Question

In a 49 s interval, 595 hailstones strike a glass window of an area of 0.954 m at an angle of 25° to the window surface. Each hailstone has a mass of 7 g and a speed of 4.5 m/s. If the collisions are elastic, what is the average force on the window?

Answer 1

Average  force on the window: 0.32 N

Explanation:

The average force exerted on the window is given by Newton's second law

$F=\frac{\Delta p}{\Delta t}$

where

$\Delta p$ is the net change in momentum of the hailstones in a time interval of $\Delta t$

In order to find the change in momentum, we have to consider only the component of the hailstone's momentum perpendicular to the window, therefore:

$p_i =m u sin \theta$ is the initial momentum of one hailstone, with

m = 7 g = 0.007 kg is the mass

$u=4.5 m/s$ is the initial speed

$\theta=25^{\circ}$ is the angle with the window

The final momentum is

$p_f = mv sin \theta$

where

v = 4.5 m/s is the final speed (the  collision is elastic so the speed is equal, while the direction changes)

$\theta=-25^{\circ}$ (after the rebound, the direction has changed)

So the change in momentum of 1 hailstone is

$\Delta p = mv sin(-25^{\circ})-mu sin(25^{\circ})=-2mu sin(25^{\circ})=-0.0266 kg m/s$

We are interested only in the magnitude, so

$\Delta p = 0.0266 kg m/s$

There are 595 hailstones hitting the window in 49 s, so the total change in momentum is

$\Delta p = 595\cdot 0.0266 = 15.8 kg m/s$

And therefore, the average force on the window is

$F=\frac{\Delta p}{\Delta t}=\frac{15.8}{49}=0.32 N$

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