<p>A child\'s game consists of a block that attaches to a table with a sunction

Question

<p>A child's game consists of a block that attaches to a table with a sunction cup, a spring connected to that block, a ball, and a launching ramp. By compressing the spring, the child can launch the ball up the ramp. the spring has a spring constant k, the ball has a mass m, and the ramp raises a height h. the spring is compressed a distance s in order to launch the ball. when the ball leaves the launching ramp its velocity makes an angle THETA with respect to the horizontal. Calculate the velocity of the ball when it just leaves the launching ramp (both magnitude and direction. Be sure to specify your coordinate system.)</p>

Explanation / Answer

Conservation Energy (on the table h = 0): Start. Only spring potential energy E1 = 0.5*K*S² = 0.5*1000*(0.04)² E1 = 0.8 J (4 cm = 0.04 m) Top of the ramp. Here the ball has potential and kinetic energy: E2 = m*g*h + 0.5*m*V² m = 0.055 kg g = 10 m/s² h = 15 cm = 0.15 m. The conservation gives E1 = E2 0.8 = m*g*h + 0.5*m*V² 0.8 = 0.055*10*0.15 + 0.5*0.055*V² 0.7175 = 0.5*0.055*V² V = 5.1 m/s. X direction (horizontal): natural ball direction parallel to the table. Y direction (vertical): from bottom to top. Vx = V*cos(?) Vy = V*sin(?) On the floor the ball has energy E3 = m*g*h + 0.5*m*Vf² (Vf = final velocity). Now I have the zero of height on the table and the floor has h = -1.2 m. E3 = E2 = E1 0.8 = -1.2*10*0.055 + 0.5*m*Vf² 1.46 = 0.5*0.055*Vf² Vf = 7.28 m/s.

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