Answer to Question #138782 in Physics for Francine

Question

A boy stands on a storey building 3m high and jumps to the ground, he did this twice,the first time he took 0.4s to land on the ground with his legs bend while 0.02s to land with his legs straight but this time he developed a fracture. Assuming he

a) Has a mass of 40 kg, explain with calculations why he developed a fracture in the second jump.

b) using Newton's second and third law of motion prove the law of conservation of linear momentum

Answer 1

a) Let's first find the velocity of a boy just before he lands on the ground. We can do this with the help of Law of Conservation of Energy:

$$PE = KE,$$ $$mgh=\dfrac{1}{2}mv^2,$$ $$v = \sqrt{2gh} = \sqrt{2 \cdot 9.8 \ \dfrac{m}{s^2} \cdot 3 \ m}= 7.7 \ \dfrac{m}{s}.$$

Then, we can calculate the average force exerted on his legs from the impulse-momentum change equation:

$$F_{avg}\Delta t = m\Delta v,$$

here, $F_{avg}$ is the average force exerted on his legs, $m = 40 \ kg$ is the mass of the boy , $\Delta v = 7.7 \ \dfrac{m}{s}$ is the change in velocity, $\Delta t$ is the time he takes to land the ground.

From this equation we can calculate the average force exerted on his legs in both cases of jump:

$$F_{avg1} = \dfrac{m\Delta v}{\Delta t_1} = \dfrac{40 \ kg \cdot 7.7 \ \dfrac{m}{s}}{0.4 \ s} = 770 \ N,$$ $$F_{avg2} = \dfrac{m\Delta v}{\Delta t_2} = \dfrac{40 \ kg \cdot 7.7 \ \dfrac{m}{s}}{0.02 \ s} = 15400 \ N.$$

As we can see from the calculations, in the second jump the average force exerted on his legs much more greater (20 times greater) then in the first one, that's why this time he developed a fracture.

b) Let's consider the collision of two balls which move in the same direction. Let the first ball has the mass $m_1$and the second one - $m_2$. Let the velocities of two balls before the collision be $u_1$and $u_2$, and after the collision - $v_1$and $v_2$, respectively. The Law of Conservation of Momentum states that the total initial momentum of two balls before the collision is equal to the total final momentum of two balls after the collision. The Law of Conservation of Momentum can be written mathematically as follows:

$$p_i = p_f,$$ $$m_1u_1+m_2u_2=m_1v_1+m_2v_2.$$

Let's prove the last equation. During the collision two balls interact with each other. Let's apply the Newton's Third Law of Motion. It states that the forces acting between the two balls are equal in magnitude and opposite in direction:

$$F_1=-F_2.$$

Let's apply the Newton's Second Law of Motion. It states that the acceleration of a ball is directly proportional to the net force acting on it and is inversely proportional to its mass:

$$m_1a_1=-m_2a_2.$$

Since the acceleration is the rate of change of velocity, we can write:

$$m_1(\dfrac{v_1-u_1}{t})=-m_2(\dfrac{v_2-u_2}{t}),$$

here, $t$ is the time of interaction of two balls.

Then, we get:

$$m_1v_1-m_1u_1=-m_2v_2+m_2u_2,$$ $$m_1u_1+m_2u_2=m_1v_1+m_2v_2.$$

That is the equation of the Law of Conservation of Momentum. Therefore, we prove it, using Newton's Third and Second Laws of Motion.