A mass \(m\) is attached to a mass \(M\) by a light string that passes over a frictionless pulley of negligible mass, so that it can pull \(M\) up the incline. The incline is frictionless and tilted at an angle of \(\theta=30^{\circ}\) from the horizontal. \(M\) is initially at rest at the bottom of the incline; then \(m\) is released. When \(M\) reaches the top it hits a stop and launches a small ball of negligible mass from a height \(h=1 \mathrm{~m}\) above the starting position. The ball lands on a shelf at the same height at a range \(R=1.8 \sqrt{3} \mathrm{~m}\) from its launch point. Assume the acceleration of gravity is \(g=10 \mathrm{~m} / \mathrm{s}^2\). cant copy graph a) What is the launch speed of the ball? b) What is the acceleration of \(M\) ? c) What is the mass \(M\) in terms of \(m\) ? d) If the ball is given a larger mass, will \(R\) increase, decrease, or stay the same? Explain your reasoning. e) The shelf is removed and the experiment repeated. This time the ball falls to the floor. If \(H=1 \mathrm{~m}\), what is the time the ball spends in the air? f) What is the new range, \(R\) ?

To find the launch speed of the ball, we must first find the time it spends in the air. We know that the ball lands at a height \(h = 1\) m. Since we only need the time it takes to move vertically and the motion is symmetric (up and down), we can use the following equation to find the time it takes to fall: \( h = \frac{1}{2}gt^2 \) Solving for \(t\), we get: \(t^2 = \frac{2h}{g}\) \(t = \sqrt{\frac{2(1)}{10}}\) \(t = \sqrt{\frac{2}{10}}\) \(t = \frac{\sqrt{2}}{√{10}}\) s Now that we have the time it takes, we can find the launch speed of the ball using the range \(R = 1.8\sqrt{3}\) m: \( R = v_{launch} \times t \times cos(θ) \) Solving for \(v_{launch}\), we get: \(v_{launch} = \frac{R}{t \times cos(θ)}\) \(v_{launch} = \frac{1.8\sqrt{3}}{(\frac{\sqrt{2}}{\sqrt{10}}) \times \sqrt{3}/2}\) \(v_{launch} = \sqrt{20}\) m/s

We know that \(m\) pulls \(M\) up the incline, so when \(M\) reaches the top, the tension force in the string must balance the gravitational force acting on \(m\). Using Newton's second law, we have: \(ma = T - mg\) Adding free-body equations for \(m\) and \(M\), we get: \(m \times \frac{du}{dt} = T - mg \quad \text{and} \quad M \times \frac{dv}{dt} = mg - T \quad (1)\) Where \(a\) represents the acceleration of \(M\), \(u\) and \(v\) are the velocities of \(m\) and \(M\), and \(T\) is the tension force in the string. Note that these two equations describe the motion of the masses. We can solve these simultaneous equations to find \(M\)'s acceleration using elimination. First, we need to write the equations in terms of the common variables and parameters. To accomplish this, we'll differentiate the first equation with respect to time. This yields: \(\frac{d^2u}{dt^2} = \frac{d^2v}{dt^2} - g \quad (2)\) To eliminate \(T\), we'll subtract the second equation from the first, which gives us: \(M \times \frac{d^2v}{dt^2} - m \times \frac{d^2u}{dt^2} = Mg\) Now we know, from equation (2), that \(\frac{d^2u}{dt^2} = \frac{d^2v}{dt^2} - g\). Then, \(M \times \left(\frac{d^2v}{dt^2}\right) - m\left(\frac{d^2v}{dt^2} - g\right) = Mg\) Now we term \(\frac{d^2v}{dt^2}\) as the acceleration of \(M\), i.e., \(a\). \(Ma - ma + mg = Mg\) Solving for \(a\), we get: \(a = \frac{mg}{M+m}\)

We are given that the total height \(H\) of the setup is 1 meter. As mass \(M\) moves some distance up the incline, mass \(m\) must move the same distance vertically downward (\(d\)). The relationship between \(H\), \(d\), and \(θ\) using the sine function is: \(H=d \times sin(θ)\) We will now use conservation of energy with both masses \(m\) and \(M\) to find the mass \(M\) in terms of \(m\) using the potential energy gained by \(M\) equals the potential energy lost by \(m\): \(Mgd = mgh\) Substituting the expression for \(H\) we got earlier, we obtain: \(Mgd = mgd \times sin(θ)\) Dividing both sides by \(gd\), we get: \(M = m \times sin(θ)\) Since \(θ = 30°\), we have: \(M = m \times \frac{1}{2}\)

The range of the projectile, \(R\), depends on the launch speed of the ball and the angle of launch. The mass of the ball is not a factor in determining the range. Therefore, if the mass of the ball is increased, the range \(R\) will not change, it will stay the same.

When the ball falls directly to the floor, the total height it falls is \(H\). Using similar analysis as in part a, the time the ball spends in the air is given by the following equation: \( H = \frac{1}{2}gt^2\) Solving for \(t\), we get: \(t^2 = \frac{2H}{g}\) \(t = \sqrt{\frac{4}{10}}\) \(t = 2\sqrt{\frac{1}{5}}\) s

To find the new range, \(R\), we can use the formula we used in part a: \( R = v_{launch} \times t \times cos(θ) \) Now, we substitute the values for \(v_{launch}\) and \(t\) found in previous parts: \(R = \sqrt{20} \times (2\sqrt{\frac{1}{5}}) \times (\frac{\sqrt{3}}{2})\) \(R = \frac{4\sqrt{3}}{\sqrt{5}} = 1.8\sqrt{3}\) m The new range of the ball, when it lands directly on the floor, is \(1.8\sqrt{3}\) m.