A ball of mass 0.140 kg is dropped from rest from a height of 1.25 m. It rebounds from the floor to reach a height of 0.875 m. What impulse was given to the ball by the floor?magnitude
Question
A ball of mass 0.140 kg is dropped from rest from a height of 1.25 m. It rebounds from the floor to reach a height of 0.875 m. What impulse was given to the ball by the floor?magnitude
Solution
To solve this problem, we need to use the principle of conservation of momentum and the formula for kinetic energy.
Step 1: Calculate the velocity of the ball just before it hits the floor. We can use the equation of motion: v² = u² + 2gs, where v is the final velocity, u is the initial velocity (which is 0 because the ball is dropped), g is the acceleration due to gravity (9.8 m/s²), and s is the distance (1.25 m).
v² = 0 + 29.81.25 v = sqrt(29.81.25) v = 4.95 m/s
Step 2: Calculate the velocity of the ball just after it rebounds. We can use the same equation of motion, but this time the distance s is the rebound height (0.875 m).
v'² = 0 + 29.80.875 v' = sqrt(29.80.875) v' = 4.14 m/s
Step 3: Calculate the change in momentum (which is the impulse given to the ball by the floor). The momentum before the ball hits the floor is (mass * velocity) = 0.140 kg * 4.95 m/s = 0.693 kgm/s. The momentum after the ball rebounds is (mass * velocity) = 0.140 kg * 4.14 m/s = 0.579 kgm/s.
The change in momentum (impulse) is the difference between these two momenta, but we must take into account that the direction of the velocity changes by 180 degrees upon impact, so the impulse is:
Impulse = momentum_after - (-momentum_before) Impulse = 0.579 kgm/s - (-0.693 kgm/s) Impulse = 1.272 kg*m/s
So, the impulse given to the ball by the floor is 1.272 kg*m/s.
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