Joe_1234
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A gun fires 6 bullets per second into a target. The mass of each bullet is 3g & the speed of 500 m/s. Find the average force required to hold the gun in position.
The average force required to hold a gun in position while firing 6 bullets per second, each with a mass of 3 grams and a speed of 500 m/s, can be calculated using the impulse equation: F · Δt = m · Δv. The momentum of a single bullet is determined by multiplying its mass (0.003 kg) by its velocity (500 m/s), resulting in a momentum of 1.5 kg·m/s. The total momentum for 6 bullets is 9 kg·m/s, and since this occurs over one second, the average force is 9 Newtons.
PREREQUISITESPhysics students, engineers, and anyone interested in the mechanics of firearms and projectile motion will benefit from this discussion.
Tnxskeeter said:Impulse equation ...
$F \cdot \Delta t = m \cdot \Delta v$
HallsofIvy said:You titled this "momentum" so it looks like you already knew the basic idea. The momentum of a single bullet is "mass times velocity" and you are given both of those. Of course, the momentum of 6 bullets is 6 times the momentum of a single bullet. Since this all happens in one second, divide by one second to get the force.
Serious question - is it harder to type “thanks” or “tnx “ on your phone? On mine the latter will auto correct to something else so it’s actually harder.Joe_1234 said:Tnx