A boy jumps from a wall 3m high. What is an estimate of the change in momentum of the boy when he lands without rebounding?
the answer is 5x10^2kg/ms, but how?
what's his mass?
momentum is ... m * v
his velocity is ... √(2gh) = √(2 * 9.8 * 3)
it didn't say:(
so the question is "estimating" his mass to be about 65 kg (143 lbs)
... a reasonable number
but a very poor question
To estimate the change in momentum of the boy when he lands without rebounding, we can use the principle of conservation of energy. This principle states that the total mechanical energy of a system remains constant if no external forces are acting on it.
When the boy jumps from a wall, his potential energy is converted into kinetic energy as he falls. The potential energy is given by the formula mgh, where m is the mass of the boy, g is the acceleration due to gravity, and h is the height of the wall.
Given that the height of the wall is 3m, we can assume the boy's mass is 50kg, and the acceleration due to gravity is approximately 9.8 m/s^2. Therefore, the initial potential energy of the boy is:
Potential Energy = mgh = (50 kg)(9.8 m/s^2)(3 m) = 1470 kg m^2/s^2
As the boy falls, this potential energy is converted entirely into kinetic energy. The kinetic energy of an object is given by the formula (1/2)mv^2, where m is the mass and v is the velocity of the object.
To estimate the velocity of the boy when he lands, we can use the equation of motion:
v^2 = u^2 + 2as
Where u is the initial velocity (which is approximately 0 since the boy jumps from rest), a is the acceleration due to gravity, and s is the distance fallen (which is also 3m).
Plugging in the values, we get:
v^2 = 0^2 + 2(9.8 m/s^2)(3 m) = 58.8 m^2/s^2
Now, using the equation for kinetic energy, we can calculate the final kinetic energy of the boy:
Kinetic Energy = (1/2)mv^2 = (1/2)(50 kg)(58.8 m^2/s^2) = 1470 kg m^2/s^2
Since the potential energy is equal to the final kinetic energy, the change in momentum is equal to the change in kinetic energy. Therefore, the estimate of the change in momentum of the boy is:
Change in Momentum = 1470 kg m^2/s^2 - 0 kg m^2/s^2 = 1470 kg m^2/s^2
In scientific notation, this is written as 1.47 x 10^3 kg m^2/s^2 or 1.47 x 10^3 Ns.