A rubber ball of mass 19.5 g is dropped from a height of 1.85 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 mechanical energy. The change in mechanical energy of the ball is equal to the impulse applied.
First, let's calculate the initial velocity of the ball just before hitting the floor. We can use the equation of motion:
v^2 = u^2 + 2as
Where:
v = final velocity (0 m/s, as the ball changes direction)
u = initial velocity (unknown)
a = acceleration due to gravity (-9.8 m/s^2)
s = vertical displacement (1.85 m)
Rearranging the equation, we have:
0^2 = u^2 + 2(-9.8)(1.85)
0 = u^2 - 2(9.8)(1.85)
0 = u^2 - 35.88
u^2 = 35.88
u ≈ 5.99 m/s (taking the positive value)
Next, let's calculate the final velocity of the ball just after rebounding. We can use the same equation of motion:
v^2 = u^2 + 2as
Where:
v = final velocity (unknown)
u = initial velocity (5.99 m/s, as the ball changes direction)
a = acceleration due to gravity (-9.8 m/s^2)
s = vertical displacement (1.85 m - 1.55 m = 0.3 m)
Rearranging the equation, we have:
v^2 = (5.99)^2 + 2(-9.8)(0.3)
v^2 = 35.88 - 5.88
v^2 = 30
v ≈ √30 ≈ 5.48 m/s (taking the positive value)
The change in velocity during the collision is the difference between the final and initial velocities:
Δv = v - u
Δv ≈ 5.48 - 5.99 ≈ -0.51 m/s
The negative sign indicates a reversal in velocity after the collision.
Finally, the impulse applied to the ball is given by the product of its mass and the change in velocity:
Impulse = mass × Δv
Impulse ≈ 0.0195 kg × (-0.51 m/s) ≈ -0.010 kgs
Since impulse is a vector quantity, the negative sign indicates a change in direction.
Therefore, the impulse applied to the ball during the collision is approximately -0.010 kgs.