Newton's laws of motion: A gun firing bullets....

Since weight is a force, I take this to mean the force is equal to the weight of a 5g mass. This is where the value for g comes in.O yes, now i get it.n* 0.01= 10 *(5/1000) (mass in kg=5/1000kg)n comes out to be 5, which is the correct answer.

Homework Statement

A gun fires bullets each of mass 1g with velocity of 10m/s by exerting a constant force of 5 g weight. Find the number of bullets fired per second (take g=10 m/s2).

Homework Equations

F=ma
M1U1 + M2U2 = M1V1 + M2V2 (conservation of linear momentum)

The Attempt at a Solution

I am not sure why and how the concept of acceleration due to gravity is to be used here, are the bullets being fired vertically upwards? Any help is appreciated.

Last edited by a moderator:

Homework Statement

A gun fires bullets each of mass 1g with velocity of 10m/s by exerting a constant force of 5 g weight. Find the number of bullets fired per second (take g=10 m/s2).

Homework Equations

F=ma
M1U1 + M2U2 = M1V1 + M2V2 (conservation of linear momentum)

The Attempt at a Solution

I am not sure why and how the concept of acceleration due to gravity is to be used here, are the bullets being fired vertically upwards? Any help is appreciated.
Welcome to the PF.

There are some issues with the problem statement. Is there a figure that goes with it? As you say, for a gun firing bullets horizontally, it kind of makes no sense. Also, "5g" is not a weight, it's a mass. Did you copy the problem statement exactly?

by exerting a constant force of 5 g weight
I assume this is the average force (assumed constant) that the gun exerts on the stream of bullets. (To get an actual force, convert the mass to its equivalent weight using W = mg.)

Yes, i have copied it correctly and there are no diagrams or figures.

Doc Al said:
I assume this is the average force (assumed constant) that the gun exerts on the stream of bullets. (To get an actual force, convert the mass to its equivalent weight using W = mg.)
Will i assume g weight as g force?

Please tell me the method to do it

How does force relate to momentum?

Force=rate of change of momentum
F = m(v-u)/t

Force=rate of change of momentum
Good. And each bullet fired has how much momentum? So how many bullets per second are required to produce the required force?

momentum of each bullet = 1(10-0) g m/s
rate of change of momentum of each bullet = 10/1 or simply 10 g/ms2 or equivalently 0.01kg m/s2
rate of change of momentum of n bullets = n*0.01 kg m/s2
now this is equal to some force
but what will be that force?

now this is equal to some force
but what will be that force?
That force is given in the problem statement:
a constant force of 5 g weight

5 g weight ?
do we have to convert it to Newton or can we directly use it?
and is my rest of the answer correct?

" ... exerting a constant force of 5 g weight. "
Since weight is a force, I take this to mean the force is equal to the weight of a 5g mass. This is where the value for g comes in.

O yes, now i get it.
n* 0.01= 10 *(5/1000) (mass in kg=5/1000kg)
n comes out to be 5, which is the correct answer.
Thanks Merlin3189 and Doc Al for helping me out. :)
thank you very much

berkeman
berkeman said:
Welcome to the PF.

There are some issues with the problem statement. Is there a figure that goes with it? As you say, for a gun firing bullets horizontally, it kind of makes no sense. Also, "5g" is not a weight, it's a mass. Did you copy the problem statement exactly?
Thanks berkeman.

berkeman said:
for a gun firing bullets horizontally, it kind of makes no sense
No, there's no issue here. The average force exerted by the gun to accelerate the bullets is independent of orientation.
berkeman said:
Also, "5g" is not a weight, it's a mass.
This highlights an ambiguity around "g". Is it here standing for "grams" or gravitational acceleration?
If the latter, the units are missing. E.g. if the mass were just m then we could write mg as a force, but substituting m=5g (grams now) leads to the frightful 5g g, i.e. a force of 5g grams.
But I would say here it means grams, and although that makes "5g" a mass it is actually ok to write "5g weight" because that is a common way of expressing the force experienced by a mass in a given gravitational field.

berkeman