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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:

  1. 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 are

a) 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

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  • $\begingroup$ Yes to the typo in b) $W=\frac{1}{2} m v^2$ $\endgroup$ – ja72 Apr 9 '15 at 18:30
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    $\begingroup$ Note that the problem statement says that the pellets "are stopped by a rigid wall". So $v_f = 0$. $\endgroup$ – garyp Apr 9 '15 at 18:37
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c) the magnitude of the average force on the wall from the stream of the pellets?

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.

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?

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.

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  • $\begingroup$ to be clear, you agree with the quoted answers. $\endgroup$ – innisfree Apr 9 '15 at 18:52
  • $\begingroup$ Well b) is wrong (obviously) and I disagree with c). The instructor has F=10 and I have F=1 $\endgroup$ – ja72 Apr 9 '15 at 18:58
  • $\begingroup$ @ja oops that was careless of me. $\endgroup$ – innisfree Apr 9 '15 at 18:59
  • $\begingroup$ Thanks for the edits. I don't like $\times$ for multiplication because I used it as cross product often. And $*$ is clunky, so that leaves $\cdot$ or nothing for implicit multiplication. $\endgroup$ – ja72 Apr 9 '15 at 19:01
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    $\begingroup$ No, the total time is one second, during which 10 bullets are fired. $\endgroup$ – innisfree Apr 9 '15 at 19:04

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