Question

In a 51 s interval, 889 hailstones strike aglass window of area 0.987 m^2 at an angle 35°to the window surface. Each hailstone has amass of 6 g and speed of 6.9 m/s.If the collisions are elastic, find the averageforce on the window.Part A. Answer in units of N.Part B. Find the pressure on the window

Answer 1

We are give that at an interval of 51 seconds, 889 hailstones hit a window. To determine the force we need to do a balance of momentum, like this:

`m_1v_(1h)+Ft=m_2v_(2h)`

Where:

`\begin{gathered} m_1,m_2=\text{ mass} \\ F=\text{ force} \\ t=\text{ time} \\ v_(1h),v_(2h)=horizontal-components-of-velocities_{} \end{gathered}`

Assuming that the window is completely vertical, then we have:

Therefore, we have that the horizontal component of the velocity when hitting the window is:

`v_(1h)=v_1\sin 35`

And the horizontal component of the velocity after hitting the window is:

`v_(2h)=v_(1h)\sin y`

Where the angle "y" is unknown.

Substituting in the balance of momentum we get:

`m_1v_1\sin 35+Ft=m_2v_2\sin y`

Since the masses are equal for each hailstone, we have:

`m_1=m_2=m`

substituting we get:

`mv_1\sin 35+Ft=mv_2\sin y`

Now, we need to determine the value of the angle after the hailstone hits the window. we are given that the collision is elastic, therefore, we have:

`e=1`

Where "e" is the coefficient of restitution.

The coefficient of restitution is the quotient between the final and initial velocities, therefore, we have:

`(v_(1h))/(v_(2h))=1`

Substituting the values in terms of sines:

`(v_2\sin y)/(v_1\sin 35)=1`

Multiplying both sides by the denominator we get:

`v_2\sin y=v_1\sin 35`

Now, we divide both sides by "sin(y)":

`v_2=(v_1\sin35)/(\sin y)`

Now, if we consider the balance of momentum in the vertical direction, we have that:

`mv_(1v)=m_{}v_(2v)`

In this case, we don't have a term associated with a force since only the horizontal component of the momentum is associated with the impulse. The verticals components of the velocities are:

`\begin{gathered} v_(1v)=v_1\cos 35 \\ v_(2v)=v_2\cos y \end{gathered}`

Substituting we get:

`mv_1\cos 35=m_{}v_2\cos y`

We can cancel out the masses:

`v_1\cos 35=v_2\cos y`

Now, we substitute the value we determine of the second velocity in terms of sines, we get:

`v_1\cos 35=((v_1\sin35)/(\sin y))_{}\cos y`

Now, we cancel out the velocity 1:

`\cos 35=((\sin 35)/(\sin y))_{}\cos y`

Now, we divide both sides by sin(35):

`(\cos35)/(\sin35)=(\cos y)/(\sin y)`

We have that:

`(\cos\theta)/(\sin\theta)=\cot \theta`

Therefore, we have:

`\cot 35=\cot y`

Therefore:

`35=y`

This means that the hailstones have the same angle after they hit the window.

Now, we go back to the equation for the second velocity and substitute the value of "y":

`v_2=(v_1\sin 35)/(\sin 35)=v_1`

Therefore, the velocities have the same magnitude, but they have opposite directions, therefore.

This means that:

`v_1=-v_2`

Substituting in the balance of momentum:

`-mv_{}\sin 35+Ft=mv_{}\sin 35`

We have taken the velocities in the direction of the window as negative and the direction leaving the window as positive

Now, we solve for the force "F":

`Ft=mv_{}\sin 35+mv_{}\sin 35`

Adding like terms:

`Ft=2mv\sin 35`

Now, we divide both sides by "t":

`F=\frac{2mv_{}\sin35}{t}`

This is the force for one single hailstone, to determine the total force we need to multiply the mass by the 889 hailstones, therefore:

`F=\frac{2(889m)v_{}\sin 35}{t}`

Now, we plug in the values:

`F=\frac{2(889)(0.006kg)(6.9(m)/(s))_{}\sin 35}{51}`

Solving the operations:

`F=0.83N`

Therefore, the average force on the window is 0.83 Newtons.

Part B. We are asked to determine the pressure on the window. To do that we will use the following formula:

`P=(F)/(A)`

Where:

`\begin{gathered} P=\text{ pressure} \\ A=\text{ area} \end{gathered}`

Now, we plug in the values:

`P=(0.83N)/(0.987m^2)`

Solving the operations:

`P=0.84(N)/(m^2)`

Therefore, the pressure is 0.84 N/m^2