Answer to Question #192645 in Physics for larisa

Question

From point A a mass body m1 is left free connected by an ideal wire of length L to the suspension point O. When the wire reaches a vertical position the body m1 collides perfectly elastically a mass body m 2 larger than m1 placed on a horizontal plane. Find: a) the maximum angle at which it deviates from the vertical, after the collision, the wire from which the body m1 is suspended.
b) the space "s" traveled by the body m2 to the stop, the body-plane friction coefficient being u. Values: m1 = 2 kilograms, L = 1.5 meters, m2 = 10 kilograms, u = 0.12.

Answer 1

From point A a mass body m1 is left free connected by an ideal wire of length L to the suspension point O. When the wire reaches a vertical position the body m1 collides perfectly elastically a mass body m 2 larger than m1 placed on a horizontal plane. Find:

a) the maximum angle at which it deviates from the vertical, after the collision, the wire from which the body m1 is suspended.

b) the space "s" traveled by the body m2 to the stop, the body-plane friction coefficient being u. Values: m1 = 2 kilograms, L = 1.5 meters, m2 = 10 kilograms, u = 0.12.

Solution:

a) Find the initial potential energy of m1:

$$E_p=m_1gh=m_1gL(1-\cos\theta).$$

At the bottom of the trajectory, when the thread is vertical, this potential energy converts into potential energy:

$$E_k=\frac12m_1v^2,\\\space\\ E_p=E_k,\\\space\\ v=\sqrt{2gL\cos\theta}.$$

Then m1 performs an absolutely elastic collision with m2 and m2 acquires kinetic energy, m1 loses kinetic energy, but some potential energy is still left and it is used to rise m1 up:

$$\frac12m_1v^2=\frac12m_1v_1^2+\frac12m_2v_2^2,\\\space\\ m_1v=m_1v_1+m_2v_2,\\\space\\ v_1=v\bigg(\frac{2m_1}{m_1+m_2}-1\bigg)=v\bigg(\frac{m_1-m_2}{m_1+m_2}\bigg),\\\space\\ v_1=\bigg(\frac{m_1-m_2}{m_1+m_2}\bigg)\sqrt{2gL(1-\cos\theta)},\\\space\\ v_2=v\frac{2m_1}{m_1+m_2}=\frac{2m_1}{m_1+m_2}\sqrt{2gL(1-\cos\theta)}.$$

After this interaction, m1 rebounds and the thread deflects for angle $\gamma$:

$$E_{k2}=\frac12m_1v_1^2,\\\space\\ E_{p2}=m_1gL(1-\cos\gamma),\\\space\\ E_{k2}=E_{p2},\\\space\\ \gamma=\arccos\bigg[1-\frac{v_1^2}{2gL}\bigg],\\\space\\ \gamma=\arccos\bigg[1-\bigg(\frac{m_1-m_2}{m_1+m_2}\bigg)^2\cos^2\theta\bigg].$$

To solve this problem, we need to know the initial angle of deflection from the vertical. Is the thread was horizontal initially, the final deflection is $\gamma=70.5°.$

b) The work done against friction is equal to the kinetic energy acquired by m2:

$$\mu m_2gs=\frac12m_2v_2^2,\\\space\\ s=\frac{v_2^2}{2\mu g}=\bigg(\frac{2m_1}{m_1+m_2}\bigg)^2·\frac{L}{\mu}(1-\cos\theta).$$

Again, we need to know the initial deflection. If it is 90°, then $s=22.2\text{ m}.$