If I understand correctly, the word average implies the average of forces acting during a certain time.
$$F_{avg}=\frac{\int_{t_i}^{t_f}{F(t)\,dt}}{t_f-t_i}$$
The teacher gave us tasks with answers to prepare for tomorrow test from his book. Here is one of them:
- A pellet gun fires 10 pellets of 2.0 g, per second with a speed of 500 m/s. The pellets are
stopped by a rigid wall. What area) the magnitude of the linear momentum of each pellet?
b) the kinetic energy of each pellet?
c) the magnitude of the average force on the wall from the stream of the pellets?
d) If each pellet is in contact with the wall for 0.55 ms, what is the magnitude of the
average force on the wall from each pellet during contact?e) Why these two average forces calculated here are so different?
Answer:
a) Linear momentum of each pellet
$$p=mv=0.002*500=1kg*m/s$$
b) Kinetic energy of each pellet is
$$W=0.5*0.002*500=0.5J$$
c) Ten pellets are stuck into the wall in a second, so the average force is
$$F=\frac{\Delta p}{\Delta t}=\frac{10*1}{1}=10N$$ d) The force during the contact is
$$F=\frac{\Delta p}{\Delta t}=\frac{1}{0.00055}=1820N$$e) The difference is due of the long (0.1 s) delay between the bullets
I think, there's a typo in b. $K=\frac{mv^2}{2}=0.002*500^2*0.5=250J$
$\int_{t_i}^{t_f}{F(t)\,dt}$ is all force exerted on wall by pellets. By third Newton's law it equals to all force, the wall exerted on pellets.
$|\int_{t_i}^{t_f}{F(t)\,dt}|=m_p|\int_{t_i}^{t_f} a(t) \, dt|=|m_{p}(v_{p_i}-v_{p_f})|=|\Delta p|$
I think that $v_{p_i}=-v_{p_f}$ because nothing is said about energy loss.
Then, under (If $F_{avg}$ is calculated per second) c): $|F|=\frac{10|F_p|}{1}=\frac{10*0.002*2*500}{1}=20N$
Conditions under d imply that $t_f-t_i=0.00055$
It does not affect $\Delta p$
Then, previous answer can be divided by 10 (to calculate result for one pellet) and by 0.00055
Answer is 3640N
Then, under e delay between shots can't affect the result. It changed because we changed $\Delta t$ we consider
Best Answer
It should be $$ \begin{align} \text{average force} \times \text{total time} &= \text{total change in momentum}\\ F_{ave} \times (1\, \mathrm{s}) &= 10 \times (1 \,\mathrm{kg\, m/s}) \end{align} $$ So $F_{ave}=10 \,\mathrm{N}$, in agreement with your instructor's answer.
Each impact has $$ F\times(0.00055 \,\mathrm{s}) = (1 \,\mathrm{kg\, m/s}) $$ So $F = 1818.2 \,\mathrm{N}$, in agreement with your instructor's answer.
The average force is explained as a constant force applied over the total time that results in the same change in momentum as all the individual varying forces.