A rubber ball of mass 41.5 g is dropped from a height of 1.75 m onto a floor. The velocity of the ball is reversed by the collision with the floor, and the ball rebounds to a height of 1.55 m. What impulse was applied to the ball during the collision?
To find the impulse applied to the ball during the collision, we can use the principle of conservation of energy. The initial potential energy of the ball when it is dropped is converted into kinetic energy right before it hits the floor, and then it is converted back into potential energy as the ball rebounds.
The potential energy of an object at height h is given by the formula: PE = mgh, where m is the mass of the object, g is the acceleration due to gravity, and h is the height.
The initial potential energy of the ball is: PE_initial = m * g * h_initial = 0.0415 kg * 9.8 m/s^2 * 1.75 m
The final potential energy of the ball after rebounding is: PE_final = m * g * h_final = 0.0415 kg * 9.8 m/s^2 * 1.55 m
Since energy is conserved, the difference in potential energy is equal to the impulse applied during the collision:
Impulse = PE_initial - PE_final = (0.0415 kg * 9.8 m/s^2 * 1.75 m) - (0.0415 kg * 9.8 m/s^2 * 1.55 m)
Therefore, the impulse applied to the ball during the collision is equal to the quantity calculated above.