To determine how many bullets the man can fire per second without exceeding the maximum force he can exert, we need to compute the force applied due to the momentum change of the bullets being fired.
The momentum
p of a single bullet is given by:
p=mâ‹…vwhere
m is the mass of the bullet and
v is the velocity of the bullet. Here,
m=30g=0.03‌kg (since
1g=0.001‌kg ) and
v=1000m∕ s.
Now we calculate the momentum of a single bullet:
p=0.03‌kg⋅1000m∕ s=30‌kg⋅m∕ sThe rate of change of momentum gives us the force exerted by the bullets being fired. If
F is the force and
n is the number of bullets fired per second, the force is equal to the rate of change of momentum:
F=n⋅‌Since
∆p is the change in momentum per bullet and
∆t=1 second (because
n is measured in bullets per second), the force exerted by
n bullets is:
F=n⋅‌=n⋅30N We know the man can exert a maximum force of
Fmax=300N. Therefore, the maximum number of bullets (
nmax ) that can be fired per second without exceeding the maximum force is obtained by equating the two:
300N=nmaxâ‹…30NSolving for
nmax :
nmax=‌=10‌ bullets ‌∕ sThus, the man can fire at most 10 bullets per second. Hence, the correct answer is:
Option C: 10