Question

two bullets of the same size, mass and horizontal velocity are fired at identical blocks, only one is made of steel and the other is made of rubber. the steel bullet has a perfectly inelastic collision with the block, while the rubber bullet has an elastic collision. which bullet is more likely to knock over the block, or are both equally likely to do so? justify your choice based on physics principles. make an initial posting before replying to other threads. reply to at least one of your classmates.

Explanation:

According to the principle of conservation of momentum, in a collision, the change in momentum of the bullet is related to the force exerted on the block. For a perfectly inelastic collision (steel bullet), the bullet and block move together after the collision. The change in momentum of the steel bullet is \(\Delta p_{steel}=m(v - V)\) (where \(v\) is the initial velocity of the bullet and \(V\) is the common velocity of the bullet - block system after the collision). For an elastic collision (rubber bullet), the bullet rebounds. The change in momentum of the rubber bullet is \(\Delta p_{rubber}=m(v-(-v'))\) (where \(v'\) is the velocity of the rubber bullet after the collision, and if we assume the block is much more massive than the bullet and the collision is elastic, approximately \(v' = v\) in magnitude for a head - on collision). So \(\Delta p_{rubber}=m(v + v)=2mv\) (approximate if the block is very massive), while for the steel bullet \(\Delta p_{steel}=mv\) (if the block is very massive, \(V\approx0\)). According to Newton's third law, the force exerted on the block is related to the change in momentum of the bullet (\(F=\frac{\Delta p}{\Delta t}\)). A larger change in momentum of the bullet (for the rubber bullet) implies a larger force exerted on the block.

Answer:

The rubber bullet is more likely to knock over the block.