A dispenser releases small identical uniform disks...

The vertical force exerted on the wedge (and thus the reading on the scale) after $t = 150 \, \text{seconds}$ is approximately $1044 \, \text{N}$. This force is downward, which is why the reading on the scale is negative, indicating a net downward force from the wedge.

The force measured by the scale is vertical and acts along the vertical direction, meaning it’s in the vertical direction (perpendicular to the ground), not normal to the wedge or horizontal. This is because the scale is oriented vertically to measure the force exerted on it by the wedge and the disks. The normal force on the wedge due to the disks is resolved into vertical and horizontal components, but the scale only measures the vertical component of the force.

Let me explain the steps involved in solving this problem:

Breakdown of the Problem:

1. Disks' motion:

   • The disks fall from rest at height $h = 50 \, \text{m}$, so their velocity upon hitting the wedge can be determined by the equation of motion under gravity.
   • The velocity of each disk just before impact with the wedge is $v = \sqrt{2gh}$, where $g = 9.81 \, \text{m/s}^2$ is the acceleration due to gravity.

2. Collision with the wedge:

   • The disks hit the wedge at an angle $\theta = 10^\circ$, with the velocity decomposed into components parallel and perpendicular to the surface of the wedge.
   • The velocity component normal (perpendicular) to the wedge will change based on the coefficient of restitution $\alpha = 0.95$, which gives the ratio of the relative speed after collision to the relative speed before the collision along the normal direction.
   • The velocity parallel to the wedge remains unchanged as there is no friction.

3. Force from each disk:

   • After each disk hits the wedge, there is an impulse that changes the velocity of the wedge. The velocity change in the normal direction will contribute to a force on the wedge, which is transferred to the scale.

4. Rate of disks falling:

   • The dispenser releases 50 disks per second, so in 150 seconds, $50 \times 150 = 7500$ disks will be released.

5. Calculation of total force on the wedge:

   • The net vertical force on the wedge is the sum of the forces imparted by each disk's collision over time. This is where the horizontal and vertical components of the velocity after the collisions come into play.

Summary:

The force that the scale measures is vertical because the scale is set up to measure the force perpendicular to the ground, not the slope of the wedge. The vertical component of the force after each disk's collision, due to the change in momentum of the wedge, is what is reflected on the scale.

Yes, you are absolutely correct! The mass of the wedge ($M = 500 \, \text{kg}$) plays an important role in this problem. The total force on the scale is affected by both the disks and the wedge. The interaction between the falling disks and the wedge will cause the wedge to move, and that movement contributes to the force measured by the scale.

Let’s go through the steps again with a focus on the wedge's mass, $M = 500 \, \text{kg}$.

Step-by-Step Solution:

1. Velocity of the disk just before impact:

   • The disk falls from a height $h = 50 \, \text{m}$. Using the equation for free fall: $$v = \sqrt{2gh}$$ where $g = 9.81 \, \text{m/s}^2$ is the acceleration due to gravity: $$v = \sqrt{2 \times 9.81 \, \text{m/s}^2 \times 50 \, \text{m}} = \sqrt{981} \approx 31.3 \, \text{m/s}$$ This is the velocity of each disk just before it strikes the wedge.

2. Velocity components when the disk hits the wedge:

   • The disk strikes the wedge at an angle $\theta = 10^\circ$. We decompose the velocity into components:
     • The normal component of the velocity (relative to the wedge surface) is: $$v_{\perp} = v \sin \theta = 31.3 \times \sin(10^\circ) \approx 31.3 \times 0.1736 \approx 5.43 \, \text{m/s}$$
     • The parallel component of the velocity (relative to the wedge surface) is: $$v_{\parallel} = v \cos \theta = 31.3 \times \cos(10^\circ) \approx 31.3 \times 0.9848 \approx 30.8 \, \text{m/s}$$

3. Coefficient of restitution $\alpha = 0.95$:

   • The coefficient of restitution tells us how the normal velocity changes after the collision. The normal velocity after the collision will be: $$v_{\perp}' = -\alpha v_{\perp} = -0.95 \times 5.43 \, \text{m/s} \approx -5.16 \, \text{m/s}$$ This negative sign indicates that the disk rebounds in the opposite direction along the normal.

4. Impulse on the wedge:

   • When each disk collides with the wedge, it exerts an impulse on the wedge. The change in momentum of the disk in the normal direction is:

     $$\Delta p_{\text{disk}} = m(v_{\perp}' - v_{\perp}) = 2 \times (-5.16 - 5.43) = 2 \times (-10.59) \approx -21.18 \, \text{kg} \cdot \text{m/s}$$

     This is the momentum change of one disk in the normal direction, which is transferred to the wedge.

   • The total momentum imparted to the wedge is the same, but it results in the wedge moving slightly. Since the wedge is fixed on a scale, the scale measures the vertical force exerted on it, which is related to the total momentum transfer from all disks.

5. Wedge's motion and force on the scale:

   • The force on the wedge causes it to move, but the scale only measures the vertical component of the force.
   • The total rate of change of momentum delivered to the wedge per second due to the 50 disks falling per second is: $$\text{Total momentum per second} = 50 \times (-21.18) \, \text{kg} \cdot \text{m/s} = -1059 \, \text{kg} \cdot \text{m/s}$$
   • The wedge has mass $M = 500 \, \text{kg}$, so the acceleration of the wedge $a$ due to the momentum transfer is: $$a = \frac{\text{Total momentum per second}}{M} = \frac{-1059}{500} \approx -2.12 \, \text{m/s}^2$$
   • The vertical force exerted on the wedge due to this acceleration is given by: $$F_{\text{vertical}} = M \times a = 500 \times (-2.12) = -1060 \, \text{N}$$
   • The negative sign indicates that the force is downward, which is how the scale detects it.

6. Scale reading after 150 seconds:

   • The reading on the scale measures the net vertical force over time. Over a period of 150 seconds, the scale will register the total vertical force caused by all the disks falling: $$F_{\text{total}} = 1060 \, \text{N} \, (\text{downward force})$$

Thus, the reading on the scale after 150 seconds will be approximately 1060 N downward.

You’re right to ask—my earlier answer was wrong because I used the wrong model.

I (incorrectly) treated the scale reading like “$M a$” from the disks’ momentum flux and ignored two big things:

1. the wedge’s own weight $Mg$, and
2. the fact that no disks leave the (very long) incline, so the number of disks being supported grows with time, adding their (projected) weight to the scale.

What the scale actually measures (time-averaged over a few impacts) is the vertical force:

where $R=50\,\text{s}^{-1}$, $t=150\,\text{s}$, $m=2\,\text{kg}$, $M=500\,\text{kg}$, $\theta=10^\circ$, $\alpha=0.95$, and $v=\sqrt{2gh}$ with $h=50\,\text{m}$.

Numerically:

• $Mg \approx 4.91\times 10^3\ \text{N}$
• $(Rt)\,m g \cos^2\theta \approx 1.43\times 10^5\ \text{N}$ (from the 7,500 disks now on the incline)
• impact term $\approx 5.9\times 10^3\ \text{N}$

So the correct averaged reading at $t=150\ \text{s}$ is

That’s why the answer changed: the first one omitted the dominant contribution (the accumulated disks’ supported weight) and misused $M$. My bad!