A boy climbs to the roof of his house and drops a 0.40 kg rubber ball from the 7.5-meter peak of the house.
How much energy was lost in the process?
The bounce takes place over a 170 ms time interval. What is the impulse given to the ball during this time?
To calculate the energy lost in the process, we need to determine the potential energy of the ball at the peak of the roof and subtract the kinetic energy of the ball just before it hits the ground.
The potential energy of an object is given by the formula: PE = mgh, where m is the mass of the object, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height from which the object is dropped.
In this case, the mass of the rubber ball is 0.40 kg, and it is dropped from a height of 7.5 m. Thus, the potential energy at the peak is:
PE = (0.40 kg) * (9.8 m/s^2) * (7.5 m) = 29.4 J
Now, let's calculate the kinetic energy just before the ball hits the ground. The kinetic energy of an object is given by the formula: KE = 0.5 * m * v^2, where v is the velocity of the object.
To find the velocity of the ball just before it hits the ground, we can use the kinematic equation: v = u + gt, where u is the initial velocity (which is 0 as the ball is dropped), g is the acceleration due to gravity, and t is the time it takes to hit the ground.
Since we don't have the time the ball takes to hit the ground, let's first calculate it using the equation: h = ut + 0.5gt^2, where h is the height and u is the initial velocity.
Rearranging the equation, we have: t^2 + (2u/g) * t - (2h/g) = 0.
Substituting the given values, we have:
t^2 + 0 - (2 * 7.5 / 9.8) = 0
t^2 - 1.53 = 0
Solving this quadratic equation, we find that t is approximately 1.24 s (taking the positive root).
Now, using the equation v = u + gt, and substituting the values, we find:
v = 0 + (9.8 m/s^2) * (1.24 s) = 12.15 m/s
Finally, let's calculate the kinetic energy just before the ball hits the ground:
KE = 0.5 * (0.40 kg) * (12.15 m/s)^2 = 29.5 J
To find the energy lost in the process, we subtract the kinetic energy from the potential energy:
Energy lost = PE - KE = 29.4 J - 29.5 J ≈ -0.1 J
Therefore, approximately 0.1 Joules of energy were lost during the process.
Now, let's calculate the impulse given to the ball during the 170 ms time interval.
Impulse is defined as the change in momentum of an object. The formula for impulse is: Impulse = Change in momentum = m * (v2 - v1), where m is the mass of the object, v2 is the final velocity, and v1 is the initial velocity.
In this case, the mass of the ball is 0.40 kg, and the initial velocity (u) is 0 m/s (as the ball is dropped), and the final velocity (v2) is the velocity just before the bounce.
Since the question mentions that the bounce takes place over a 170 ms time interval, we can assume that the ball comes to rest after the bounce, so the final velocity is 0 m/s.
Therefore, the impulse given to the ball during this time can be calculated as:
Impulse = (0.40 kg) * (0 m/s - 12.15 m/s) = - 4.86 kg·m/s
Therefore, the impulse given to the ball during this time is approximately -4.86 kg·m/s.