Physics Answer Note #27

Question 1 At the top of a building 125 meters high, a dense ball is shot horizontally such that its initial velocity is 20 m/s. How far from the base of the building will the ball strike the ground? Answer According to physics principles, when an object is projected horizontally with a certain initial velocity, it will continue to move horizontally at a constant velocity until it hits the ground. To find the distance the ball will strike the ground from the base of the building, we can use the equation d = vt, where d is the distance, v is the initial velocity, and t is the time taken to reach the ground. First, we need to determine the time it takes for the ball to reach the ground. This can be found using the equation t = √(2h/g), where h is the height of the building and g is the acceleration due to gravity. In this case, h = 125m and g = 9.8 m/s². Substituting these values, we get t = √(2*125/9.8) = √25.51 = 5.05 seconds. Next, we can use the equation d = vt to find the distance. Since the initial velocity is given as 20 m/s, we have d = (20)(5.05) = 101 meters. This means that the ball will strike the ground 101 meters away from the base of the building. In-depth explanation: When an object is in motion, it experiences two types of forces - a horizontal force and a vertical force. In this scenario, the ball experiences a horizontal force due to its initial velocity, but no vertical force as it is not moving up or down. This means that the ball's horizontal velocity remains constant throughout its motion. When the ball is shot at an angle, its horizontal velocity is affected by gravity, resulting in a curved path. However, in this case, the ball is shot horizontally, which means that gravity only affects the vertical component of its motion. Therefore, the ball will continue to move horizontally at a constant velocity and will strike the ground at the same horizontal distance from where it was shot. Question 2 A .03kg marble is launched from a spring launcher. The marble travels through a series of tracks that extends for 3 meters. It takes the marble approximately 9 seconds to travel the length of track. The marble eventually then rolls off a ramp and becomes a projectile. The angle of the ramp is 60o above the horizontal table. The table is 2 meters above the ground. The marble leaves the end of the ramp with a velocity of 2m/s. At the same time, a constant velocity buggy is traveling toward where the marble should land, with a cup attached to catch the marble. The buggy can travel 10 meters in 4 seconds. The buggy has a mass of 2.2kg. The buggy has a height of 10cm and the cup has a height of 15cm. Answer To accurately answer this question, we will need to apply principles of mechanics and projectile motion. First, let us analyze the motion of the marble as it travels through the tracks and rolls off the ramp. The force that is responsible for launching the marble from the spring launcher is the spring force. According to Hooke's Law, the force exerted by a spring is directly proportional to the displacement of the object (in this case, the marble) from its equilibrium position. Therefore, as the spring is compressed and the marble is launched, the force acting on the marble is also increasing. To calculate the spring constant, we can use the formula k = F/x, where k is the spring constant, F is the force applied by the spring, and x is the displacement of the object. In this case, we know that the mass of the marble is 0.03kg and the displacement is 0.03m (measured from the compressed length of the spring to its equilibrium length), therefore the spring constant is determined to be 1N/m. Next, we need to determine the work done by the spring force on the marble. We can use the formula W = 1/2kx^2, where W is the work done, k is the spring constant, and x is the displacement of the object. In this case, it is important to note that the displacement used in this formula is the distance the object travels, which is 3 meters in this case. Therefore, the work done by the spring force is calculated to be 4.5J. Now, let us consider the motion of the marble as it rolls off the ramp and becomes a projectile. The angle of the ramp is 60o, which means the horizontal component of the marble's velocity is 2cos60o = 1 m/s and the vertical component is 2sin60o = 1.73 m/s. Since the marble has a velocity of 2 m/s and we know the horizontal component, we can use the formula v = u + at to calculate the time it takes for the marble to reach the end of the ramp. Therefore, t = (2 - 1) / 9.8 m/s^2 = 0.1s. Applying the formula v = u + at again, we can calculate the initial velocity of the marble as it rolls off the ramp, which is v = 1 + (9.8)(0.1) = 2.98 m/s. Using this initial velocity, we can further calculate the horizontal range (distance traveled) of the marble, which is R = (v^2)(sin2θ) / g, where θ is the angle of projection, in this case 90o. The range is then calculated to be 2.94m. Now, let us consider the motion of the buggy as it travels towards where the marble will land. We know that the buggy can travel 10 meters in 4 seconds, therefore its velocity is 2.5m/s. This is important in determining when the buggy will catch the marble, as we need to consider its speed relative to the marble's speed. Since the buggy is moving towards the direction of the marble, the relative velocity will be calculated as the sum of their velocities, which is 2 + 2.5 = 4.5m/s. To determine the time it takes for the marble to be caught by the buggy, we can use the formula R = vt, where R is the range calculated previously and t is the time. Therefore, t = 2.94 / 4.5 = 0.65s. This is the time needed for the buggy to reach where the marble will land. To determine the height at which the marble will land, we can use the formula h = uyt + 1/2gt^2, where u is the initial vertical velocity (1.73 m/s), y is the displacement in the direction of motion (equal to the height of the ramp, which is 2m), and t is the time we calculated previously. Therefore, h = (1.73 Question 3 A .03kg marble is launched from a spring launcher. The marble travels through a series of tracks that extends for 3 meters. It takes the marble approximately 9 seconds to travel the length of track. The marble eventually then rolls off a ramp and becomes a projectile. The angle of the ramp is 60o above the horizontal table. The table is 2 meters above the ground. The marble leaves the end of the ramp with a velocity of 2m/s. At the same time, a constant velocity buggy is traveling toward where the marble should land, with a cup attached to catch the marble. The buggy can travel 10 meters in 4 seconds. The buggy has a mass of 2.2kg. Determine how high the marble will go Answer To determine the maximum height reached by the marble, we need to consider the different stages of its motion - from being launched from the spring launcher to becoming a projectile off the ramp. First, let's calculate the initial velocity of the marble as it leaves the spring launcher. Using the equation v= u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time, we can plug in the given values to get: v = 0 + (0.03kg)(9.8m/s^2)(9s) = 2.646m/s This is the velocity of the marble as it reaches the end of the track. Next, we need to calculate the velocity of the marble as it rolls off the ramp and becomes a projectile. We can use the formula vf^2 = vi^2 + 2ad, where vf is the final velocity, vi is the initial velocity, a is the acceleration, and d is the distance. In this case, the distance is the length of the ramp, which is 3 meters. Plugging in the values, we get: vf^2 = (2.646)^2 + 2(9.8)(3) vf = 7.104m/s Now, we can use the equation v= u + at to find the time it takes for the marble to reach its maximum height. The initial velocity is vf from the previous calculation, the final velocity is 0 (since the marble will reach its peak and then fall back down), and the acceleration is -9.8m/s^2 (due to gravity). We get: 0 = 7.104 + (-9.8)t t = 0.72s Therefore, it will take the marble 0.72 seconds to reach its maximum height. Now, we can use the equation s = ut + 0.5at^2 to find the maximum height h. The initial velocity is 7.104m/s, the acceleration is -9.8m/s^2, and the time is 0.72s. We get: h = (7.104)(0.72) + 0.5(-9.8)(0.72)^2 h = 2.54 meters Question 4 Consider swimmers A and B. Both of them begin swimming at t=0s. Swimmer A has a velocity of +2.0 m/s and begins at -10 m, while swimmer B has a velocity of -0.50 m/s and begins at +40 m. Calculate the position of each swimmer after ten seconds. Answer To calculate the position of swimmers A and B after ten seconds, we can use the formula: x = xo + v*t, where x is the final position, xo is the initial position, v is the velocity, and t is the time. For swimmer A, x = -10 m + (2.0 m/s * 10 s) = -10 m + 20 m = 10 m. For swimmer B, x = 40 m + (-0.50 m/s * 10 s) = 40 m - 5 m = 35 m. Therefore, after ten seconds, the position of swimmer A would be 10 m and the position of swimmer B would be 35 m. Question 5 Consider swimmers A and B. Both of them begin swimming at t=0s. Swimmer A has a velocity of +2.0 m/s and begins at -10 m, while swimmer B has a velocity of -0.50 m/s and begins at +40 m. How long will it take each swimmer to reach the +20m mark? Answer According to the given conditions, both Swimmer A and B are starting from different positions and moving at different velocities. To calculate the time it takes for each swimmer to reach the +20m mark, we will need to use the formula of displacement (s) = initial velocity (u) * time (t) + 1/2 * acceleration (a) * time (t)^2. In this case, we can assume that acceleration is constant as both swimmers are moving with a constant velocity. For Swimmer A: Given initial velocity (u) = +2.0 m/s and initial displacement (s) = -10 m, we can rearrange the above formula as t = (s - u*t)/(-1/2*a). Substituting the values, we get t = (20-(-10))/(1/2* 2) = 10/1 = 10 seconds. For Swimmer B: Given initial velocity (u) = -0.50 m/s and initial displacement (s) = +40 m, we can use the same formula as t = (s - u*t)/(-1/2*a). Substituting the values, we get t = (20-40)/(1/2*0.50) = -20/-0.25 = 80 seconds. Therefore, it would take Swimmer A 10 seconds and Swimmer B 80 seconds to reach the +20m mark.