.006(435) + 1.25(V4)and solving for "V4"V4 = 2.21 m/sec.Hope this helps.

Mr. Toni Zulauf Sr. · Answer

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.

Mr. Toni Zulauf Sr. · Answer

conservation of momentumFind 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.

Mr. Toni Zulauf Sr. · Answer

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^2Doug

Mr. Toni Zulauf Sr. · Answer

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.

Mr. Toni Zulauf Sr. · Answer

Impulse = m(vf - vo) = 0.006(435 - 896) = - 2.766 kg-m/sv_block = 2.2128 m/s

Sigrid O'Kon

Bullet through block problem?

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