A ball of mass 0.120 kg is dropped from rest from a height of 1.25 m. It rebounds from the floor to reach a height of 0.600 m. What impulse was given to the ball by the floor?
answer in kg·m/s
figure the velocity at impact, and leaving impact.
at impact, velocity= sqrt 2gh
returning, velocity= sqrt 2gh*.6 opposite direction
change velocity= Vi-Vf=sqrt2gh+sqrt(2gh*.6)
impulse= mass*changevelocity, upward direction.
8.78
To find the impulse given to the ball by the floor, we can use the principle of conservation of mechanical energy. The total mechanical energy of the ball-floor system remains constant throughout the motion.
The initial mechanical energy of the ball is given by:
E_initial = m * g * h_initial
where m is the mass of the ball (0.120 kg), g is the acceleration due to gravity (approximated to 9.8 m/s²), and h_initial is the initial height (1.25 m).
The final mechanical energy of the ball is given by:
E_final = m * g * h_final
where h_final is the final height reached by the ball after rebounding (0.600 m).
Since energy is conserved, the impulse given to the ball by the floor can be calculated by the change in mechanical energy:
Impulse = ΔE = E_final - E_initial
Substituting the values, we get:
Impulse = (m * g * h_final) - (m * g * h_initial)
= m * g * (h_final - h_initial)
Now let's calculate the impulse:
Impulse = 0.120 kg * 9.8 m/s² * (0.600 m - 1.25 m)
= 0.120 kg * 9.8 m/s² * (-0.65 m)
= -0.7644 kg·m/s
Therefore, the impulse given to the ball by the floor is -0.7644 kg·m/s. The negative sign indicates a change in momentum in the opposite direction.