91Ó°ÊÓ

First, we need to find the time taken for the ball to reach 1m above the ground, using its vertical motion. The initial vertical velocity of the ball is 0, and the vertical distance to be covered is 2.5m (since the ball is projected horizontally from a height of 3.5m). Use the equation of motion: \(h = v_0 t + \frac{1}{2}at^2\) Here, \(h = -2.5 \mathrm{~m}\) (downwards), \(v_0 = 0\), and \(a = 9.8 \mathrm{~m/s^2} \) (acceleration due to gravity). Plugging the values in the equation, we get: \(-2.5 = 0 - \frac{1}{2}(9.8)t^2\) Solve for t: \( t^2 = \frac{(-2.5)(2)}{9.8}\) \( t \approx 0.714\mathrm{s}\)

Using the horizontal motion of the ball, calculate the displacement of the ball when it collides with the wedge. The horizontal component of the velocity of the ball remains constant throughout the motion since there is no air resistance. So, we can simply use the formula: \(x = vt\), where \(x\) is the horizontal displacement, \(v\) is the horizontal velocity (7 m/s) and \(t\) is the time calculated in step 1. Plugging the values, we get: \(x = (7)(0.714)\mathrm{m}\) \(x \approx 5\mathrm{m}\) (to the nearest meter)

Before the collision, the wedge is at rest, so its velocity is 0. The velocity of the ball just before the collision can be found using the Pythagorean theorem. We have the horizontal (\(v_x = 7\mathrm{m/s}\)) and vertical (\(v_y = gt\)) velocities. Plugging the values, we get: \(v_y = (9.8)(0.714)\mathrm{m/s}\) \(v_y \approx 7\mathrm{m/s}\) (to the nearest m/s, downwards) Now, we can find the total velocity of the ball before the collision: \(v_{ball} = \sqrt{v_x^2 + v_y^2}\) \(v_{ball} = \sqrt{(7^2) + (7^2)}\mathrm{m/s}\) \(v_{ball} = 7\sqrt{2}\mathrm{m/s}\)

As the collision is elastic, we can now apply conservation of momentum and conservation of kinetic energy to find the velocities of the wedge and the ball after the collision. Since there is no friction, the momentum of the system before and after the collision will be conserved only in the horizontal direction: Before collision: \(m_1v_{1x} + m_2v_{2x} = (1)(7) + (3)(0) = 7\mathrm{kg \, m/s}\) After collision: \(m_1v'_{1x} + m_2v'_{2x} = 7\mathrm{kg \, m/s}\) Since the collision is elastic, the coefficient of restitution is 1, which means: \(e = \frac{(v'_{2x} - v'_{1x})}{(v_{1x} - v_{2x})} = 1\) Since there are 2 unknowns, \(v'_{1x}\) and \(v'_{2x}\), we need to apply the conservation of kinetic energy: Before collision: \(\frac{1}{2}m_1v_{1}^2 + \frac{1}{2}m_2v_{2}^2 = \frac{1}{2}(1)(7\sqrt{2})^2 + \frac{1}{2}(3)(0)^2 = 49\mathrm{J}\) After collision: \(\frac{1}{2}m_1(v'_{1x})^2 + \frac{1}{2}m_2(v'_{2x})^2 = 49\mathrm{J}\) Solving these equations simultaneously, we get: \(v'_{1x} \approx 1.75\mathrm{m/s}\) \(v'_{2x} \approx 1.75\mathrm{m/s}\)

Since the velocities of the ball and the wedge after the collision have the same magnitude in the horizontal direction and they are moving along the plane of the wedge, we can now present the final answer: The velocity of the ball after the collision is \(1.75\mathrm{m/s}\) along the incline of the wedge (upwards). The velocity of the wedge after the collision is \(1.75\mathrm{m/s}\) horizontally.