91Ó°ÊÓ

1. Number of pellets is, n = 10
2. Mass r of pellets is, m = 2g = 0.002 kg
3. Speed of pellets is,$v=500\frac{\mathrm{m}}{\mathrm{s}}$

Use the basic formula of momentum, which is the product of mass and velocity. Then to find kinetic energy, use the basic formula of mass and square of velocity. Then, use the basic formula of the average force, which is ratio of momentum and time.

Formulae are as follow:

$$\begin{gathered} P=mv \\ KE=\frac{1}{2}m{v}^2 \\ F=\frac{P}{t} \end{gathered}$$

Here, P is momentum, m is mass, v is velocity, KE is kinetic energy, F is force and t is time.

To calculate magnitude of momentum of each pellet, use the following formula,

$$\begin{gathered} P=mv \\ P=0.002×500 \\ P=1\frac{\mathrm{kg}.\mathrm{m}}{\mathrm{s}} \end{gathered}$$

Hence,the magnitude of momentum of each pellet is $1\frac{\mathrm{kg}.\mathrm{m}}{\mathrm{s}}$.

Kinetic energy of each pellet,

$$\begin{gathered} KE=\frac{1}{2}×m×v^2 \\ KE=0.5×0.002×{500}^2 \\ KE=250\mathrm{J} \end{gathered}$$

Hence, the kinetic energy of each pellet is 250J.

Average force on n pallet is,

$$\begin{gathered} F_{avg}=\frac{nP}{t} \\ {F}_{avg}=\frac{10×1}{1} \\ {F}_{avg}=10\mathrm{N} \end{gathered}$$

Hence,the magnitude of average force on the wall from stream of pellets is 10N.

Average force when t = 0.6 s:

$F_{avg}=\frac{P}{t}$

Substitute the values and solve as:

$$\begin{gathered} F_{avg}=\frac{1}{0.6×{10}^{-3}} \\ {F}_{avg}=1666.67 \end{gathered}$$

In two significant figures,

$F_{avg}=1700\mathrm{N}$

Hence,the magnitude of average force if t = 0.6 sec is 1700 N.

Average forces are different because the contact time is different. For case 1, it isand for case 2, it is 0,6 s.

Therefore, the momentum, K.E and average force on an object can be found using the corresponding formulae.